electric charge

7/17/2019 Electric Charge

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CHAPTER -1 ELECTRIC CHARGES AND FIELDS

1.An electric dipole with dipole moment 4 × 10−9 C m is aligned at 30° with the direction of a

uniform electric field of magnitude 5 × 104 C−1. Calculate the magnitude of the tor!ue acting

on the dipole.

"lectric dipole moment# p $ 4 × 10−9 C m

Angle made %& p with a uniform electric field# θ $ 30°

"lectric field# E $ 5 × 104 C−1

'or!ue acting on the dipole is gi(en %& the relation#

τ $ pE sinθ

'herefore# the magnitude of the tor!ue acting on the dipole is 10−4 m.

). *hat is the force %etween two small charged spheres ha(ing charges of ) × 10−+ C and 3 ×

10−+ C placed 30 cm apart in air

,epulsi(e force of magnitude - × 10−3

Charge on the first sphere# q1 $ ) × 10−+ C

Charge on the second sphere# q) $ 3 × 10−+ C

istance %etween the spheres# r $ 30 cm $ 0.3 m

"lectrostatic force %etween the spheres is gi(en %& the relation#

*here# ∈0 $ /ermitti(it& of free space



ence# force %etween the two small charged spheres is - × 10−3 . 'he charges are of samenature. ence# force %etween them will %e repulsi(e.

3. A pol&thene piece ru%%ed with wool is found to ha(e a negati(e charge of 3 × 10−+

C.

(a) "stimate the num%er of electrons transferred from which to which2

(b) s there a transfer of mass from wool to pol&thene2

(a) *hen pol&thene is ru%%ed against wool# a num%er of electrons get transferred from woolto pol&thene. ence# wool %ecomes positi(el& charged and pol&thene %ecomes negati(el&charged.

Amount of charge on the pol&thene piece# q $ −3 × 10−+ C

Amount of charge on an electron# e $ −1.- × 10−19 C

um%er of electrons transferred from wool to pol&thene $ n

n can %e calculated using the relation#

q $ ne

$ 1.+ × 101)

'herefore# the num%er of electrons transferred from wool to pol&thene is 1.+ × 101).

(b) 6es.

'here is a transfer of mass ta7ing place. 'his is %ecause an electron has mass#

me $ 9.1 × 10−3 7g

'otal mass transferred to pol&thene from wool#

m $ me × n

$ 9.1 × 10−31 × 1.5 × 101)



$ 1.+0- × 10−1 7g

ence# a negligi%le amount of mass is transferred from wool to pol&thene.

4. (a) 'wo insulated charged copper spheres A and 8 ha(e their centers separated %& a

distance of 50 cm. *hat is the mutual force of electrostatic repulsion if the charge on each is

-.5 × 10−+ C2 'he radii of A and 8 are negligi%le compared to the distance of separation.

(b) *hat is the force of repulsion if each sphere is charged dou%le the a%o(e amount# and the

distance %etween them is hal(ed2

(a) Charge on sphere A# qA $ Charge on sphere 8# q8 $ -.5 × 10−+ C


orce of repulsion %etween the two spheres#

*here#

∈0 $ ree space permitti(it&

$ 9 × 109 m) C−)

∴

$ 1.5) × 10−)

'herefore# the force %etween the two spheres is 1.5) × 10−)

.

(b) After dou%ling the charge# charge on sphere A# qA $ Charge on sphere 8# q8 $ ) × -.5 ×10−+ C $ 1.3 × 10−- C

'he distance %etween the spheres is hal(ed.



∴


$ 1- × 1.5) × 10−)

$ 0.)43

'herefore# the force %etween the two spheres is 0.)43 .

5. 'he electrostatic force on a small sphere of charge 0.4 :C due to another small sphere of

charge − 0. :C in air is 0.) . a *hat is the distance %etween the two spheres2 % *hat is

the force on the second sphere due to the first2

(a) "lectrostatic force on the first sphere# F $ 0.)

Charge on this sphere# q1 $ 0.4 :C $ 0.4 × 10−- C

Charge on the second sphere# q) $ − 0. :C $ − 0. × 10−- C

"lectrostatic force %etween the spheres is gi(en %& the relation#

*here# ∈0 $ /ermitti(it& of free space



'he distance %etween the two spheres is 0.1) m.

(b) 8oth the spheres attract each other with the same force. 'herefore# the force on thesecond sphere due to the first is 0.) .

-. Chec7 that the ratio ke) ;G mem p is dimensionless. <oo7 up a 'a%le of /h&sical Constants and

determine the (alue of this ratio. *hat does the ratio signif&2

'he gi(en ratio is .

*here#

= $ =ra(itational constant

ts unit is m) 7g−).

me and mp $ >asses of electron and proton.

'heir unit is 7g.

e $ "lectric charge.

ts unit is C.

∈0 $ /ermitti(it& of free space

ts unit is m) C−).



ence# the gi(en ratio is dimensionless.

e $ 1.- × 10−19 C

= $ -.-+ × 10−11 m) 7g?)

me$ 9.1 × 10−31 7g

mp $ 1.-- × 10−)+ 7g

ence# the numerical (alue of the gi(en ratio is

'his is the ratio of electric force to the gra(itational force %etween a proton and an electron#7eeping distance %etween them constant.

+(a) "@plain the meaning of the statement electric charge of a %od& is !uantisedB.

(b) *h& can one ignore !uantisation of electric charge when dealing with macroscopic i.e.#large scale charges2

(a) "lectric charge of a %od& is !uantied. 'his means that onl& integral 1# )# D.# n num%erof electrons can %e transferred from one %od& to the other. Charges are not transferred infraction. ence# a %od& possesses total charge onl& in integral multiples of electric charge.

(b) n macroscopic or large scale charges# the charges used are huge as compared to themagnitude of electric charge. ence# !uantiation of electric charge is of no use onmacroscopic scale. 'herefore# it is ignored and it is considered that electric charge iscontinuous.

. *hen a glass rod is ru%%ed with a sil7 cloth# charges appear on %oth. A similar phenomenonis o%ser(ed with man& other pairs of %odies. "@plain how this o%ser(ation is consistent with

the law of conser(ation of charge.

,u%%ing produces charges of e!ual magnitude %ut of opposite nature on the two %odies%ecause charges are created in pairs. 'his phenomenon of charging is called charging %&



friction. 'he net charge on the s&stem of two ru%%ed %odies is ero. 'his is %ecause e!ualamount of opposite charges annihilate each other. *hen a glass rod is ru%%ed with a sil7 cloth#opposite natured charges appear on %oth the %odies. 'his phenomenon is in consistence withthe law of conser(ation of energ&. A similar phenomenon is o%ser(ed with man& other pairs of%odies.

9. our point charges qA $ ) :C# q8 $ −5 :C# qC $ ) :C# and q $ −5 :C are located at the

corners of a s!uare A8C of side 10 cm. *hat is the force on a charge of 1 :C placed at the

centre of the s!uare2

'he gi(en figure shows a s!uare of side 10 cm with four charges placed at its corners. E is thecentre of the s!uare.

*here#

Fides A8 $ 8C $ C $ A $ 10 cm

iagonals AC $ 8 $ cm

AE $ EC $ E $ E8 $ cm

A charge of amount 1:C is placed at point E.

orce of repulsion %etween charges placed at corner A and centre E is e!ual in magnitude %utopposite in direction relati(e to the force of repulsion %etween the charges placed at corner Cand centre E. ence# the& will cancel each other. Fimilarl&# force of attraction %etween chargesplaced at corner 8 and centre E is e!ual in magnitude %ut opposite in direction relati(e to theforce of attraction %etween the charges placed at corner and centre E. ence# the& will also

cancel each other. 'herefore# net force caused %& the four charges placed at the corner of thes!uare on 1 :C charge at centre E is ero.

10. (a) An electrostatic field line is a continuous cur(e. 'hat is# a field line cannot ha(e sudden

%rea7s. *h& not2



(b) "@plain wh& two field lines ne(er cross each other at an& point2

(a) An electrostatic field line is a continuous cur(e %ecause a charge e@periences a continuousforce when traced in an electrostatic field. 'he field line cannot ha(e sudden %rea7s %ecausethe charge mo(es continuousl& and does not Gump from one point to the other.

(b) f two field lines cross each other at a point# then electric field intensit& will show twodirections at that point. 'his is not possi%le. ence# two field lines ne(er cross each other.

11. 'wo point charges qA $ 3 :C and q8 $ −3 :C are located )0 cm apart in (acuum.

(a) *hat is the electric field at the midpoint E of the line A8 Goining the two charges2

(b) f a negati(e test charge of magnitude 1.5 × 10−9 C is placed at this point# what is the

force e@perienced %& the test charge2

(a) 'he situation is represented in the gi(en figure. E is the mid?point of line A8.

istance %etween the two charges# A8 $ )0 cm

∴AE $ E8 $ 10 cm

et electric field at point E $ E

"lectric field at point E caused %& H3:C charge#

E 1 $ along E8

*here#

$ /ermitti(it& of free space

>agnitude of electric field at point E caused %& −3:C charge#



E ) $ $ along E8

$ 5.4 × 10- ;C along E8

'herefore# the electric field at mid?point E is 5.4 × 10- C−1 along E8.

(b) A test charge of amount 1.5 × 10−9 C is placed at mid?point E.

q $ 1.5 × 10−9

C

orce e@perienced %& the test charge $ F

∴F = qE

$ 1.5 × 10−9 × 5.4 × 10-

$ .1 × 10−3

'he force is directed along line EA. 'his is %ecause the negati(e test charge is repelled %& thecharge placed at point 8 %ut attracted towards point A.

'herefore# the force e@perienced %& the test charge is .1 × 10−3 along EA.

1). A s&stem has two charges qA $ ).5 × 10−+ C and q8 $ −).5 × 10−+ C located at points AI

0# 0# − 15 cm and 8I 0# 0# H 15 cm# respecti(el&. *hat are the total charge and electric

dipole moment of the s&stem2

8oth the charges can %e located in a coordinate frame of reference as shown in the gi(enfigure.



At A# amount of charge# qA $ ).5 × 10−+C

At 8# amount of charge# q8 $ −).5 × 10−+ C

'otal charge of the s&stem#

q $ qA H q8

$ ).5 × 10+ C − ).5 × 10−+ C

$ 0

istance %etween two charges at points A and 8#

d $ 15 H 15 $ 30 cm $ 0.3 m

"lectric dipole moment of the s&stem is gi(en %&#

p $ qA × d = q8 × d

$ ).5 × 10−+ × 0.3

$ +.5 × 10− C m along positi(e z ?a@is

'herefore# the electric dipole moment of the s&stem is +.5 × 10− C m along positi(e z −a@is.

13. An electric dipole with dipole moment 4 × 10−9

C m is aligned at 30° with the direction of auniform electric field of magnitude 5 × 104 C−1. Calculate the magnitude of the tor!ue acting

on the dipole.

"lectric dipole moment# p $ 4 × 10−9 C m

Angle made %& p with a uniform electric field# θ $ 30°



"lectric field# E $ 5 × 104 C−1

'or!ue acting on the dipole is gi(en %& the relation#

τ $ pE sinθ

'herefore# the magnitude of the tor!ue acting on the dipole is 10−4 m.

14. A pol&thene piece ru%%ed with wool is found to ha(e a negati(e charge of 3 × 10−+ C.

(a) "stimate the num%er of electrons transferred from which to which2

(b) s there a transfer of mass from wool to pol&thene2

(a) *hen pol&thene is ru%%ed against wool# a num%er of electrons get transferred from woolto pol&thene. ence# wool %ecomes positi(el& charged and pol&thene %ecomes negati(el&charged.

Amount of charge on the pol&thene piece# q $ −3 × 10−+ C

Amount of charge on an electron# e $ −1.- × 10−19 C

um%er of electrons transferred from wool to pol&thene $ n

n can %e calculated using the relation#

q $ ne

$ 1.+ × 101)

'herefore# the num%er of electrons transferred from wool to pol&thene is 1.+ × 101).



(b) 6es.

'here is a transfer of mass ta7ing place. 'his is %ecause an electron has mass#

me $ 9.1 × 10−3 7g

'otal mass transferred to pol&thene from wool#

m $ me × n

$ 9.1 × 10−31 × 1.5 × 101)

$ 1.+0- × 10−1 7g

ence# a negligi%le amount of mass is transferred from wool to pol&thene.

15. (a) 'wo insulated charged copper spheres A and 8 ha(e their centers separated %& a

distance of 50 cm. *hat is the mutual force of electrostatic repulsion if the charge on each is

-.5 × 10−+ C2 'he radii of A and 8 are negligi%le compared to the distance of separation.

(b) *hat is the force of repulsion if each sphere is charged dou%le the a%o(e amount# and the

distance %etween them is hal(ed2

(a) Charge on sphere A# qA $ Charge on sphere 8# q8 $ -.5 × 10−+ C



*here#

∈0 $ ree space permitti(it&

$ 9 × 109 m) C−)

∴

$ 1.5) × 10−)



'herefore# the force %etween the two spheres is 1.5) × 10−) .

(b) After dou%ling the charge# charge on sphere A# qA $ Charge on sphere 8# q8 $ ) × -.5 ×10−+ C $ 1.3 × 10−- C

'he distance %etween the spheres is hal(ed.

∴


$ 1- × 1.5) × 10−)

$ 0.)43

'herefore# the force %etween the two spheres is 0.)43 .

1-. Fuppose the spheres A and 8 in "@ercise 1.1) ha(e identical sies. A third sphere of the

same sie %ut uncharged is %rought in contact with the first# then %rought in contact with the

second# and finall& remo(ed from %oth. *hat is the new force of repulsion %etween A and 82

istance %etween the spheres# A and 8# r $ 0.5 m

nitiall&# the charge on each sphere# q $ -.5 × 10−+ C

*hen sphere A is touched with an uncharged sphere C# amount of charge from A will

transfer to sphere C. ence# charge on each of the spheres# A and C# is .

*hen sphere C with charge is %rought in contact with sphere 8 with charge q# total chargeson the s&stem will di(ide into two e!ual hal(es gi(en as#



1. Consider a uniform electric field E $ 3 × 103 J;C. a *hat is the flu@ of this field through

a s!uare of 10 cm on a side whose plane is parallel to the yz plane2 % *hat is the flu@

through the same s!uare if the normal to its plane ma7es a -0° angle with the x ?a@is2

(a) "lectric field intensit&# $ 3 × 103 J ;C

>agnitude of electric field intensit&# $ 3 × 103 ;C

Fide of the s!uare# s $ 10 cm $ 0.1 m

Area of the s!uare# A $ s) $ 0.01 m)

'he plane of the s!uare is parallel to the y-z plane. ence# angle %etween the unit (ector

normal to the plane and electric field# θ $ 0°

lu@ Φ through the plane is gi(en %& the relation#

K $

$ 3 × 103 × 0.01 × cos0°

$ 30 m) ;C

(b) /lane ma7es an angle of -0° with the x ?a@is. ence# θ $ -0°

lu@# Φ $

$ 3 × 103 × 0.01 × cos-0°

$ 15 m) ;C

19. *hat is the net flu@ of the uniform electric field of "@ercise 1.15 through a cu%e of side )0

cm oriented so that its faces are parallel to the coordinate planes2

All the faces of a cu%e are parallel to the coordinate a@es. 'herefore# the num%er of field linesentering the cu%e is e!ual to the num%er of field lines piercing out of the cu%e. As a result# netflu@ through the cu%e is ero.



)0. Careful measurement of the electric field at the surface of a %lac7 %o@ indicates that the

net outward flu@ through the surface of the %o@ is .0 × 103 m) ;C. a *hat is the net

charge inside the %o@2 % f the net outward flu@ through the surface of the %o@ were ero#

could &ou conclude that there were no charges inside the %o@2 *h& or *h& not2

(a) et outward flu@ through the surface of the %o@# Φ $ .0 × 103 m) ;C

or a %od& containing net charge q# flu@ is gi(en %& the relation#


$ .54 × 10−1) −1C) m−)

q $ ∈0Φ

$ .54 × 10−1) × .0 × 103

$ +.0 × 10−

$ 0.0+ :C

'herefore# the net charge inside the %o@ is 0.0+ :C.

(b) o

et flu@ piercing out through a %od& depends on the net charge contained in the %od&. f netflu@ is ero# then it can %e inferred that net charge inside the %od& is ero. 'he %od& ma& ha(ee!ual amount of positi(e and negati(e charges.

)1. A point charge H10 :C is a distance 5 cm directl& a%o(e the centre of a s!uare of side 10

cm# as shown in ig. 1.34. *hat is the magnitude of the electric flu@ through the s!uare2

Hint: 'hin7 of the s!uare as one face of a cu%e with edge 10 cm.



'he s!uare can %e considered as one face of a cu%e of edge 10 cm with a centre where chargeq is placed. According to =aussBs theorem for a cu%e# total electric flu@ is through all its si@faces.

ence# electric flu@ through one face of the cu%e i.e.# through the s!uare#

*here#


$ .54 × 10−1) −1C) m−)

q $ 10 :C $ 10 × 10−- C

∴

$ 1. × 105 m) C−1

'herefore# electric flu@ through the s!uare is 1. × 105 m) C−1.

)). A point charge of ).0 :C is at the centre of a cu%ic =aussian surface 9.0 cm on edge. *hat

is the net electric flu@ through the surface2



et electric flu@ Φet through the cu%ic surface is gi(en %&#

*here#


$ .54 × 10−1) −1C) m−)

q $ et charge contained inside the cu%e $ ).0 :C $ ) × 10−- C

∴

$ ).)- × 105 m) C−1

'he net electric flu@ through the surface is ).)- ×105 m)C−1.

)3. A point charge causes an electric flu@ of −1.0 × 103 m) ;C to pass through a spherical

=aussian surface of 10.0 cm radius centered on the charge. a f the radius of the =aussian

surface were dou%led# how much flu@ would pass through the surface2 % *hat is the (alue of

the point charge2

(a) "lectric flu@# Φ $ −1.0 × 103 m) ;C

,adius of the =aussian surface#

r $ 10.0 cm

"lectric flu@ piercing out through a surface depends on the net charge enclosed inside a %od&.

t does not depend on the sie of the %od&. f the radius of the =aussian surface is dou%led#then the flu@ passing through the surface remains the same i.e.# −103 m) ;C.

(b) "lectric flu@ is gi(en %& the relation#



*here#

q $ et charge enclosed %& the spherical surface

∈0 $ /ermitti(it& of free space $ .54 × 10 −1) −1C) m−)

∴

$ −1.0 × 103 × .54 × 10−1)

$ −.54 × 10−9 C

$ −.54 nC

'herefore# the (alue of the point charge is −.54 nC.

)4. A conducting sphere of radius 10 cm has an un7nown charge. f the electric field )0 cm

from the centre of the sphere is 1.5 × 103 ;C and points radiall& inward# what is the net

charge on the sphere2

"lectric field intensit& E at a distance d from the centre of a sphere containing net charge qis gi(en %& the relation#

*here#

q $ et charge $ 1.5 × 103 ;C

d $ istance from the centre $ )0 cm $ 0.) m


And# $ 9 × 109 m) C−)

∴



$ -.-+ × 109 C

$ -.-+ nC

'herefore# the net charge on the sphere is -.-+ nC.

)5. A uniforml& charged conducting sphere of ).4 m diameter has a surface charge densit& of

0.0 :C;m). a ind the charge on the sphere. % *hat is the total electric flu@ lea(ing the

surface of the sphere2

(a) iameter of the sphere# d $ ).4 m

,adius of the sphere# r $ 1.) m

Furface charge densit&# $ 0.0 :C;m) $ 0 × 10−- C;m)

'otal charge on the surface of the sphere#

Q $ Charge densit& × Furface area

$

$ 0 × 10−- × 4 × 3.14 × 1.))

$ 1.44+ × 10−3 C

'herefore# the charge on the sphere is 1.44+ × 10−3

C.

(b) 'otal electric flu@ lea(ing out the surface of a sphere containing net charge Q isgi(en %& the relation#

*here#


$ .54 × 10−1) −1C) m−)

Q $ 1.44+ × 10−3 C



$ 1.-3 × 10 C−1 m)

'herefore# the total electric flu@ lea(ing the surface of the sphere is 1.-3 × 10 C−1 m).

)-.An infinite line charge produces a field of 9 × 104 ;C at a distance of ) cm. Calculate the

linear charge densit&.

"lectric field produced %& the infinite line charges at a distance d ha(ing linear charge densit& λ is gi(en %& the relation#

*here#

d $ ) cm $ 0.0) m

E $ 9 × 104 ;C


$ 9 × 109 m) C−)

$ 10 :C;m

'herefore# the linear charge densit& is 10 :C;m.

)+. 'wo large# thin metal plates are parallel and close to each other. En their inner faces# the

plates ha(e surface charge densities of opposite signs and of magnitude 1+.0 × 10−)) C;m).

*hat is EI a in the outer region of the first plate# % in the outer region of the second plate#and c %etween the plates2

'he situation is represented in the following figure.



A and 8 are two parallel plates close to each other. Euter region of plate A is la%elled as I#outer region of plate 8 is la%elled as III# and the region %etween the plates# A and 8# isla%elled as II.

Charge densit& of plate A# σ $ 1+.0 × 10−)) C;m)

Charge densit& of plate 8# σ $ −1+.0 × 10−)) C;m)

n the regions# I and III# electric field E is ero. 'his is %ecause charge is not enclosed %& therespecti(e plates.

"lectric field E in region II is gi(en %& the relation#

*here#

∈0 $ /ermitti(it& of free space $ .54 × 10 −1) −1C) m−)

∴

$ 1.9) × 10−10 ;C

'herefore# electric field %etween the plates is 1.9) × 10−10 ;C.

). An oil drop of 1) e@cess electrons is held stationar& under a constant electric field of ).55

× 104 C−1 in >illi7anBs oil drop e@periment. 'he densit& of the oil is 1.)- g cm−3. "stimate the

radius of the drop. g $ 9.1 m s−)L e $ 1.-0 × 10−19 C.

"@cess electrons on an oil drop# n $ 1)

"lectric field intensit&# E $ ).55 × 104 C−1

ensit& of oil# ρ $ 1.)- gm;cm3 $ 1.)- × 103 7g;m3

Acceleration due to gra(it&# g $ 9.1 m s−)



Charge on an electron# e $ 1.- × 10−19 C

,adius of the oil drop $ r

orce F due to electric field E is e!ual to the weight of the oil drop

F =

Eq $ mg

Ene

*here#

q $ et charge on the oil drop $ ne

m $ >ass of the oil drop

$ Molume of the oil drop × ensit& of oil

$ 9.) × 10−4 mm

'herefore# the radius of the oil drop is 9.) × 10−4 mm.

)9. *hich among the cur(es shown in ig. 1.35 cannot possi%l& represent electrostatic field

lines2

(a)



(b)

(c)

(d)

(e)



(a) 'he field lines showed in a do not represent electrostatic field lines %ecause field linesmust %e normal to the surface of the conductor.

(b) 'he field lines showed in % do not represent electrostatic field lines %ecause the fieldlines cannot emerge from a negati(e charge and cannot terminate at a positi(e charge.

(c) 'he field lines showed in c represent electrostatic field lines. 'his is %ecause the field

lines emerge from the positi(e charges and repel each other.

(d) 'he field lines showed in d do not represent electrostatic field lines %ecause the fieldlines should not intersect each other.

(e) 'he field lines showed in e do not represent electrostatic field lines %ecause closed loopsare not formed in the area %etween the field lines.

30. n a certain region of space# electric field is along the ?direction throughout. 'he

magnitude of electric field is# howe(er# not constant %ut increases uniforml& along the positi(e

z ?direction# at the rate of 105 C−1 per metre. *hat are the force and tor!ue e@perienced %& a

s&stem ha(ing a total dipole moment e!ual to 10−+

Cm in the negati(e z ?direction2

ipole moment of the s&stem# p $ q × d! $ −10−+ C m

,ate of increase of electric field per unit length#

orce F e@perienced %& the s&stem is gi(en %& the relation#

F $ qE

$ −10−+ × 10−5



$ −10−)

'he force is −10−) in the negati(e ?direction i.e.# opposite to the direction of electric field.ence# the angle %etween electric field and dipole moment is 10°.

'or!ue τ is gi(en %& the relation#

τ $ pE sin10°

$ 0

'herefore# the tor!ue e@perienced %& the s&stem is ero.

31. a A conductor A with a ca(it& as shown in ig. 1.3-a is gi(en a charge Q. Fhow that

the entire charge must appear on the outer surface of the conductor. % Another conductor 8

with charge q is inserted into the ca(it& 7eeping 8 insulated from A. Fhow that the total

charge on the outside surface of A is Q H q Nig. 1.3-%O. c A sensiti(e instrument is to %e

shielded from the strong electrostatic fields in its en(ironment. Fuggest a possi%le wa&.

(a) <et us consider a =aussian surface that is l&ing wholl& within a conductor and enclosingthe ca(it&. 'he electric field intensit& E inside the charged conductor is ero.

<et q is the charge inside the conductor and is the permitti(it& of free space.

According to =aussBs law#

lu@#

ere# E $ 0



'herefore# charge inside the conductor is ero.

'he entire charge Q appears on the outer surface of the conductor.

(b) 'he outer surface of conductor A has a charge of amount Q. Another conductor 8 ha(ingcharge Hq is 7ept inside conductor A and it is insulated from A. ence# a charge of amount −qwill %e induced in the inner surface of conductor A and Hq is induced on the outer surface ofconductor A. 'herefore# total charge on the outer surface of conductor A is Q H q.

(c) A sensiti(e instrument can %e shielded from the strong electrostatic field in itsen(ironment %& enclosing it full& inside a metallic surface. A closed metallic %od& acts as anelectrostatic shield.

3). A hollow charged conductor has a tin& hole cut into its surface. Fhow that the electric field

in the hole is # where is the unit (ector in the outward normal direction# and is

the surface charge densit& near the hole.

<et us consider a conductor with a ca(it& or a hole. "lectric field inside the ca(it& is ero.

<et E is the electric field Gust outside the conductor# q is the electric charge# is the charge

densit&# and is the permitti(it& of free space.

Charge

According to =aussBs law#

'herefore# the electric field Gust outside the conductor is . 'his field is a superposition of

field due to the ca(it& and the field due to the rest of the charged conductor . 'hese



fields are e!ual and opposite inside the conductor# and e!ual in magnitude and directionoutside the conductor.

'herefore# the field due to the rest of the conductor is .

ence# pro(ed.

33. E%tain the formula for the electric field due to a long thin wire of uniform linear charge

densit& λ without using =aussBs law. NHint: Pse Coulom%Bs law directl& and e(aluate thenecessar& integral.O

'a7e a long thin wire Q6 as shown in the figure of uniform linear charge densit& .

Consider a point A at a perpendicular distance ! from the mid?point E of the wire# as shown inthe following figure.

<et E %e the electric field at point A due to the wire# Q6.

Consider a small length element dx on the wire section with ER $ x

<et q %e the charge on this piece.

"lectric field due to the piece#



'he electric field is resol(ed into two rectangular components. is the perpendicular

component and is the parallel component.

*hen the whole wire is considered# the component is cancelled.

Enl& the perpendicular component affects point A.

ence# effecti(e electric field at point A due to the element dx is dE 1.

En differentiating e!uation )# we o%tain

rom e!uation )#

/utting e!uations 3 and 4 in e!uation 1# we o%tain



'he wire is so long that tends from to .

8& integrating e!uation 5# we o%tain the (alue of field E 1 as#

'herefore# the electric field due to long wire is .

33. t is now %elie(ed that protons and neutrons which constitute nuclei of ordinar& matter

are themsel(es %uilt out of more elementar& units called !uar7s. A proton and a neutron

consist of three !uar7s each. 'wo t&pes of !uar7s# the so called upB !uar7 denoted %& u of

charge H);3 e# and the downB !uar7 denoted %& d of charge −1;3 e# together with

electrons %uild up ordinar& matter. Suar7s of other t&pes ha(e also %een found which gi(e

rise to different unusual (arieties of matter. Fuggest a possi%le !uar7 composition of a proton

and neutron.

A proton has three !uar7s. <et there %e n up !uar7s in a proton# each ha(ing a charge of

.

Charge due to n up !uar7s

um%er of down !uar7s in a proton $ 3 − n

"ach down !uar7 has a charge of .

Charge due to 3 − n down !uar7s



'otal charge on a proton $ H e

um%er of up !uar7s in a proton# n $ )

um%er of down !uar7s in a proton $ 3 − n $ 3 − ) $ 1

'herefore# a proton can %e represented as uudB.

A neutron also has three !uar7s. <et there %e n up !uar7s in a neutron# each ha(ing a charge

of .

Charge on a neutron due to n up !uar7s

um%er of down !uar7s is 3 − n#each ha(ing a charge of .

Charge on a neutron due to down !uar7s $

'otal charge on a neutron $ 0

um%er of up !uar7s in a neutron# n $ 1

um%er of down !uar7s in a neutron $ 3 − n $ )

'herefore# a neutron can %e represented as uddB.



34. (a) Consider an ar%itrar& electrostatic field configuration. A small test charge is placed at

a null point i.e.# where E $ 0 of the configuration. Fhow that the e!uili%rium of the test

charge is necessaril& unsta%le.

(b) Merif& this result for the simple configuration of two charges of the same magnitude and

sign placed a certain distance apart

(a) <et the e!uili%rium of the test charge %e sta%le. f a test charge is in e!uili%rium anddisplaced from its position in an& direction# then it e@periences a restoring force towards a nullpoint# where the electric field is ero. All the field lines near the null point are directed inwardstowards the null point. 'here is a net inward flu@ of electric field through a closed surfacearound the null point. According to =aussBs law# the flu@ of electric field through a surface#which is not enclosing an& charge# is ero. ence# the e!uili%rium of the test charge can %esta%le.

(b) 'wo charges of same magnitude and same sign are placed at a certain distance. 'he mid?point of the Goining line of the charges is the null point. *hen a test charged is displaced along

the line# it e@periences a restoring force. f it is displaced normal to the Goining line# then thenet force ta7es it awa& from the null point. ence# the charge is unsta%le %ecause sta%ilit& ofe!uili%rium re!uires restoring force in all directions.

35. A particle of mass m and charge −q enters the region %etween the two charged plates

initiall& mo(ing along x ?a@is with speed "x li7e particle 1 in ig. 1.33. 'he length of plate is #

and an uniform electric field E is maintained %etween the plates. Fhow that the (ertical

deflection of the particle at the far edge of the plate is qE#) ; )m .

$%mp&re t'is m%ti%n (it' m%ti%n %) & pr%*e+ti!e in gr&"it&ti%n&! )ie!d dis+,ssed in e+ti%n ./01

%) $!&ss 23 4ext5%%k %) 6'ysi+s

Charge on a particle of mass m $ − q

Melocit& of the particle $ " x

<ength of the plates $ #

>agnitude of the uniform electric field %etween the plates $ E

>echanical force# F $ >ass m × Acceleration &



'herefore# acceleration#

'ime ta7en %& the particle to cross the field of length # is gi(en %&#

t

n the (ertical direction# initial (elocit&# , $ 0

According to the third e!uation of motion# (ertical deflection s of the particle can %e o%tainedas#

ence# (ertical deflection of the particle at the far edge of the plate is

. 'his is similar to the motion of horiontal proGectiles under gra(it&.

3-.Fuppose that the particle in "@ercise in 1.33 is an electron proGected with (elocit& " x $ ).0

× 10- m s−1. f E %etween the plates separated %& 0.5 cm is 9.1 × 10) ;C# where will the

electron stri7e the upper plate2 T e T $1.- × 10−19 C# me $ 9.1 × 10−31 7g.

Melocit& of the particle# " x $ ).0 × 10- m;s

Feparation of the two plates# d $ 0.5 cm $ 0.005 m

"lectric field %etween the two plates# E $ 9.1 × 10) ;C

Charge on an electron# q $ 1.- × 10−19 C

>ass of an electron# me $ 9.1 × 10−31

7g

<et the electron stri7e the upper plate at the end of plate ## when deflection is s.

'herefore#



'herefore# the electron will stri7e the upper plate after tra(elling 1.- cm.

CHAPTER -2 ELECTROSTATIC POTENTIAL AND CAPACITANCE

Question 2.1:

'wo charges 5 × 10− C and −3 × 10− C are located 1- cm apart. At what points on the line

Goining the two charges is the electric potential ero2 'a7e the potential at infinit& to %e ero.

'here are two charges#

istance %etween the two charges# d $ 1- cm $ 0.1- m

Consider a point / on the line Goining the two charges# as shown in the gi(en figure.

r $ istance of point / from charge q1

<et the electric potential 7 at point / %e ero.

/otential at point / is the sum of potentials caused %& charges q1 and q) respecti(el&.



*here#


or 7 $ 0# e!uation i reduces to

'herefore# the potential is ero at a distance of 10 cm from the positi(e charge %etween thecharges.

Fuppose point / is outside the s&stem of two charges at a distance s from the negati(echarge# where potential is ero# as shown in the following figure.

or this arrangement# potential is gi(en %&#

or 7 $ 0# e!uation ii reduces to



'herefore# the potential is ero at a distance of 40 cm from the positi(e charge outside thes&stem of charges.

Question 2.2:

A regular he@agon of side 10 cm has a charge 5 UC at each of its (ertices. Calculate the

potential at the centre of the he@agon.

'he gi(en figure shows si@ e!ual amount of charges# q# at the (ertices of a regular he@agon.

*here#

Charge# q $ 5 UC $ 5 × 10−- C

Fide of the he@agon# ! $ A8 $ 8C $ C $ " $ " $ A $ 10 cm

istance of each (erte@ from centre E# d $ 10 cm



"lectric potential at point E#

*here#


'herefore# the potential at the centre of the he@agon is ).+ × 10- M.

Question 2.3:

'wo charges ) :C and −) UC are placed at points A and 8 - cm apart.

(a) dentif& an e!uipotential surface of the s&stem.

(b) *hat is the direction of the electric field at e(er& point on this surface2

(a) 'he situation is represented in the gi(en figure.

An e!uipotential surface is the plane on which total potential is ero e(er&where. 'his plane isnormal to line A8. 'he plane is located at the mid?point of line A8 %ecause the magnitude ofcharges is the same.

(b) 'he direction of the electric field at e(er& point on this surface is normal to the plane inthe direction of A8.

Question 2.4:

A spherical conductor of radius 1) cm has a charge of 1.- × 10−+C distri%uted uniforml& on its

surface. *hat is the electric field

(a) nside the sphere



(b) Vust outside the sphere

(c) At a point 1 cm from the centre of the sphere2

(a) ,adius of the spherical conductor# r $ 1) cm $ 0.1) m

Charge is uniforml& distri%uted o(er the conductor# q $ 1.- × 10−+ C

"lectric field inside a spherical conductor is ero. 'his is %ecause if there is field inside theconductor# then charges will mo(e to neutralie it.

(b) "lectric field E Gust outside the conductor is gi(en %& the relation#

*here#


'herefore# the electric field Gust outside the sphere is .

(c) "lectric field at a point 1 m from the centre of the sphere $ E 1

istance of the point from the centre# d $ 1 cm $ 0.1 m

'herefore# the electric field at a point 1 cm from the centre of the sphere is



.

Question 2.5:

A parallel plate capacitor with air %etween the plates has a capacitance of p 1p $ 10−1)

. *hat will %e the capacitance if the distance %etween the plates is reduced %& half# and the

space %etween them is filled with a su%stance of dielectric constant -2

Capacitance %etween the parallel plates of the capacitor# C $ p

nitiall&# distance %etween the parallel plates was d and it was filled with air. ielectric constantof air# k $ 1

Capacitance# $ # is gi(en %& the formula#

*here#

A $ Area of each plate


f distance %etween the plates is reduced to half# then new distance# d B $

ielectric constant of the su%stance filled in %etween the plates# $ -

ence# capacitance of the capacitor %ecomes

'a7ing ratios of e!uations i and ii# we o%tain

'herefore# the capacitance %etween the plates is 9- p.



Question 2.6:

'hree capacitors each of capacitance 9 p are connected in series.

(a) *hat is the total capacitance of the com%ination2

(b) *hat is the potential difference across each capacitor if the com%ination is connected to a

1)0 M suppl&2

(a) Capacitance of each of the three capacitors# $ $ 9 p

"!ui(alent capacitance $ B of the com%ination of the capacitors is gi(en %& the relation#

'herefore# total capacitance of the com%ination is .

(b) Fuppl& (oltage# 7 $ 100 M

/otential difference 7 B across each capacitor is e!ual to one?third of the suppl& (oltage.

'herefore# the potential difference across each capacitor is 40 M.

Question 2.7:

'hree capacitors of capacitances ) p# 3 p and 4 p are connected in parallel.

(a) *hat is the total capacitance of the com%ination2

(b) etermine the charge on each capacitor if the com%ination is connected to a 100 M suppl&.

(a) Capacitances of the gi(en capacitors are



or the parallel com%ination of the capacitors# e!ui(alent capacitor is gi(en %& the alge%raicsum#

'herefore# total capacitance of the com%ination is 9 p.

(b) Fuppl& (oltage# 7 $ 100 M

'he (oltage through all the three capacitors is same $ 7 $ 100 M

Charge on a capacitor of capacitance $ and potential difference 7 is gi(en %& the relation#

q $ 7$ D i

or C $ ) p#

or C $ 3 p#

or C $ 4 p#

Question 2.8:

n a parallel plate capacitor with air %etween the plates# each plate has an area of - × 10−3 m)

and the distance %etween the plates is 3 mm. Calculate the capacitance of the capacitor. f this

capacitor is connected to a 100 M suppl&# what is the charge on each plate of the capacitor2

Area of each plate of the parallel plate capacitor# A $ - × 10−3 m)

istance %etween the plates# d $ 3 mm $ 3 × 10−3 m

Fuppl& (oltage# 7 $ 100 M

Capacitance $ of a parallel plate capacitor is gi(en %&#

*here#




$ .54 × 10−1) −1 m−) C−)

'herefore# capacitance of the capacitor is 1+.+1 p and charge on each plate is 1.++1 × 10−9

C.

Question 2.9:

"@plain what would happen if in the capacitor gi(en in "@ercise ).# a 3 mm thic7 mica sheet

of dielectric constant $ - were inserted %etween the plates#

(a) *hile the (oltage suppl& remained connected.

(b) After the suppl& was disconnected.

(a) ielectric constant of the mica sheet# k $ -

nitial capacitance# $ $ 1.++1 × 10−11


/otential across the plates remains 100 M.

(b) ielectric constant# k $ -

nitial capacitance# $ $ 1.++1 × 10−11



f suppl& (oltage is remo(ed# then there will %e no effect on the amount of charge in theplates.

Charge $ 1.++1 × 10−9 C

/otential across the plates is gi(en %&#

Question 2.10:

A 1) p capacitor is connected to a 50M %atter&. ow much electrostatic energ& is stored in

the capacitor2

Capacitor of the capacitance# $ $ 1) p $ 1) × 10−1)

/otential difference# 7 $ 50 M

"lectrostatic energ& stored in the capacitor is gi(en %& the relation#

'herefore# the electrostatic energ& stored in the capacitor is

Question 2.11:

A -00 p capacitor is charged %& a )00 M suppl&. t is then disconnected from the suppl& and

is connected to another uncharged -00 p capacitor. ow much electrostatic energ& is lost inthe process2

Capacitance of the capacitor# $ $ -00 p

/otential difference# 7 $ )00 M

"lectrostatic energ& stored in the capacitor is gi(en %&#



f suppl& is disconnected from the capacitor and another capacitor of capacitance $ $ -00 pis connected to it# then e!ui(alent capacitance $ B of the com%ination is gi(en %&#

ew electrostatic energ& can %e calculated as

'herefore# the electrostatic energ& lost in the process is .

Question 2.12:

A charge of mC is located at the origin. Calculate the wor7 done in ta7ing a small charge of

−) × 10−9 C from a point / 0# 0# 3 cm to a point S 0# 4 cm# 0# (ia a point , 0# - cm# 9

cm.

Charge located at the origin# q $ mC$ × 10−3 C

>agnitude of a small charge# which is ta7en from a point / to point , to point S# q1 $ − ) ×10−9 C

All the points are represented in the gi(en figure.



/oint / is at a distance# d 1 $ 3 cm# from the origin along z ?a@is.

/oint S is at a distance# d ) $ 4 cm# from the origin along y ?a@is.

/otential at point /#

/otential at point S#

*or7 done %& the electrostatic force is independent of the path.

'herefore# wor7 done during the process is 1.)+ V.

Question 2.13:



A cu%e of side 5 has a charge q at each of its (ertices. etermine the potential and electric

field due to this charge arra& at the centre of the cu%e.

<ength of the side of a cu%e $ 5

Charge at each of its (ertices $ q

A cu%e of side 5 is shown in the following figure.

d $ iagonal of one of the si@ faces of the cu%e

! $ <ength of the diagonal of the cu%e

'he electric potential 7 at the centre of the cu%e is due to the presence of eight charges atthe (ertices.



'herefore# the potential at the centre of the cu%e is .

'he electric field at the centre of the cu%e# due to the eight charges# gets cancelled. 'his is%ecause the charges are distri%uted s&mmetricall& with respect to the centre of the cu%e.ence# the electric field is ero at the centre.

Question 2.14:

'wo tin& spheres carr&ing charges 1.5 :C and ).5 :C are located 30 cm apart. ind the

potential and electric fieldI

(a) at the mid?point of the line Goining the two charges# and

(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the

mid?point.

'wo charges placed at points A and 8 are represented in the gi(en figure. E is the mid?point of the line Goining the two charges.

>agnitude of charge located at A# q1 $ 1.5 :C

>agnitude of charge located at 8# q) $ ).5 :C

istance %etween the two charges# d $ 30 cm $ 0.3 m

(a) <et 7 1 and E 1 are the electric potential and electric field respecti(el& at E.

7 1 $ /otential due to charge at A H /otential due to charge at 8



*here#


E 1 $ "lectric field due to q) − "lectric field due to q1

'herefore# the potential at mid?point is ).4 × 105 M and the electric field at mid?point is 4× 105

M m−1. 'he field is directed from the larger charge to the smaller charge.

(b) Consider a point R such that normal distanceER $ 10 cm $ 0.1 m# as shown in thefollowing figure.

7 ) and E ) are the electric potential and electric field respecti(el& at R.

t can %e o%ser(ed from the figure that distance#

7 )$ "lectric potential due to A H "lectric /otential due to 8



"lectric field due to q at R#

"lectric field due to q) at 8 #

'he resultant field intensit& at R#

*here# )θis the angle# ∠ A8 8

rom the figure# we o%tain



'herefore# the potential at a point 10 cm perpendicular to the mid?point is ).0 × 105 M andelectric field is -.- ×105 M m−1.

Question 2.15:

A spherical conducting shell of inner radius r 1 and outer radius r ) has a charge Q.

(a) A charge q is placed at the centre of the shell. *hat is the surface charge densit& on the

inner and outer surfaces of the shell2

(b) s the electric field inside a ca(it& with no charge ero# e(en if the shell is not spherical#

%ut has an& irregular shape2 "@plain.

(a) Charge placed at the centre of a shell is Hq. ence# a charge of magnitude −q will %einduced to the inner surface of the shell. 'herefore# total charge on the inner surface of theshell is −q.

Furface charge densit& at the inner surface of the shell is gi(en %& the relation#

A charge of Hq is induced on the outer surface of the shell. A charge of magnitude Q is placedon the outer surface of the shell. 'herefore# total charge on the outer surface of the shell is Q H q. Furface charge densit& at the outer surface of the shell#

(b) 6es

'he electric field intensit& inside a ca(it& is ero# e(en if the shell is not spherical and has an&irregular shape. 'a7e a closed loop such that a part of it is inside the ca(it& along a field linewhile the rest is inside the conductor. et wor7 done %& the field in carr&ing a test charge o(era closed loop is ero %ecause the field inside the conductor is ero. ence# electric field is ero#whate(er is the shape.

Question 2.16:

(a) Fhow that the normal component of electrostatic field has a discontinuit& from one side of

a charged surface to another gi(en %&



*here is a unit (ector normal to the surface at a point and W is the surface charge densit& at

that point. 'he direction of is from side 1 to side ). ence show that Gust outside a

conductor# the electric field is W

(b) Fhow that the tangential component of electrostatic field is continuous from one side of acharged surface to another. NintI or a# use =aussBs law. or# % use the fact that wor7

done %& electrostatic field on a closed loop is ero.O

(a) "lectric field on one side of a charged %od& is E 1 and electric field on the other side of thesame %od& is E ). f infinite plane charged %od& has a uniform thic7ness# then electric field dueto one surface of the charged %od& is gi(en %&#

*here#

$ Pnit (ector normal to the surface at a point

W $ Furface charge densit& at that point

"lectric field due to the other surface of the charged %od&#

"lectric field at an& point due to the two surfaces#

Fince inside a closed conductor# $ 0#

∴

'herefore# the electric field Gust outside the conductor is .



(b) *hen a charged particle is mo(ed from one point to the other on a closed loop# the wor7done %& the electrostatic field is ero. ence# the tangential component of electrostatic field iscontinuous from one side of a charged surface to the other.

Question 2.17:

A long charged c&linder of linear charged densit& X is surrounded %& a hollow co?a@ial

conducting c&linder. *hat is the electric field in the space %etween the two c&linders2

Charge densit& of the long charged c&linder of length # and radius r is λ.

Another c&linder of same length surrounds the per(ious c&linder. 'he radius of this c&linder is9.

<et E %e the electric field produced in the space %etween the two c&linders.

"lectric flu@ through the =aussian surface is gi(en %& =aussBs theorem as#

*here# d $ istance of a point from the common a@is of the c&linders

<et q %e the total charge on the c&linder.

t can %e written as

*here#

q $ Charge on the inner sphere of the outer c&linder


'herefore# the electric field in the space %etween the two c&linders is .

Question 2.18:

n a h&drogen atom# the electron and proton are %ound at a distance of a%out 0.53 YI



(a) "stimate the potential energ& of the s&stem in eM# ta7ing the ero of the potential energ&

at infinite separation of the electron from proton.

(b) *hat is the minimum wor7 re!uired to free the electron# gi(en that its 7inetic energ& in

the or%it is half the magnitude of potential energ& o%tained in a2

(c) *hat are the answers to a and % a%o(e if the ero of potential energ& is ta7en at 1.0-

Y separation2

'he distance %etween electron?proton of a h&drogen atom#

Charge on an electron# q1 $ −1.- ×10−19 C

Charge on a proton# q) $ H1.- ×10−19 C

(a) /otential at infinit& is ero.

/otential energ& of the s&stem# p?e $ /otential energ& at infinit& − /otential energ& atdistance# d

*here#

∈0 is the permitti(it& of free space

'herefore# the potential energ& of the s&stem is −)+.) eM.

(b) Zinetic energ& is half of the magnitude of potential energ&.

'otal energ& $ 13.- − )+.) $ 13.- eM



'herefore# the minimum wor7 re!uired to free the electron is 13.- eM.

(c) *hen ero of potential energ& is ta7en#

∴/otential energ& of the s&stem $ /otential energ& at d 1 − /otential energ& at d

Question 2.19:

f one of the two electrons of a ) molecule is remo(ed# we get a h&drogen molecular ion .

n the ground state of an # the two protons are separated %& roughl& 1.5 Y# and the

electron is roughl& 1 Y from each proton. etermine the potential energ& of the s&stem.

Fpecif& &our choice of the ero of potential energ&.

'he s&stem of two protons and one electron is represented in the gi(en figure.

Charge on proton 1# q1 $ 1.- ×10−19 C

Charge on proton )# q) $ 1.- ×10−19 C

Charge on electron# q3 $ −1.- ×10−19 C

istance %etween protons 1 and )# d 1 $ 1.5 ×10−10 m

istance %etween proton 1 and electron# d ) $ 1 ×10−10 m

istance %etween proton ) and electron# d 3 $ 1 × 10−10 m



'he potential energ& at infinit& is ero.

/otential energ& of the s&stem#

'herefore# the potential energ& of the s&stem is −19.) eM.

Question 2.20:

'wo charged conducting spheres of radii & and 5 are connected to each other %& a wire. *hat

is the ratio of electric fields at the surfaces of the two spheres2 Pse the result o%tained to

e@plain wh& charge densit& on the sharp and pointed ends of a conductor is higher than on its

flatter portions.

<et & %e the radius of a sphere A# Q A %e the charge on the sphere# and $ A %e the capacitance of the sphere. <et 5 %e the radius of a sphere 8# Q %e the charge on the sphere# and $ %e the

capacitance of the sphere. Fince the two spheres are connected with a wire# their potential 7 will %ecome e!ual.

<et E A%e the electric field of sphere A and E %e the electric field of sphere 8. 'herefore# theirratio#



/utting the (alue of ) in 1# we o%tain

'herefore# the ratio of electric fields at the surface is .

Question 2.21:

'wo charges ;q and <q are located at points 0# 0# − & and 0# 0# &# respecti(el&.

(a) *hat is the electrostatic potential at the points2

(b) E%tain the dependence of potential on the distance r of a point from the origin when r ;&

[[ 1.

(c) ow much wor7 is done in mo(ing a small test charge from the point 5# 0# 0 to −+# 0#

0 along the x ?a@is2 oes the answer change if the path of the test charge %etween the same

points is not along the x ?a@is2

(a) Rero at %oth the points

Charge − q is located at 0# 0# − & and charge H q is located at 0# 0# &. ence# the& form adipole. /oint 0# 0# z is on the a@is of this dipole and point x # y # 0 is normal to the a@is ofthe dipole. ence# electrostatic potential at point x # y # 0 is ero. "lectrostatic potential at

point 0# 0# z is gi(en %&#

*here#


p $ ipole moment of the s&stem of two charges $ )q&



(b) istance r is much greater than half of the distance %etween the two charges. ence# the

potential 7 at a distance r is in(ersel& proportional to s!uare of the distance i.e.#

(c) Rero

'he answer does not change if the path of the test is not along the x ?a@is.

A test charge is mo(ed from point 5# 0# 0 to point −+# 0# 0 along the x ?a@is. "lectrostaticpotential 7 1 at point 5# 0# 0 is gi(en %&#

"lectrostatic potential# 7 )# at point − +# 0# 0 is gi(en %&#

ence# no wor7 is done in mo(ing a small test charge from point 5# 0# 0 to point −+# 0# 0along the x ?a@is.

'he answer does not change %ecause wor7 done %& the electrostatic field in mo(ing a testcharge %etween the two points is independent of the path connecting the two points.

Question 2.22:

igure ).34 shows a charge arra& 7nown as an e!e+tri+ q,&dr,p%!e. or a point on the a@is of

the !uadrupole# o%tain the dependence of potential on r for r ;& [[ 1# and contrast &our

results with that due to an electric dipole# and an electric monopole i.e.# a single charge.

our charges of same magnitude are placed at points Q# 6# 6# and R respecti(el&# as shown inthe following figure.



A point is located at /# which is r distance awa& from point 6.

'he s&stem of charges forms an electric !uadrupole.

t can %e considered that the s&stem of the electric !uadrupole has three charges.

Charge Hq placed at point Q

Charge −)q placed at point 6

Charge Hq placed at point R

Q6 $ 6R $ &

6/ $ r

/Q $ r H &

/R $ r − &

"lectrostatic potential caused %& the s&stem of three charges at point / is gi(en %&#

Fince #



is ta7en as negligi%le.

t can %e inferred that potential#

owe(er# it is 7nown that for a dipole#

And# for a monopole#

Question 2.23:

An electrical technician re!uires a capacitance of ) U in a circuit across a potential difference

of 1 7M. A large num%er of 1 U capacitors are a(aila%le to him each of which can withstand a

potential difference of not more than 400 M. Fuggest a possi%le arrangement that re!uires the

minimum num%er of capacitors.

'otal re!uired capacitance# $ $ ) U

/otential difference# 7 $ 1 7M $ 1000 M

Capacitance of each capacitor# $ 1 $ 1U

"ach capacitor can withstand a potential difference# 7 1 $ 400 M

Fuppose a num%er of capacitors are connected in series and these series circuits areconnected in parallel row to each other. 'he potential difference across each row must %e1000 M and potential difference across each capacitor must %e 400 M. ence# num%er ofcapacitors in each row is gi(en as

ence# there are three capacitors in each row.

Capacitance of each row

<et there are n rows# each ha(ing three capacitors# which are connected in parallel. ence#e!ui(alent capacitance of the circuit is gi(en as



/otential difference across $

/otential difference across $ 4 $ 7 4

Charge on

Q4$ $7

ence# potential difference# 7 1# across $ 1 is 100 M.

Charge on $ 1 is gi(en %&#

$ ) and $ 3 ha(ing same capacitances ha(e a potential difference of 100 M together. Fince $ ) and $ 3 are in series# the potential difference across $ ) and $ 3 is gi(en %&#

7 ) = 7 3 = 50 M

'herefore# charge on $ ) is gi(en %&#

And charge on $ 3 is gi(en %&#



ence# the e!ui(alent capacitance of the gi(en circuit is

Question 2.26:

'he plates of a parallel plate capacitor ha(e an area of 90 cm)

each and are separated %& ).5mm. 'he capacitor is charged %& connecting it to a 400 M suppl&.

(a) ow much electrostatic energ& is stored %& the capacitor2

(b) Miew this energ& as stored in the electrostatic field %etween the plates# and o%tain the

energ& per unit (olume ,. ence arri(e at a relation %etween , and the magnitude of electric

field E %etween the plates.

Area of the plates of a parallel plate capacitor# A $ 90 cm) $ 90 × 10−4 m)

istance %etween the plates# d $ ).5 mm $ ).5 × 10−3 m

/otential difference across the plates# 7 $ 400 M

(a) Capacitance of the capacitor is gi(en %& the relation#

"lectrostatic energ& stored in the capacitor is gi(en %& the relation#

*here#

$ /ermitti(it& of free space $ .5 × 10−1) C) −1 m−)



ence# the electrostatic energ& stored %& the capacitor is

(b) Molume of the gi(en capacitor#

"nerg& stored in the capacitor per unit (olume is gi(en %&#

*here#

$ "lectric intensit& $ E

Question 2.27:

A 4 U capacitor is charged %& a )00 M suppl&. t is then disconnected from the suppl&# and is

connected to another uncharged ) U capacitor. ow much electrostatic energ& of the first

capacitor is lost in the form of heat and electromagnetic radiation2

Capacitance of a charged capacitor#

Fuppl& (oltage# 7 1 $ )00 M

"lectrostatic energ& stored in $ 1 is gi(en %&#



Capacitance of an uncharged capacitor#

*hen $ ) is connected to the circuit# the potential ac!uired %& it is 7 ).

According to the conser(ation of charge# initial charge on capacitor $ 1 is e!ual to the finalcharge on capacitors# $ 1 and $ ).

"lectrostatic energ& for the com%ination of two capacitors is gi(en %&#

ence# amount of electrostatic energ& lost %& capacitor $ 1

$ E 1 − E )

$ 0.0 − 0.0533 $ 0.0)-+

$ ).-+ × 10−) V

Question 2.28:

Fhow that the force on each plate of a parallel plate capacitor has a magnitude e!ual to \QE # where Q is the charge on the capacitor# and E is the magnitude of electric field %etween

the plates. "@plain the origin of the factor \.

<et F %e the force applied to separate the plates of a parallel plate capacitor %& a distance of] x . ence# wor7 done %& the force to do so $ F>x

As a result# the potential energ& of the capacitor increases %& an amount gi(en as ,A>x .



*here#

, $ "nerg& densit&


d $ istance %etween the plates

7 $ /otential difference across the plates

'he wor7 done will %e e!ual to the increase in the potential energ& i.e.#

"lectric intensit& is gi(en %&#

owe(er# capacitance#

Charge on the capacitor is gi(en %&#

Q $ $7

'he ph&sical origin of the factor# # in the force formula lies in the fact that Gust outside the

conductor# field is E and inside it is ero. ence# it is the a(erage (alue# # of the field thatcontri%utes to the force.

Question 2.29:



A spherical capacitor consists of two concentric spherical conductors# held in position %&

suita%le insulating supports ig. ).3-. Fhow

that the capacitance of a spherical capacitor is gi(en %&

where r 1 and r ) are the radii of outer and inner spheres# respecti(el&.

,adius of the outer shell $ r 1

,adius of the inner shell $ r )

'he inner surface of the outer shell has charge HQ.

'he outer surface of the inner shell has induced charge −Q.

/otential difference %etween the two shells is gi(en %&#

*here#




ence# pro(ed.

Question 2.30:

A spherical capacitor has an inner sphere of radius 1) cm and an outer sphere of radius 13

cm. 'he outer sphere is earthed and the inner sphere is gi(en a charge of ).5 UC. 'he space

%etween the concentric spheres is filled with a li!uid of dielectric constant 3).

(a) etermine the capacitance of the capacitor.

(b) *hat is the potential of the inner sphere2

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 1) cm.

"@plain wh& the latter is much smaller.

,adius of the inner sphere# $ 1) cm $ 0.1) m

,adius of the outer sphere# $ 13 cm $ 0.13 m

Charge on the inner sphere#

ielectric constant of a li!uid#

(a)

*here#



(c) A small test charge is released at rest at a point in an electrostatic field configuration. *ill

it tra(el along the field line passing through that point2

(d) *hat is the wor7 done %& the field of a nucleus in a complete circular or%it of the electron2

*hat if the or%it is elliptical2

(e) *e 7now that electric field is discontinuous across the surface of a charged conductor. s

electric potential also discontinuous there2

(f) *hat meaning would &ou gi(e to the capacitance of a single conductor2

(g) =uess a possi%le reason wh& water has a much greater dielectric constant $ 0 than

sa&# mica $ -.

(a) 'he force %etween two conducting spheres is not e@actl& gi(en %& the e@pression# Q0

Q) ;4^ r )# %ecause there is a non?uniform charge distri%ution on the spheres.

(b) =aussBs law will not %e true# if Coulom%Bs law in(ol(ed 1;r 3 dependence# instead of1;r )# onr .

(c) 6es#

f a small test charge is released at rest at a point in an electrostatic field configuration# then itwill tra(el along the field lines passing through the point# onl& if the field lines are straight.'his is %ecause the field lines gi(e the direction of acceleration and not of (elocit&.

(d) *hene(er the electron completes an or%it# either circular or elliptical# the wor7 done %&

the field of a nucleus is ero.

(e) o

"lectric field is discontinuous across the surface of a charged conductor. owe(er# electricpotential is continuous.

(f) 'he capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinit&.

(g) *ater has an uns&mmetrical space as compared to mica. Fince it has a permanent dipolemoment# it has a greater dielectric constant than mica.

Question 2.32:

A c&lindrical capacitor has two co?a@ial c&linders of length 15 cm and radii 1.5 cm and 1.4 cm.

'he outer c&linder is earthed and the inner c&linder is gi(en a charge of 3.5 UC. etermine the

capacitance of the s&stem and the potential of the inner c&linder. eglect end effects i.e.#

%ending of field lines at the ends.



<ength of a co?a@ial c&linder# ! $ 15 cm $ 0.15 m

,adius of outer c&linder# r 1 $ 1.5 cm $ 0.015 m

,adius of inner c&linder# r ) $ 1.4 cm $ 0.014 m

Charge on the inner c&linder# q $ 3.5 UC $ 3.5 × 10−- C

*here#

$ /ermitti(it& of free space $

/otential difference of the inner c&linder is gi(en %&#

Question 2.33:

A parallel plate capacitor is to %e designed with a (oltage rating 1 7M# using a material of

dielectric constant 3 and dielectric strength a%out 10+ Mm−1. ielectric strength is the

ma@imum electric field a material can tolerate without %rea7down# i.e.# without starting to

conduct electricit& through partial ionisation. or safet&# we should li7e the field ne(er to

e@ceed# sa& 10_ of the dielectric strength. *hat minimum area of the plates is re!uired to

ha(e a capacitance of 50 p2

/otential rating of a parallel plate capacitor# 7 $ 1 7M $ 1000 M

ielectric constant of a material# $3



ielectric strength $ 10+ M;m

or safet&# the field intensit& ne(er e@ceeds 10_ of the dielectric strength.

ence# electric field intensit&# E $ 10_ of 10+ $ 10- M;m

Capacitance of the parallel plate capacitor# $ $ 50 p $ 50 × 10−1)

istance %etween the plates is gi(en %&#

*here#


$ /ermitti(it& of free space $

ence# the area of each plate is a%out 19 cm).

Question 2.34:

escri%e schematicall& the e!uipotential surfaces corresponding to

(a) a constant electric field in the z ?direction#

(b) a field that uniforml& increases in magnitude %ut remains in a constant sa&# z direction#

(c) a single positi(e charge at the origin# and

(d) a uniform grid consisting of long e!uall& spaced parallel charged wires in a plane.

a) "!uidistant planes parallel to the x ?y plane are the e!uipotential surfaces.



(b) /lanes parallel to the x-y plane are the e!uipotential surfaces with the e@ception thatwhen the planes get closer# the field increases.

(c) Concentric spheres centered at the origin are e!uipotential surfaces.

(d) A periodicall& (ar&ing shape near the gi(en grid is the e!uipotential surface. 'his shape

graduall& reaches the shape of planes parallel to the grid at a larger distance.

Question 2.35:

n a Man de =raaff t&pe generator a spherical metal shell is to %e a 15 × 10- M electrode. 'he

dielectric strength of the gas surrounding the electrode is 5 × 10+ Mm−1. *hat is the minimum

radius of the spherical shell re!uired2 6ou will learn from this e@ercise wh& one cannot %uild

an electrostatic generator using a (er& small shell which re!uires a small charge to ac!uire a

high potential.

/otential difference# 7 $ 15 × 10- M

ielectric strength of the surrounding gas $ 5 × 10+ M;m

"lectric field intensit&# E $ ielectric strength $ 5 × 10+ M;m

>inimum radius of the spherical shell re!uired for the purpose is gi(en %&#

ence# the minimum radius of the spherical shell re!uired is 30 cm.

Question 2.36:

A small sphere of radius r 1 and charge q1 is enclosed %& a spherical shell of radius r ) and

charge q). Fhow that if q1 is positi(e# charge will necessaril& flow from the sphere to the shell

when the two are connected %& a wire no matter what the charge q) on the shell is.

According to =aussBs law# the electric field %etween a sphere and a shell is determined %& thecharge q1 on a small sphere. ence# the potential difference# 7 # %etween the sphere and the

shell is independent of charge q). or positi(e charge q1# potential difference 7 is alwa&spositi(e.

Question 2.37:

Answer the followingI



(a) 'he top of the atmosphere is at a%out 400 7M with respect to the surface of the earth#

corresponding to an electric field that decreases with altitude. ear the surface of the earth#

the field is a%out 100 Mm−1. *h& then do we not get an electric shoc7 as we step out of our

house into the open2 Assume the house to %e a steel cage so there is no field inside`

(b) A man fi@es outside his house one e(ening a two metre high insulating sla% carr&ing on its

top a large aluminium sheet of area 1m). *ill he get an electric shoc7 if he touches the metal

sheet ne@t morning2

(c) 'he discharging current in the atmosphere due to the small conducti(it& of air is 7nown to

%e 100 A on an a(erage o(er the glo%e. *h& then does the atmosphere not discharge itself

completel& in due course and %ecome electricall& neutral2 n other words# what 7eeps the

atmosphere charged2

(d) *hat are the forms of energ& into which the electrical energ& of the atmosphere isdissipated during a lightning2 intI 'he earth has an electric field of a%out 100 Mm−1 at its

surface in the downward direction# corresponding to a surface charge densit& $ −10−9 C m−).

ue to the slight conducti(it& of the atmosphere up to a%out 50 7m %e&ond which it is good

conductor# a%out H 100 C is pumped e(er& second into the earth as a whole. 'he earth#

howe(er# does not get discharged since thunderstorms and lightning occurring continuall& all

o(er the glo%e pump an e!ual amount of negati(e charge on the earth.

(a) *e do not get an electric shoc7 as we step out of our house %ecause the originale!uipotential surfaces of open air changes# 7eeping our %od& and the ground at the samepotential.

(b) 6es# the man will get an electric shoc7 if he touches the metal sla% ne@t morning. 'hestead& discharging current in the atmosphere charges up the aluminium sheet. As a result# its(oltage rises graduall&. 'he raise in the (oltage depends on the capacitance of the capacitorformed %& the aluminium sla% and the ground.

(c) 'he occurrence of thunderstorms and lightning charges the atmosphere continuousl&.ence# e(en with the presence of discharging current of 100 A# the atmosphere is notdischarged completel&. 'he two opposing currents are in e!uili%rium and the atmosphereremains electricall& neutral.

(d) uring lightning and thunderstorm# light energ&# heat energ&# and sound energ& aredissipated in the atmosphere.

CHAPTER 3 –ELECTRICITYQuestion 3.1:

'he storage %atter& of a car has an emf of 1) M. f the internal resistance of the %atter& is

0.4# what is the ma@imum current that can %e drawn from the %atter&2



"mf of the %atter&# E $ 1) M

nternal resistance of the %atter&# r $ 0.4

>a@imum current drawn from the %atter& $ 3

According to EhmBs law#

'he ma@imum current drawn from the gi(en %atter& is 30 A.

Question 3.2:

A %atter& of emf 10 M and internal resistance 3 is connected to a resistor. f the current in

the circuit is 0.5 A# what is the resistance of the resistor2 *hat is the terminal (oltage of the

%atter& when the circuit is closed2

"mf of the %atter&# E $ 10 M

nternal resistance of the %atter&# r $ 3

Current in the circuit# 3 $ 0.5 A

,esistance of the resistor $ 9

'he relation for current using EhmBs law is#

'erminal (oltage of the resistor $ 7

According to EhmBs law#

7 $ 39

$ 0.5 × 1+



$ .5 M

'herefore# the resistance of the resistor is 1+ and the terminal (oltage is

.5 M.

Question 3.3:

(a) 'hree resistors 1 # ) # and 3 are com%ined in series. *hat is the total resistance of

the com%ination2

(b) f the com%ination is connected to a %atter& of emf 1) M and negligi%le internal resistance#

o%tain the potential drop across each resistor.

(a) 'hree resistors of resistances 1 # ) # and 3 are com%ined in series. 'otal resistance of the com%ination is gi(en %& the alge%raic sum of indi(idual resistances.

'otal resistance $ 1 H ) H 3 $ -

(b) Current flowing through the circuit $ 3

"mf of the %atter&# E $ 1) M

'otal resistance of the circuit# 9 $ -

'he relation for current using EhmBs law is#

/otential drop across 1 resistor $ 7 1

rom EhmBs law# the (alue of 7 1 can %e o%tained as

7 1 $ ) × 1$ ) M D i

/otential drop across ) resistor $ 7 )

Again# from EhmBs law# the (alue of 7 ) can %e o%tained as

7 ) $ ) × )$ 4 M D ii

/otential drop across 3 resistor $ 7 3

Again# from EhmBs law# the (alue of 7 3 can %e o%tained as

7 3 $ ) × 3$ - M D iii



'herefore# the potential drop across 1 # ) # and 3 resistors are ) M# 4 M# and - Mrespecti(el&.

Question 3.4:

(a) 'hree resistors ) # 4 and 5 are com%ined in parallel. *hat is the total resistance of

the com%ination2

(b) f the com%ination is connected to a %atter& of emf )0 M and negligi%le internal resistance#

determine the current through each resistor# and the total current drawn from the %atter&.

(a) 'here are three resistors of resistances#

91 $ ) # 9) $ 4 # and 93 $ 5

'he& are connected in parallel. ence# total resistance 9 of the com%ination is gi(en %&#

'herefore# total resistance of the com%ination is .

(b) "mf of the %atter&# 7 $ )0 M

Current 3 1 flowing through resistor 91 is gi(en %&#

Current 3 ) flowing through resistor 9) is gi(en %&#

Current 3 3 flowing through resistor 93 is gi(en %&#



'otal current# 3 $ 3 1 H 3 ) H 3 3 $ 10 H 5 H 4 $ 19 A

'herefore# the current through each resister is 10 A# 5 A# and 4 A respecti(el& and the totalcurrent is 19 A.

Question 3.5:

At room temperature )+.0 °C the resistance of a heating element is 100 . *hat is the

temperature of the element if the resistance is found to %e 11+ # gi(en that the temperature

coefficient of the material of the resistor is

,oom temperature# 4 $ )+°C

,esistance of the heating element at 4 # 9 $ 100

<et 4 1 is the increased temperature of the filament.

,esistance of the heating element at 4 1# 91 $ 11+

'emperature co?efficient of the material of the filament#

'herefore# at 10)+°C# the resistance of the element is 11+.

Question 3.6:



A negligi%l& small current is passed through a wire of length 15 m and uniform cross?section

-.0 × 10−+ m)# and its resistance is measured to %e 5.0 . *hat is the resisti(it& of the

material at the temperature of the e@periment2

<ength of the wire# ! $15 m

Area of cross?section of the wire# & $ -.0 × 10−+ m)

,esistance of the material of the wire# 9 $ 5.0

,esisti(it& of the material of the wire $ ρ

,esistance is related with the resisti(it& as

'herefore# the resisti(it& of the material is ) × 10−+ m.

Question 3.7:

A sil(er wire has a resistance of ).1 at )+.5 °C# and a resistance of ).+ at 100 °C.

etermine the temperature coefficient of resisti(it& of sil(er.

'emperature# 4 1 $ )+.5°C

,esistance of the sil(er wire at 4 1# 91 $ ).1

'emperature# 4 ) $ 100°C

,esistance of the sil(er wire at 4 )# 9) $ ).+

'emperature coefficient of sil(er $ α

t is related with temperature and resistance as

'herefore# the temperature coefficient of sil(er is 0.0039°C−1.



Question 3.8:

Aheating element using nichrome connected to a )30 M suppl& draws an initial current of 3.) A

which settles after a few seconds toa stead& (alue of ). A. *hat is the stead& temperature of

the heating element if the room temperature is )+.0 °C2 'emperature coefficient of resistance

of nichrome a(eraged o(er the temperature range in(ol(ed is 1.+0 × 10−4 °C −1.

Fuppl& (oltage# 7 $ )30 M

nitial current drawn# 3 1 $ 3.) A

nitial resistance $ 91# which is gi(en %& the relation#

Ftead& state (alue of the current# 3 ) $ ). A

,esistance at the stead& state $ 9)# which is gi(en as

'emperature co?efficient of nichrome# α $ 1.+0 × 10−4 °C −1

nitial temperature of nichrome# 4 1$ )+.0°C

Ftud& state temperature reached %& nichrome $ 4 )

4 ) can %e o%tained %& the relation for α #

'herefore# the stead& temperature of the heating element is -+.5°C

Question 3.10:



(a) n a metre %ridge Nig. 3.)+O# the %alance point is found to %e at 39.5 cm from the end A#

when the resistor ? is of 1).5 . etermine the resistance of 2 . *h& are the connections

%etween resistors in a *heatstone or meter %ridge made of thic7 copper strips2

(b) etermine the %alance point of the %ridge a%o(e if 2 and ? are interchanged.

(c) *hat happens if the gal(anometer and cell are interchanged at the %alance point of the

%ridge2 *ould the gal(anometer show an& current2

A metre %ridge with resistors 2 and ? is represented in the gi(en figure.

(a) 8alance point from end A# ! 1 $ 39.5 cm

,esistance of the resistor ? $ 1).5

Condition for the %alance is gi(en as#

'herefore# the resistance of resistor 2 is .) .

'he connection %etween resistors in a *heatstone or metre %ridge is made of thic7 copperstrips to minimie the resistance# which is not ta7en into consideration in the %ridge formula.

(b) f 2 and ? are interchanged# then ! 1 and 100−! 1 get interchanged.

'he %alance point of the %ridge will %e 100−! 1 from A.

100−! 1 $ 100 − 39.5 $ -0.5 cm

'herefore# the %alance point is -0.5 cm from A.



(c) *hen the gal(anometer and cell are interchanged at the %alance point of the %ridge# thegal(anometer will show no deflection. ence# no current would flow through the gal(anometer.

Question 3.9:

etermine the current in each %ranch of the networ7 shown in fig 3.30I

Current flowing through (arious %ranches of the circuit is represented in the gi(en figure.

3 1 $ Current flowing through the outer circuit

3 ) $ Current flowing through %ranch A8

3 3 $ Current flowing through %ranch A

3 ) − 3 4 $ Current flowing through %ranch 8C

3 3 H 3 4 $ Current flowing through %ranch C

3 4 $ Current flowing through %ranch 8

or the closed circuit A8A# potential is ero i.e.#



103 ) H 53 4 − 53 3 $ 0

)3 ) H 3 4 −3 3 $ 0

3 3 $ )3 ) H 3 4 D 1

or the closed circuit 8C8# potential is ero i.e.#

53 ) − 3 4 − 103 3 H 3 4 − 53 4 $ 0

53 ) H 53 4 − 103 3 − 103 4 − 53 4 $ 0

53 ) − 103 3 − )03 4 $ 0

3 ) $ )3 3 H 43 4 D )

or the closed circuit A8C"A# potential is ero i.e.#

−10 H 10 3 1 H 103 ) H 53 ) − 3 4 $ 0

10 $ 153 ) H 103 1 − 53 4

33 ) H )3 1 − 3 4 $ ) D 3

rom e!uations 1 and )# we o%tain

3 3 $ ))3 3 H 43 4 H 3 4

3 3 $ 43 3 H 3 4 H 3 4

− 33 3 $ 93 4

− 33 4 $ H 3 3 D 4

/utting e!uation 4 in e!uation 1# we o%tain

3 3 $ )3 ) H 3 4

− 43 4 $ )3 )

3 ) $ − )3 4 D 5

t is e(ident from the gi(en figure that#

3 1 $ 3 3 H 3 ) D -

/utting e!uation - in e!uation 1# we o%tain

33 ) H)3 3 H 3 ) − 3 4 $ )

53 ) H )3 3 − 3 4 $ ) D +



/utting e!uations 4 and 5 in e!uation +# we o%tain

5−) 3 4 H )− 3 3 4 − 3 4 $ )

− 103 4 − -3 4 − 3 4 $ )

1+3 4 $ − )

"!uation 4 reduces to

3 3 $ − 33 4

'herefore# current in %ranch

n %ranch 8C $

n %ranch C $

n %ranch A

n %ranch 8 $



'otal current $

Question 3.11:

A storage %atter& of emf .0 M and internal resistance 0.5 is %eing charged %& a 1)0 M dc

suppl& using a series resistor of 15.5 . *hat is the terminal (oltage of the %atter& during

charging2 *hat is the purpose of ha(ing a series resistor in the charging circuit2

"mf of the storage %atter&# E $ .0 M

nternal resistance of the %atter&# r $ 0.5

C suppl& (oltage# 7 $ 1)0 M

,esistance of the resistor# 9 $ 15.5

"ffecti(e (oltage in the circuit $ 7 1

9 is connected to the storage %atter& in series. ence# it can %e written as

7 1 $ 7 − E

7 1 $ 1)0 − $ 11) M

Current flowing in the circuit $ 3 # which is gi(en %& the relation#

Moltage across resistor 9 gi(en %& the product# 39 $ + × 15.5 $ 10.5 M

C suppl& (oltage $ 'erminal (oltage of %atter& H Moltage drop across 9

'erminal (oltage of %atter& $ 1)0 − 10.5 $ 11.5 M

A series resistor in a charging circuit limits the current drawn from the e@ternal source. 'hecurrent will %e e@tremel& high in its a%sence. 'his is (er& dangerous.

Question 3.12:

n a potentiometer arrangement# a cell of emf 1.)5 M gi(es a %alance point at 35.0 cm length

of the wire. f the cell is replaced %& another cell and the %alance point shifts to -3.0 cm# what

is the emf of the second cell2



"mf of the cell# E 1 $ 1.)5 M

8alance point of the potentiometer# ! 1$ 35 cm

'he cell is replaced %& another cell of emf E ).

ew %alance point of the potentiometer# ! ) $ -3 cm

'herefore# emf of the second cell is ).)5M.

Question 3.13:

'he num%er densit& of free electrons in a copper conductor estimated in "@ample 3.1 is .5 ×

10) m−3. ow long does an electron ta7e to drift from one end of a wire 3.0 m long to its

other end2 'he area of cross?section of the wire is ).0 × 10−- m) and it is carr&ing a current of

3.0 A.

um%er densit& of free electrons in a copper conductor# n $ .5 × 10) m−3 <ength of the

copper wire# ! $ 3.0 m

Area of cross?section of the wire# A $ ).0 × 10−- m)

Current carried %& the wire# 3 $ 3.0 A# which is gi(en %& the relation#

3 $ nAe7 d

*here#

e $ "lectric charge $ 1.- × 10−19 C

7 d $ rift (elocit&



'herefore# the time ta7en %& an electron to drift from one end of the wire to the other is ).+ ×104 s.

Question 3.14:

'he earthBs surface has a negati(e surface charge densit& of 10−9 C m−). 'he potential

difference of 400 7M %etween the top of the atmosphere and the surface results due to the

low conducti(it& of the lower atmosphere in a current of onl& 100 A o(er the entire glo%e. fthere were no mechanism of sustaining atmospheric electric field# how much time roughl&

would %e re!uired to neutralise the earthBs surface2 'his ne(er happens in practice %ecause

there is a mechanism to replenish electric charges# namel& the continual thunderstorms and

lightning in different parts of the glo%e. ,adius of earth $ -.3+ × 10- m.

Furface charge densit& of the earth# σ $ 10−9 C m−)

Current o(er the entire glo%e# 3 $ 100 A

,adius of the earth# r $ -.3+ × 10- m

Furface area of the earth#

A $ 4^r )

$ 4^ × -.3+ × 10-)

$ 5.09 × 1014 m)

Charge on the earth surface#

q $ σ × A

$ 10−9 × 5.09 × 1014

$ 5.09 × 105 C

'ime ta7en to neutralie the earthBs surface $ t



Current#

'herefore# the time ta7en to neutralie the earthBs surface is )).++ s.

Question 3.15:

(a) Fi@ lead?acid t&pe of secondar& cells each of emf ).0 M and internal resistance 0.015 are

Goined in series to pro(ide a suppl& to a resistance of .5 . *hat are the current drawn from

the suppl& and its terminal (oltage2

(b) A secondar& cell after long use has an emf of 1.9 M and a large internal resistance of 30

. *hat ma@imum current can %e drawn from the cell2 Could the cell dri(e the starting motor

of a car2

(a) um%er of secondar& cells# n $ -

"mf of each secondar& cell# E $ ).0 M

nternal resistance of each cell# r $ 0.015

series resistor is connected to the com%ination of cells.

,esistance of the resistor# 9 $ .5

Current drawn from the suppl& $ 3 # which is gi(en %& the relation#

'erminal (oltage# 7 $ 39 $ 1.39 × .5 $ 11.+ A

'herefore# the current drawn from the suppl& is 1.39 A and terminal (oltage is

11.+ A.



(b) After a long use# emf of the secondar& cell# E $ 1.9 M

nternal resistance of the cell# r $ 30

ence# ma@imum current

'herefore# the ma@imum current drawn from the cell is 0.005 A. Fince a large current isre!uired to start the motor of a car# the cell cannot %e used to start a motor.

Question 3.16:

'wo wires of e!ual length# one of aluminium and the other of copper ha(e the same

resistance. *hich of the two wires is lighter2 ence e@plain wh& aluminium wires are preferred

for o(erhead power ca%les. ρAl $ ).-3 × 10− m# ρCu $ 1.+) × 10− m# ,elati(e densit& of

Al $ ).+# of Cu $ .9.

,esisti(it& of aluminium# ρAl $ ).-3 × 10− m

,elati(e densit& of aluminium# d 1 $ ).+

<et ! 1 %e the length of aluminium wire and m1 %e its mass.

,esistance of the aluminium wire $ 91

Area of cross?section of the aluminium wire $ A1

,esisti(it& of copper# ρCu $ 1.+) × 10− m

,elati(e densit& of copper# d ) $ .9

<et ! ) %e the length of copper wire and m) %e its mass.

,esistance of the copper wire $ 9)

Area of cross?section of the copper wire $ A)

'he two relations can %e written as

t is gi(en that#



And#

>ass of the aluminium wire#

m1 $ Molume × ensit&

$ A1! 1 × d 1 $ A1 ! 1d 1 D 3

>ass of the copper wire#

m) $ Molume × ensit&

$ A)! ) × d ) $ A) ! )d ) D 4

i(iding e!uation 3 %& e!uation 4# we o%tain

t can %e inferred from this ratio that m1 is less than m). ence# aluminium is lighter thancopper.

Fince aluminium is lighter# it is preferred for o(erhead power ca%les o(er copper.

Question 3.17:



*hat conclusion can &ou draw from the following o%ser(ations on a resistor made of allo&

manganin2

CP,,"'

A

ME<'A="

M

CP,,"'

A

ME<'A="

M

0.) 3.94 3.0 59.)

0.4 +.+ 4.0 +.

0.- 11. 5.0 9.-

0. 15.+ -.0 11.5

1.0 19.+ +.0 13.)

).0 39.4 .0 15.0

t can %e inferred from the gi(en ta%le that the ratio of (oltage with current is a constant#which is e!ual to 19.+. ence# manganin is an ohmic conductor i.e.# the allo& o%e&s EhmBslaw. According to EhmBs law# the ratio of (oltage with current is the resistance of theconductor. ence# the resistance of manganin is 19.+ .

Question 3.18:

Answer the following !uestionsI

(a) A stead& current flows in a metallic conductor of non?uniform cross? section. *hich of

these !uantities is constant along the conductorI current# current densit&# electric field# drift

speed2

(b) s EhmBs law uni(ersall& applica%le for all conducting elements2

f not# gi(e e@amples of elements which do not o%e& EhmBs law.

(c) A low (oltage suppl& from which one needs high currents must ha(e (er& low internal

resistance. *h&2

(d) A high tension ' suppl& of# sa&# - 7M must ha(e a (er& large internal resistance. *h&2

(a) *hen a stead& current flows in a metallic conductor of non?uniform cross?section# thecurrent flowing through the conductor is constant. Current densit&# electric field# and driftspeed are in(ersel& proportional to the area of cross?section. 'herefore# the& are not constant.



(b) =i(en the resistances of 1 # ) # 3 # how will %e com%ine them to get an e!ui(alent

resistance of i 11;3 ii 11;5 # iii - # i( -;11 2

(c) etermine the e!ui(alent resistance of networ7s shown in ig. 3.31.

(a) 'otal num%er of resistors $ n

,esistance of each resistor $ 9

(i) *hen n resistors are connected in series# effecti(e resistance 91is the ma@imum# gi(en %&the product n9.

ence# ma@imum resistance of the com%ination# 91 $ n9

(ii) *hen n resistors are connected in parallel# the effecti(e resistance 9) is the minimum#

gi(en %& the ratio .

ence# minimum resistance of the com%ination# 9) $

(iii) 'he ratio of the ma@imum to the minimum resistance is#

(b) 'he resistance of the gi(en resistors is#

91 $ 1 # 9) $ ) # 93 $ 3 )

i. "!ui(alent resistance#

Consider the following com%ination of the resistors.



"!ui(alent resistance of the circuit is gi(en %&#

ii. "!ui(alent resistance#

Consider the following com%ination of the resistors.


(iii) "!ui(alent resistance# 9B $ -

Consider the series com%ination of the resistors# as shown in the gi(en circuit.

"!ui(alent resistance of the circuit is gi(en %& the sum#

9B $ 1 H ) H 3 $ -

(iv) "!ui(alent resistance#

Consider the series com%ination of the resistors# as shown in the gi(en circuit.




(c) (a) t can %e o%ser(ed from the gi(en circuit that in the first small loop# two resistors ofresistance 1 each are connected in series.

ence# their e!ui(alent resistance $ 1H1 $ )

t can also %e o%ser(ed that two resistors of resistance ) each are connected in series.

ence# their e!ui(alent resistance $ ) H ) $ 4 .

'herefore# the circuit can %e redrawn as

t can %e o%ser(ed that ) and 4 resistors are connected in parallel in all the four loops.ence# e!ui(alent resistance 9B of each loop is gi(en %&#

'he circuit reduces to#

All the four resistors are connected in series.

ence# e!ui(alent resistance of the gi(en circuit is

(b) t can %e o%ser(ed from the gi(en circuit that fi(e resistors of resistance 9 each areconnected in series.

ence# e!ui(alent resistance of the circuit $ 9 H 9 H 9 H 9 H 9

$ 5 9

@

Question 3.21:

etermine the current drawn from a 1) M suppl& with internal resistance 0.5 %& the infinite

networ7 shown in ig. 3.3). "ach resistor has 1 resistance.



'he resistance of each resistor connected in the gi(en circuit# 9 $ 1

"!ui(alent resistance of the gi(en circuit $ 9B

'he networ7 is infinite. ence# e!ui(alent resistance is gi(en %& the relation#

egati(e (alue of 9B cannot %e accepted. ence# e!ui(alent resistance#

nternal resistance of the circuit# r $ 0.5

ence# total resistance of the gi(en circuit $ ).+3 H 0.5 $ 3.)3

Fuppl& (oltage# 7 $ 1) M

According to EhmBs <aw# current drawn from the source is gi(en %& the ratio# $ 3.+) A

Question 3.22:

igure 3.33 shows a potentiometer with a cell of ).0 M and internal resistance 0.40

maintaining a potential drop across the resistor wire A8. A standard cell which maintains a

constant emf of 1.0) M for (er& moderate currents up to a few mA gi(es a %alance point at

-+.3 cm length of the wire. 'o ensure (er& low currents drawn from the standard cell# a (er&

high resistance of -00 7 is put in series with it# which is shorted close to the %alance point.



'he standard cell is then replaced %& a cell of un7nown emf and the %alance point found

similarl&# turns out to %e at ).3 cm length of the wire.

(a) *hat is the (alue 2

(b) *hat purpose does the high resistance of -00 7 ha(e2

(c) s the %alance point affected %& this high resistance2

(d) s the %alance point affected %& the internal resistance of the dri(er cell2

(e) *ould the method wor7 in the a%o(e situation if the dri(er cell of the potentiometer had

an emf of 1.0 M instead of ).0 M2

(f ) *ould the circuit wor7 well for determining an e@tremel& small emf# sa& of the order of afew mM such as the t&pical emf of a thermo?couple2 f not# how will &ou modif& the circuit2

(a) Constant emf of the gi(en standard cell# E 1 $ 1.0) M

8alance point on the wire# ! 1 $ -+.3 cm

A cell of un7nown emf# #replaced the standard cell. 'herefore# new %alance point on the wire#! $ ).3 cm

'he relation connecting emf and %alance point is#

'he (alue of un7nown emfis 1.)4+ M.



(b) 'he purpose of using the high resistance of -00 7 is to reduce the current through thegal(anometer when the mo(a%le contact is far from the %alance point.

(c) 'he %alance point is not affected %& the presence of high resistance.

(d) 'he point is not affected %& the internal resistance of the dri(er cell.

(e) 'he method would not wor7 if the dri(er cell of the potentiometer had an emf of 1.0 Minstead of ).0 M. 'his is %ecause if the emf of the dri(er cell of the potentiometer is less thanthe emf of the other cell# then there would %e no %alance point on the wire.

(f) 'he circuit would not wor7 well for determining an e@tremel& small emf. As the circuitwould %e unsta%le# the %alance point would %e close to end A. ence# there would %e a largepercentage of error.

'he gi(en circuit can %e modified if a series resistance is connected with the wire A8. 'hepotential drop across A8 is slightl& greater than the emf measured. 'he percentage errorwould %e small.

Question 3.23:

igure 3.34 shows a potentiometer circuit for comparison of two resistances. 'he %alance point

with a standard resistor 9 $ 10.0 is found to %e 5.3 cm# while that with the un7nown

resistance 2 is -.5 cm. etermine the (alue of 2 . *hat might &ou do if &ou failed to find a

%alance point with the gi(en cell of emf 2

,esistance of the standard resistor# 9 $ 10.0

8alance point for this resistance# ! 1 $ 5.3 cm

Current in the potentiometer wire $ i

ence# potential drop across 9# E 1 $ i9

,esistance of the un7nown resistor $ 2

8alance point for this resistor# ! ) $ -.5 cm



ence# potential drop across 2 # E ) $ i2

'he relation connecting emf and %alance point is#

'herefore# the (alue of the un7nown resistance# 2 # is 11.+5 .

f we fail to find a %alance point with the gi(en cell of emf# # then the potential drop across 9 and 2 must %e reduced %& putting a resistance in series with it. Enl& if the potential dropacross 9 or 2 is smaller than the potential drop across the potentiometer wire A8# a %alancepoint is o%tained.

Question 3.24:

igure 3.35 shows a ).0 M potentiometer used for the determination of internal resistance of a

1.5 M cell. 'he %alance point of the cell in open circuit is +-.3 cm. *hen a resistor of 9.5 is

used in the e@ternal circuit of the cell# the %alance point shifts to -4. cm length of the

potentiometer wire. etermine the internal resistance of the cell.

nternal resistance of the cell $ r

8alance point of the cell in open circuit# ! 1 $ +-.3 cm

An e@ternal resistance 9 is connected to the circuit with 9 $ 9.5

ew %alance point of the circuit# ! ) $ -4. cm

Current flowing through the circuit $ 3



'he relation connecting resistance and emf is#

'herefore# the internal resistance of the cell is 1.-.

CA/'", 4 ? >EM= CA,="F A >A="'F>

Question 4.1:

A circular coil of wire consisting of 100 turns# each of radius .0 cm carries a current of 0.40 A.

*hat is the magnitude of the magnetic field at the centre of the coil2

um%er of turns on the circular coil# n $ 100

,adius of each turn# r $ .0 cm $ 0.0 m

Current flowing in the coil# 3 $ 0.4 A

>agnitude of the magnetic field at the centre of the coil is gi(en %& the relation#

*here#

$ /ermea%ilit& of free space

$ 4^ × 10b+ ' m Ab1

ence# the magnitude of the magnetic field is 3.14 × 10b4 '.

Question 4.2:

A long straight wire carries a current of 35 A. *hat is the magnitude of the field at a point

)0 cm from the wire2



Current in the wire# 3 $ 35 A

istance of a point from the wire# r $ )0 cm $ 0.) m

>agnitude of the magnetic field at this point is gi(en asI

*here#

$ /ermea%ilit& of free space $ 4^ × 10b+ ' m Ab1

ence# the magnitude of the magnetic field at a point )0 cm from the wire is 3.5 × 10b5 '.

Question 4.3:

A long straight wire in the horiontal plane carries a current of 50 A in north to south direction.

=i(e the magnitude and direction of at a point ).5 m east of the wire.

Current in the wire# 3 $ 50 A

A point is ).5 m awa& from the "ast of the wire.

>agnitude of the distance of the point from the wire# r $ ).5 m.

>agnitude of the magnetic field at that point is gi(en %& the relation#

*here#


'he point is located normal to the wire length at a distance of ).5 m. 'he direction of thecurrent in the wire is (erticall& downward. ence# according to the >a@wellBs right hand thum%rule# the direction of the magnetic field at the gi(en point is (erticall& upward.



Question 4.4:

A horiontal o(erhead power line carries a current of 90 A in east to west direction. *hat is

the magnitude and direction of the magnetic field due to the current 1.5 m %elow the line2

Current in the power line# 3 $ 90 A

/oint is located %elow the power line at distance# r $ 1.5 m

ence# magnetic field at that point is gi(en %& the relation#

*here#

$ /ermea%ilit& of free space $ 4^ × 10b+

' m Ab1

'he current is flowing from "ast to *est. 'he point is %elow the power line. ence# accordingto >a@wellBs right hand thum% rule# the direction of the magnetic field is towards the Fouth.

Question 4.5:

*hat is the magnitude of magnetic force per unit length on a wire carr&ing a current of A

and ma7ing an angle of 30 with the direction of a uniform magnetic field of 0.15 '2

Current in the wire# 3 $ A

>agnitude of the uniform magnetic field# $ 0.15 '

Angle %etween the wire and magnetic field# θ $ 30°.

>agnetic force per unit length on the wire is gi(en asI

) $ 3 sinθ

$ 0.15 × ×1 × sin30°

$ 0.- mb1

ence# the magnetic force per unit length on the wire is 0.- mb1.

Question 4.6:



A 3.0 cm wire carr&ing a current of 10 A is placed inside a solenoid perpendicular to its a@is.

'he magnetic field inside the solenoid is gi(en to %e 0.)+ '. *hat is the magnetic force on the

wire2

<ength of the wire# ! $ 3 cm $ 0.03 m

Current flowing in the wire# 3 $ 10 A

>agnetic field# $ 0.)+ '

Angle %etween the current and magnetic field# θ $ 90°

>agnetic force e@erted on the wire is gi(en asI

F $ 3! sinθ

$ 0.)+ × 10 × 0.03 sin90°

$ .1 × 10b)

ence# the magnetic force on the wire is .1 × 10b) . 'he direction of the force can %eo%tained from lemingBs left hand rule.

Question 4.7:

'wo long and parallel straight wires A and 8 carr&ing currents of .0 A and 5.0 A in the same

direction are separated %& a distance of 4.0 cm. "stimate the force on a 10 cm section of wire

A.

Current flowing in wire A# 3 A $ .0 A

Current flowing in wire 8# 3 8 $ 5.0 A

istance %etween the two wires# r $ 4.0 cm $ 0.04 m

<ength of a section of wire A# ! $ 10 cm $ 0.1 m

orce e@erted on length ! due to the magnetic field is gi(en asI

*here#




'he magnitude of force is ) × 10b5 . 'his is an attracti(e force normal to A towards 8 %ecause

the direction of the currents in the wires is the same.

Question 4.8:

A closel& wound solenoid 0 cm long has 5 la&ers of windings of 400 turns each. 'he diameter

of the solenoid is 1. cm. f the current carried is .0 A# estimate the magnitude of inside

the solenoid near its centre.

<ength of the solenoid# ! $ 0 cm $ 0. m

'here are fi(e la&ers of windings of 400 turns each on the solenoid.

'otal num%er of turns on the solenoid# B $ 5 × 400 $ )000

iameter of the solenoid# C $ 1. cm $ 0.01 m

Current carried %& the solenoid# 3 $ .0 A

>agnitude of the magnetic field inside the solenoid near its centre is gi(en %& the relation#

*here#


ence# the magnitude of the magnetic field inside the solenoid near its centre is ).51) × 10b) '.

Question 4.9:

A s!uare coil of side 10 cm consists of )0 turns and carries a current of 1) A. 'he coil is

suspended (erticall& and the normal to the plane of the coil ma7es an angle of 30 with the

direction of a uniform horiontal magnetic field of magnitude 0.0 '. *hat is the magnitude of

tor!ue e@perienced %& the coil2



<ength of a side of the s!uare coil# ! $ 10 cm $ 0.1 m

Current flowing in the coil# 3 $ 1) A

um%er of turns on the coil# n $ )0

Angle made %& the plane of the coil with magnetic field# θ $ 30°

Ftrength of magnetic field# $ 0.0 '

>agnitude of the magnetic tor!ue e@perienced %& the coil in the magnetic field is gi(en %& therelation#

τ $ n 3A sinθ

*here#

A $ Area of the s!uare coil

! × ! $ 0.1 × 0.1 $ 0.01 m)

∴ τ $ )0 × 0. × 1) × 0.01 × sin30°

$ 0.9- m

ence# the magnitude of the tor!ue e@perienced %& the coil is 0.9- m.

Question 4.10:

'wo mo(ing coil meters# >1 and >) ha(e the following particularsI

91 $ 10 # B 1 $ 30#

A1 $ 3.- × 10b3 m) D 1 $ 0.)5 '

9) $ 14 # B ) $ 4)#

A) $ 1. × 10b3 m)# ) $ 0.50 '

'he spring constants are identical for the two meters.

etermine the ratio of a current sensiti(it& and % (oltage sensiti(it& of >) and >1.

or mo(ing coil meter >1I

,esistance# 91 $ 10

um%er of turns# B 1 $ 30



Area of cross?section# A1 $ 3.- × 10b3 m)

>agnetic field strength# 1 $ 0.)5 '

Fpring constant 1 $

or mo(ing coil meter >)I

,esistance# 9) $ 14

um%er of turns# B ) $ 4)

Area of cross?section# A) $ 1. × 10b3 m)

>agnetic field strength# ) $ 0.50 '

Fpring constant# ) $

(a) Current sensiti(it& of >1 is gi(en asI

And# current sensiti(it& of >) is gi(en asI

,atio

ence# the ratio of current sensiti(it& of >) to >1 is 1.4.

(b) Moltage sensiti(it& for >) is gi(en asI

And# (oltage sensiti(it& for >1 is gi(en asI



,atio

ence# the ratio of (oltage sensiti(it& of >) to >1 is 1.

Question 4.11:

n a cham%er# a uniform magnetic field of -.5 = 1 = $ 10 b4 ' is maintained. An electron is

shot into the field with a speed of 4. × 10- m sb1 normal to the field. "@plain wh& the path of

the electron is a circle. etermine the radius of the circular or%it. e $ 1.- × 10b19 C# me$

9.1×10b31 7g

>agnetic field strength# $ -.5 = $ -.5 × 10b4 '

Fpeed of the electron# " $ 4. × 10- m;s

Charge on the electron# e $ 1.- × 10b19 C

>ass of the electron# me $ 9.1 × 10b31 7g

Angle %etween the shot electron and magnetic field# θ $ 90°

>agnetic force e@erted on the electron in the magnetic field is gi(en asI

F $ e" sinθ

'his force pro(ides centripetal force to the mo(ing electron. ence# the electron starts mo(ingin a circular path of radius r .

ence# centripetal force e@erted on the electron#

n e!uili%rium# the centripetal force e@erted on the electron is e!ual to the magnetic force i.e.#



ence# the radius of the circular or%it of the electron is 4.) cm.

Question 4.12:

n "@ercise 4.11 o%tain the fre!uenc& of re(olution of the electron in its circular or%it. oes

the answer depend on the speed of the electron2 "@plain.

>agnetic field strength# $ -.5 × 10−4 '

Charge of the electron# e $ 1.- × 10−19 C

>ass of the electron# me $ 9.1 × 10−31 7g

Melocit& of the electron# " $ 4. × 10- m;s

,adius of the or%it# r $ 4.) cm $ 0.04) m

re!uenc& of re(olution of the electron $ ν

Angular fre!uenc& of the electron $ $ )^ν

Melocit& of the electron is related to the angular fre!uenc& asI

" $ r

n the circular or%it# the magnetic force on the electron is %alanced %& the centripetal force.ence# we can writeI

'his e@pression for fre!uenc& is independent of the speed of the electron.



En su%stituting the 7nown (alues in this e@pression# we get the fre!uenc& asI

ence# the fre!uenc& of the electron is around 1 > and is independent of the speed of theelectron.

Question 4.13:

(a) A circular coil of 30 turns and radius .0 cm carr&ing a current of -.0 A is suspended

(erticall& in a uniform horiontal magnetic field of magnitude 1.0 '. 'he field lines ma7e an

angle of -0 with the normal of the coil. Calculate the magnitude of the counter tor!ue that

must %e applied to pre(ent the coil from turning.

(b) *ould &our answer change# if the circular coil in a were replaced %& a planar coil of

some irregular shape that encloses the same area2 All other particulars are also unaltered.

(a) um%er of turns on the circular coil# n $ 30

,adius of the coil# r $ .0 cm $ 0.0 m

Area of the coil

Current flowing in the coil# 3 $ -.0 A

>agnetic field strength# $ 1 '

Angle %etween the field lines and normal with the coil surface#

θ $ -0°

'he coil e@periences a tor!ue in the magnetic field. ence# it turns. 'he counter tor!ue appliedto pre(ent the coil from turning is gi(en %& the relation#

τ $ n 3A sinθ D i

$ 30 × - × 1 × 0.0)01 × sin-0°

$ 3.133 m

(b) t can %e inferred from relation i that the magnitude of the applied tor!ue is notdependent on the shape of the coil. t depends on the area of the coil. ence# the answerwould not change if the circular coil in the a%o(e case is replaced %& a planar coil of someirregular shape that encloses the same area.



Question 4.14:

'wo concentric circular coils Q and 6 of radii 1- cm and 10 cm# respecti(el&# lie in the same

(ertical plane containing the north to south direction. Coil Q has )0 turns and carries a current

of 1- AL coil 6 has )5 turns and carries a current of 1 A. 'he sense of the current in Q is

anticloc7wise# and cloc7wise in 6# for an o%ser(er loo7ing at the coils facing west. =i(e the

magnitude and direction of the net magnetic field due to the coils at their centre.

,adius of coil Q# r 1 $ 1- cm $ 0.1- m

,adius of coil 6# r ) $ 10 cm $ 0.1 m

um%er of turns of on coil Q# n1 $ )0

um%er of turns of on coil 6# n) $ )5

Current in coil Q# 3 1 $ 1- A

Current in coil 6# 3 ) $ 1 A

>agnetic field due to coil Q at their centre is gi(en %& the relation#

*here#

$ /ermea%ilit& of free space $

>agnetic field due to coil 6 at their centre is gi(en %& the relation#

ence# net magnetic field can %e o%tained asI



Question 4.15:

A magnetic field of 100 = 1 = $ 10−4 ' is re!uired which is uniform in a region of linear

dimension a%out 10 cm and area of cross?section a%out 10−3 m). 'he ma@imum current?

carr&ing capacit& of a gi(en coil of wire is 15 A and the num%er of turns per unit length that

can %e wound round a core is at most 1000 turns m−1. Fuggest some appropriate design

particulars of a solenoid for the re!uired purpose. Assume the core is not ferromagnetic

>agnetic field strength# $ 100 = $ 100 × 10−4 '

um%er of turns per unit length# n $ 1000 turns m−1

Current flowing in the coil# 3 $ 15 A

/ermea%ilit& of free space# $

>agnetic field is gi(en %& the relation#

f the length of the coil is ta7en as 50 cm# radius 4 cm# num%er of turns 400# and current 10A# then these (alues are not uni!ue for the gi(en purpose. 'here is alwa&s a possi%ilit& ofsome adGustments with limits.

Question 4.16:

or a circular coil of radius 9 and B turns carr&ing current 3 # the magnitude of the magnetic

field at a point on its a@is at a distance x from its centre is gi(en %&#



(a) Fhow that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel co?a@ial circular coils of e!ual radius 9# and num%er of turns B #

carr&ing e!ual currents in the same direction# and separated %& a distance 9. Fhow that the

field on the a@is around the mid?point %etween the coils is uniform o(er a distance that is

small as compared to 9# and is gi(en %&#

# appro@imatel&.

NFuch an arrangement to produce a nearl& uniform magnetic field o(er a small region is 7nown

as He!m'%!tz +%i!s.O

,adius of circular coil $ 9

um%er of turns on the coil $ B

Current in the coil $ 3

>agnetic field at a point on its a@is at distance x is gi(en %& the relation#

*here#


(a) f the magnetic field at the centre of the coil is considered# then x $ 0.

'his is the familiar result for magnetic field at the centre of the coil.

(b) ,adii of two parallel co?a@ial circular coils $ 9

um%er of turns on each coil $ B

Current in %oth coils $ 3

istance %etween %oth the coils $ 9

<et us consider point S at distance d from the centre.



'hen# one coil is at a distance of from point S.

>agnetic field at point S is gi(en asI

Also# the other coil is at a distance of from point S.

>agnetic field due to this coil is gi(en asI

'otal magnetic field#



(c) >agnetic field in the empt& space surrounded %& the toroid is ero.

Question 4.18:


(a) A magnetic field that (aries in magnitude from point to point %ut has a constant direction

east to west is set up in a cham%er. A charged particle enters the cham%er and tra(els

undeflected along a straight path with constant speed. *hat can &ou sa& a%out the initial

(elocit& of the particle2

(b) A charged particle enters an en(ironment of a strong and non?uniform magnetic field

(ar&ing from point to point %oth in magnitude and direction# and comes out of it following a

complicated traGector&. *ould its final speed e!ual the initial speed if it suffered no collisions

with the en(ironment2

(c) An electron tra(elling west to east enters a cham%er ha(ing a uniform electrostatic field in

north to south direction. Fpecif& the direction in which a uniform magnetic field should %e set

up to pre(ent the electron from deflecting from its straight line path.

(a) 'he initial (elocit& of the particle is either parallel or anti?parallel to the magnetic field.ence# it tra(els along a straight path without suffering an& deflection in the field.

(b) 6es# the final speed of the charged particle will %e e!ual to its initial speed. 'his is %ecausemagnetic force can change the direction of (elocit&# %ut not its magnitude.

(c) An electron tra(elling from *est to "ast enters a cham%er ha(ing a uniform electrostaticfield in the orth?Fouth direction. 'his mo(ing electron can remain undeflected if the electricforce acting on it is e!ual and opposite of magnetic field. >agnetic force is directed towardsthe Fouth. According to lemingBs left hand rule# magnetic field should %e applied in a(erticall& downward direction.

Question 4.19:

An electron emitted %& a heated cathode and accelerated through a potential difference of ).0

7M# enters a region with uniform magnetic field of 0.15 '. etermine the traGector& of the

electron if the field a is trans(erse to its initial (elocit&# % ma7es an angle of 30 with the

initial (elocit&.

>agnetic field strength# $ 0.15 '

Charge on the electron# e $ 1.- × 10−19 C

>ass of the electron# m $ 9.1 × 10−31 7g

/otential difference# 7 $ ).0 7M $ ) × 103 M



'hus# 7inetic energ& of the electron $ e7

*here#

" $ (elocit& of the electron

(a) >agnetic force on the electron pro(ides the re!uired centripetal force of the electron.ence# the electron traces a circular path of radius r .

>agnetic force on the electron is gi(en %& the relation#

e"

Centripetal force

rom e!uations 1 and )# we get

ence# the electron has a circular traGector& of radius 1.0 mm normal to the magnetic field.

(b) *hen the field ma7es an angle θ of 30° with initial (elocit&# the initial (elocit& will %e#

rom e!uation )# we can write the e@pression for new radius asI



ence# the electron has a helical traGector& of radius 0.5 mm along the magnetic fielddirection.

Question 4.20:

A magnetic field set up using elmholt coils descri%ed in "@ercise 4.1- is uniform in a small

region and has a magnitude of 0.+5 '. n the same region# a uniform electrostatic field is

maintained in a direction normal to the common a@is of the coils. A narrow %eam of single

species charged particles all accelerated through 15 7M enters this region in a direction

perpendicular to %oth the a@is of the coils and the electrostatic field. f the %eam remains

undeflected when the electrostatic field is 9.0 × 10−5 M m−1# ma7e a simple guess as to what

the %eam contains. *h& is the answer not uni!ue2

>agnetic field# $ 0.+5 '

Accelerating (oltage# 7 $ 15 7M $ 15 × 103 M

"lectrostatic field# E $ 9 × 105 M m−1

>ass of the electron $ m

Charge of the electron $ e

Melocit& of the electron $ "

Zinetic energ& of the electron $ e7



Fince the particle remains undeflected %& electric and magnetic fields# we can infer that theelectric field is %alancing the magnetic field.

/utting e!uation ) in e!uation 1# we get

'his (alue of specific charge e ;m is e!ual to the (alue of deuteron or deuterium ions. 'his isnot a uni!ue answer. Ether possi%le answers are eHH# <iHH# etc.

Question 4.21:

A straight horiontal conducting rod of length 0.45 m and mass -0 g is suspended %& two

(ertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.

(a) *hat magnetic field should %e set up normal to the conductor in order that the tension in

the wires is ero2

(b) *hat will %e the total tension in the wires if the direction of current is re(ersed 7eeping

the magnetic field same as %efore2 gnore the mass of the wires. g $ 9. m s−).

<ength of the rod# ! $ 0.45 m

>ass suspended %& the wires# m $ -0 g $ -0 × 10−3 7g

Acceleration due to gra(it&# g $ 9. m;s)

Current in the rod flowing through the wire# 3 $ 5 A

(a) >agnetic field is e!ual and opposite to the weight of the wire i.e.#



A horiontal magnetic field of 0.)- ' normal to the length of the conductor should %e set up inorder to get ero tension in the wire. 'he magnetic field should %e such that lemingBs lefthand rule gi(es an upward magnetic force.

(b) f the direction of the current is re(ered# then the force due to magnetic field and theweight of the wire acts in a (erticall& downward direction.

∴'otal tension in the wire $ 3! H mg

Question 4.22:

'he wires which connect the %atter& of an automo%ile to its starting motor carr& a current of

300 A for a short time. *hat is the force per unit length %etween the wires if the& are +0 cm

long and 1.5 cm apart2 s the force attracti(e or repulsi(e2

Current in %oth wires# 3 $ 300 A

istance %etween the wires# r $ 1.5 cm $ 0.015 m

<ength of the two wires# ! $ +0 cm $ 0.+ m

orce %etween the two wires is gi(en %& the relation#

*here#


Fince the direction of the current in the wires is opposite# a repulsi(e force e@ists %etween

them.

Question 4.23:

A uniform magnetic field of 1.5 ' e@ists in a c&lindrical region of radius10.0 cm# its direction

parallel to the a@is along east to west. A wire carr&ing current of +.0 A in the north to south



direction passes through this region. *hat is the magnitude and direction of the force on the

wire if#

(a) the wire intersects the a@is#

(b) the wire is turned from ?F to northeast?northwest direction#

(c) the wire in the ?F direction is lowered from the a@is %& a distance of -.0 cm2


,adius of the c&lindrical region# r $ 10 cm $ 0.1 m

Current in the wire passing through the c&lindrical region# 3 $ + A

(a) f the wire intersects the a@is# then the length of the wire is the diameter of the c&lindrical

region.

'hus# ! $ )r $ 0.) m

Angle %etween magnetic field and current# θ $ 90°

>agnetic force acting on the wire is gi(en %& the relation#

F $ 3! sin θ

$ 1.5 × + × 0.) × sin 90°

$ ).1

ence# a force of ).1 acts on the wire in a (erticall& downward direction.

(b) ew length of the wire after turning it to the ortheast?orthwest direction can %e gi(enasI I

Angle %etween magnetic field and current# θ $ 45°

orce on the wire#

F $ 3! 1 sin θ



ence# a force of ).1 acts (erticall& downward on the wire. 'his is independent ofangleθ%ecause ! sinθ is fi@ed.

(c) 'he wire is lowered from the a@is %& distance# d $ -.0 cm

<et ! ) %e the new length of the wire.

>agnetic force e@erted on the wire#

ence# a force of 1.- acts in a (erticall& downward direction on the wire.

Question 4.24:

A uniform magnetic field of 3000 = is esta%lished along the positi(e z ?direction. A rectangular

loop of sides 10 cm and 5 cm carries a current of 1) A. *hat is the tor!ue on the loop in the

different cases shown in ig. 4.)2 *hat is the force on each case2 *hich case corresponds to

sta%le e!uili%rium2

>agnetic field strength# $ 3000 = $ 3000 × 10−4 ' $ 0.3 '



<ength of the rectangular loop# ! $ 10 cm

*idth of the rectangular loop# 5 $ 5 cm

Area of the loop#

A $ ! × 5 $ 10 × 5 $ 50 cm) $ 50 × 10−4 m)

Current in the loop# 3 $ 1) A

ow# ta7ing the anti?cloc7wise direction of the current as positi(e and (ise?(ersaI

(a) 'or!ue#

rom the gi(en figure# it can %e o%ser(ed that A is normal to the y ? z plane and is directedalong the z ?a@is.

'he tor!ue is m along the negati(e y ?direction. 'he force on the loop is ero%ecause the angle %etween A and is ero.

(b) 'his case is similar to case a. ence# the answer is the same as a.

(c) 'or!ue

rom the gi(en figure# it can %e o%ser(ed that A is normal to the x ? z plane and is directedalong the z ?a@is.

'he tor!ue is m along the negati(e x direction and the force is ero.

(d) >agnitude of tor!ue is gi(en asI

'or!ue is m at an angle of )40° with positi(e x direction. 'he force is ero.

(e) 'or!ue



ence# the tor!ue is ero. 'he force is also ero.

(f) 'or!ue

ence# the tor!ue is ero. 'he force is also ero.

n case e# the direction of and is the same and the angle %etween them is ero. fdisplaced# the& come %ac7 to an e!uili%rium. ence# its e!uili%rium is sta%le.

*hereas# in case f# the direction of and is opposite. 'he angle %etween them is 10°.f distur%ed# it does not come %ac7 to its original position. ence# its e!uili%rium is unsta%le.

Question 4.25:

A circular coil of )0 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 '

normal to the plane of the coil. f the current in the coil is 5.0 A# what is the

(a) total tor!ue on the coil#

(b) total force on the coil#

(c) a(erage force on each electron in the coil due to the magnetic field2

'he coil is made of copper wire of cross?sectional area 10−5 m)# and the free electron densit&

in copper is gi(en to %e a%out 10)9 m−3.

um%er of turns on the circular coil# n $ )0

,adius of the coil# r $ 10 cm $ 0.1 m


Current in the coil# 3 $ 5.0 A

(a) 'he total tor!ue on the coil is ero %ecause the field is uniform.

(b) 'he total force on the coil is ero %ecause the field is uniform.

(c) Cross?sectional area of copper coil# A $ 10−5 m)



um%er of free electrons per cu%ic meter in copper# B $ 10)9 ;m3

Charge on the electron# e $ 1.- × 10−19 C

>agnetic force# F $ e" d

*here#

" d $ rift (elocit& of electrons

ence# the a(erage force on each electron is

Question 4.26:

A solenoid -0 cm long and of radius 4.0 cm has 3 la&ers of windings of 300 turns each. A ).0

cm long wire of mass ).5 g lies inside the solenoid near its centre normal to its a@isL %oth

the wire and the a@is of the solenoid are in the horiontal plane. 'he wire is connected through

two leads parallel to the a@is of the solenoid to an e@ternal %atter& which supplies a current of

-.0 A in the wire. *hat (alue of current with appropriate sense of circulation in the windings

of the solenoid can support the weight of the wire2 g $ 9. m s−)

<ength of the solenoid# # $ -0 cm $ 0.- m

,adius of the solenoid# r $ 4.0 cm $ 0.04 m

t is gi(en that there are 3 la&ers of windings of 300 turns each.

'otal num%er of turns# n $ 3 × 300 $ 900

<ength of the wire# ! $ ) cm $ 0.0) m

>ass of the wire# m $ ).5 g $ ).5 × 10−3

7g

Current flowing through the wire# i $ - A


>agnetic field produced inside the solenoid#



*here#


3 $ Current flowing through the windings of the solenoid

>agnetic force is gi(en %& the relation#

Also# the force on the wire is e!ual to the weight of the wire.

ence# the current flowing through the solenoid is 10 A.

Question 4.27:

A gal(anometer coil has a resistance of 1) and the metre shows full scale deflection for a

current of 3 mA. ow will &ou con(ert the metre into a (oltmeter of range 0 to 1 M2

,esistance of the gal(anometer coil# G $ 1)

Current for which there is full scale deflection# $ 3 mA $ 3 × 10−3 A

,ange of the (oltmeter is 0# which needs to %e con(erted to 1 M.

7 $ 1 M

<et a resistor of resistance 9 %e connected in series with the gal(anometer to con(ert it into a

(oltmeter. 'his resistance is gi(en asI



ence# a resistor of resistance is to %e connected in series with the gal(anometer.

Question 4.28:

A gal(anometer coil has a resistance of 15 and the metre shows full scale deflection for a

current of 4 mA. ow will &ou con(ert the metre into an ammeter of range 0 to - A2

,esistance of the gal(anometer coil# G $ 15

Current for which the gal(anometer shows full scale deflection#

$ 4 mA $ 4 × 10−3 A

,ange of the ammeter is 0# which needs to %e con(erted to - A.

Current# 3 $ - A

A shunt resistor of resistance is to %e connected in parallel with the gal(anometer to con(ertit into an ammeter. 'he (alue of is gi(en asI

ence# a shunt resistor is to %e connected in parallel with the gal(anometer.

CA/'", ?5.>A="'F> A >A''",

Question 5.1:

Answer the following !uestions regarding earthBs magnetismI

(a) A (ector needs three !uantities for its specification. ame the three independent

!uantities con(entionall& used to specif& the earthBs magnetic field.

(b) 'he angle of dip at a location in southern ndia is a%out 1.



*ould &ou e@pect a greater or smaller dip angle in 8ritain2

(c) f &ou made a map of magnetic field lines at >el%ourne in Australia# would the lines seem

to go into the ground or come out of the ground2

(d) n which direction would a compass free to mo(e in the (ertical plane point to# if located

right on the geomagnetic north or south pole2

(e) 'he earthBs field# it is claimed# roughl& appro@imates the field due to a dipole of magnetic

moment × 10)) V '−1 located at its centre. Chec7 the order of magnitude of this num%er in

some wa&.

(f ) =eologists claim that %esides the main magnetic ?F poles# there are se(eral local poles

on the earthBs surface oriented in different directions. ow is such a thing possi%le at all2

(a) 'he three independent !uantities con(entionall& used for specif&ing earthBs magnetic fieldareI

i >agnetic declination#

ii Angle of dip# and

iii oriontal component of earthBs magnetic field

(b)'he angle of dip at a point depends on how far the point is located with respect to theorth /ole or the Fouth /ole. 'he angle of dip would %e greater in 8ritain it is a%out +0° thanin southern ndia %ecause the location of 8ritain on the glo%e is closer to the magnetic orth

/ole.

(c)t is h&potheticall& considered that a huge %ar magnet is dipped inside earth with its northpole near the geographic Fouth /ole and its south pole near the geographic orth /ole.

>agnetic field lines emanate from a magnetic north pole and terminate at a magnetic southpole. ence# in a map depicting earthBs magnetic field lines# the field lines at >el%ourne#Australia would seem to come out of the ground.

(d)f a compass is located on the geomagnetic orth /ole or Fouth /ole# then the compasswill %e free to mo(e in the horiontal plane while earthBs field is e@actl& (ertical to themagnetic poles. n such a case# the compass can point in an& direction.

(e)>agnetic moment# $ × 10)) V '−1

,adius of earth D r $ -.4 × 10- m

>agnetic field strength#

*here#




'his !uantit& is of the order of magnitude of the o%ser(ed field on earth.

(f)6es# there are se(eral local poles on earthBs surface oriented in different directions. Amagnetised mineral deposit is an e@ample of a local ?F pole.

Question 5.2:


(a) 'he earthBs magnetic field (aries from point to point in space.

oes it also change with time2 f so# on what time scale does it change apprecia%l&2

(b) 'he earthBs core is 7nown to contain iron. 6et geologists do not regard this as a source of

the earthBs magnetism. *h&2

(c) 'he charged currents in the outer conducting regions of the earthBs core are thought to %e

responsi%le for earthBs magnetism. *hat might %e the %atter&B i.e.# the source of energ& to

sustain these currents2

(d) 'he earth ma& ha(e e(en re(ersed the direction of its field se(eral times during its histor&

of 4 to 5 %illion &ears. ow can geologists 7now a%out the earthBs field in such distant past2

(e) 'he earthBs field departs from its dipole shape su%stantiall& at large distances greater

than a%out 30#000 7m. *hat agencies ma& %e responsi%le for this distortion2

(f ) nterstellar space has an e@tremel& wea7 magnetic field of the order of 10−1) '. Can such

a wea7 field %e of an& significant conse!uence2 "@plain.

NN!"e# "@ercise 5.) is meant mainl& to arouse &our curiosit&. Answers to some !uestions

a%o(e are tentati(e or un7nown. 8rief answers where(er possi%le are gi(en at the end. or

details# &ou should consult a good te@t on geomagnetism.O

(a) "arthBs magnetic field changes with time. t ta7es a few hundred &ears to change %& anapprecia%le amount. 'he (ariation in earthBs magnetic field with the time cannot %e neglected.



(b)"arthBs core contains molten iron. 'his form of iron is not ferromagnetic. ence# this is notconsidered as a source of earthBs magnetism.

(c)'heradioacti(it& in earthBs interior is the source of energ& that sustains the currents in theouter conducting regions of earthBs core. 'hese charged currents are considered to %eresponsi%le for earthBs magnetism.

(d)"arth re(ersed the direction of its field se(eral times during its histor& of 4 to 5 %illion&ears. 'hese magnetic fields got wea7l& recorded in roc7s during their solidification. Ene canget clues a%out the geomagnetic histor& from the anal&sis of this roc7 magnetism.

(e)"arthBs field departs from its dipole shape su%stantiall& at large distances greater thana%out 30#000 7m %ecause of the presence of the ionosphere. n this region# earthBs field getsmodified %ecause of the field of single ions. *hile in motion# these ions produce the magneticfield associated with them.

(f)An e@tremel& wea7 magnetic field can %end charged particles mo(ing in a circle. 'his ma&not %e noticea%le for a large radius path. *ith reference to the gigantic interstellar space# thedeflection can affect the passage of charged particles.

Question 5.3:

A short %ar magnet placed with its a@is at 30 with a uniform e@ternal magnetic field of 0.)5 '

e@periences a tor!ue of magnitude e!ual to 4.5 × 10−) V. *hat is the magnitude of magnetic

moment of the magnet2

>agnetic field strength# $ 0.)5 '

'or!ue on the %ar magnet# 4 $ 4.5 × 10−) V

Angle %etween the %ar magnet and the e@ternal magnetic field#θ $ 30°

'or!ue is related to magnetic moment asI

4 $ sin

ence# the magnetic moment of the magnet is 0.3- V '−1.

Question 5.4:

A short %ar magnet of magnetic moment m $ 0.3) V '−1 is placed in a uniform magnetic field

of 0.15 '. f the %ar is free to rotate in the plane of the field# which orientation would



correspond to its a sta%le# and % unsta%le e!uili%rium2 *hat is the potential energ& of the

magnet in each case2

>oment of the %ar magnet# $ 0.3) V '−1

"@ternal magnetic field# $ 0.15 '

(a)'he %ar magnet is aligned along the magnetic field. 'his s&stem is considered as %eing insta%le e!uili%rium. ence# the angle θ# %etween the %ar magnet and the magnetic field is 0°.

/otential energ& of the s&stem

(b)'he %ar magnet is oriented 10° to the magnetic field. ence# it is in unsta%le e!uili%rium.

$ 10°

/otential energ& $ − cos θ

Question 5.5:

A closel& wound solenoid of 00 turns and area of cross section ).5 × 10−4 m) carries a current

of 3.0 A. "@plain the sense in which the solenoid acts li7e a %ar magnet. *hat is its associated

magnetic moment2

um%er of turns in the solenoid# n $ 00

Area of cross?section# A $ ).5 × 10−4 m)

Current in the solenoid# 3 $ 3.0 A

A current?carr&ing solenoid %eha(es as a %ar magnet %ecause a magnetic field de(elops alongits a@is# i.e.# along its length.

'he magnetic moment associated with the gi(en current?carr&ing solenoid is calculated asI

$ n 3 A

$ 00 × 3 × ).5 × 10−4

$ 0.- V '−1



Question 5.6:

f the solenoid in "@ercise 5.5 is free to turn a%out the (ertical direction and a uniform

horiontal magnetic field of 0.)5 ' is applied# what is the magnitude of tor!ue on the solenoid

when its a@is ma7es an angle of 30° with the direction of applied field2

>agnetic field strength# $ 0.)5 '

>agnetic moment# $ 0.- '−1

'he angle θ# %etween the a@is of the solenoid and the direction of the applied field is 30°.

'herefore# the tor!ue acting on the solenoid is gi(en asI

Question 5.7:

A %ar magnet of magnetic moment 1.5 V '−1 lies aligned with the direction of a uniform

magnetic field of 0.)) '.

(a) *hat is the amount of wor7 re!uired %& an e@ternal tor!ue to turn the magnet so as to

align its magnetic momentI i normal to the field direction# ii opposite to the field direction2

(b) *hat is the tor!ue on the magnet in cases i and ii2

(a)>agnetic moment# $ 1.5 V '−1

>agnetic field strength# $ 0.)) '

(i)nitial angle %etween the a@is and the magnetic field# θ1 $ 0°

inal angle %etween the a@is and the magnetic field# θ) $ 90°

'he wor7 re!uired to ma7e the magnetic moment normal to the direction of magnetic field isgi(en asI

(ii) nitial angle %etween the a@is and the magnetic field# θ1 $ 0°



inal angle %etween the a@is and the magnetic field# θ) $ 10°

'he wor7 re!uired to ma7e the magnetic moment opposite to the direction of magnetic field isgi(en asI

(b)or case iI

∴'or!ue#

or case iiI

∴'or!ue#

Question 5.8:

A closel& wound solenoid of )000 turns and area of cross?section 1.- × 10−4 m)# carr&ing a

current of 4.0 A# is suspended through its centre allowing it to turn in a horiontal plane.

a *hat is the magnetic moment associated with the solenoid2

% *hat is the force and tor!ue on the solenoid if a uniform horiontal magnetic field of +.5 ×

10−) ' is set up at an angle of 30 with the a@is of the solenoid2

um%er of turns on the solenoid# n $ )000

Area of cross?section of the solenoid# A $ 1.- × 10−4 m)

Current in the solenoid# 3 $ 4 A

(a)'he magnetic moment along the a@is of the solenoid is calculated asI

$ nA3

$ )000 × 1.- × 10−4 × 4



$ 1.) Am)

(b)>agnetic field# $ +.5 × 10−) '

Angle %etween the magnetic field and the a@is of the solenoid# θ $ 30°

'or!ue#

Fince the magnetic field is uniform# the force on the solenoid is ero. 'he tor!ue on the

solenoid is

Question 5.9:

A circular coil of 1- turns and radius 10 cm carr&ing a current of 0.+5 A rests with its plane

normal to an e@ternal field of magnitude 5.0 × 10−) '. 'he coil is free to turn a%out an a@is in

its plane perpendicular to the field direction. *hen the coil is turned slightl& and released# it

oscillates a%out its sta%le e!uili%rium with a fre!uenc& of ).0 s−1. *hat is the moment of

inertia of the coil a%out its a@is of rotation2

um%er of turns in the circular coil# B $ 1-

,adius of the coil# r $ 10 cm $ 0.1 m

Cross?section of the coil# A $ ^r ) $ ^ × 0.1) m)

Current in the coil# 3 $ 0.+5 A

>agnetic field strength# $ 5.0 × 10−) '

re!uenc& of oscillations of the coil# " $ ).0 s−1

∴>agnetic moment# $ B3A

$ 1- × 0.+5 × ^ × 0.1)

$ 0.3++ V '−1



*here#

3 $ >oment of inertia of the coil

ence# the moment of inertia of the coil a%out its a@is of rotation is

Question 5.10:

A magnetic needle free to rotate in a (ertical plane parallel to the magnetic meridian has its

north tip pointing down at )) with the horiontal. 'he horiontal component of the earthBs

magnetic field at the place is 7nown to %e 0.35 =. etermine the magnitude of the earthBs

magnetic field at the place.

oriontal component of earthBs magnetic field# H $ 0.35 =

Angle made %& the needle with the horiontal plane $ Angle of dip $

"arthBs magnetic field strength $

*e can relate and H asI

ence# the strength of earthBs magnetic field at the gi(en location is 0.3++ =.

Question 5.11:



At a certain location in Africa# a compass points 1) west of the geographic north. 'he north

tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points -0

a%o(e the horiontal. 'he horiontal component of the earthBs field is measured to %e 0.1- =.

Fpecif& the direction and magnitude of the earthBs field at the location.

Angle of declination#θ $ 1)°

Angle of dip#

oriontal component of earthBs magnetic field# H $ 0.1- =

"arthBs magnetic field at the gi(en location $

*e can relate and H asI

"arthBs magnetic field lies in the (ertical plane# 1)° *est of the geographic meridian# ma7ingan angle of -0° upward with the horiontal direction. ts magnitude is 0.3) =.

Question 5.12:

A short %ar magnet has a magnetic moment of 0.4 V '−1. =i(e the direction and magnitude of

the magnetic field produced %& the magnet at a distance of 10 cm from the centre of the

magnet on a the a@is# % the e!uatorial lines normal %isector of the magnet.

>agnetic moment of the %ar magnet# $ 0.4 V '−1

(a) istance# d $ 10 cm $ 0.1 m

'he magnetic field at distance d # from the centre of the magnet on the a@is is gi(en %& the

relationI

*here#




'he magnetic field is along the F − direction.

(b) 'he magnetic field at a distance of 10 cm i.e.# d $ 0.1 m on the e!uatorial line of themagnet is gi(en asI

'he magnetic field is along the − F direction.

Question 5.13:

A short %ar magnet placed in a horiontal plane has its a@is aligned along the magnetic north?

south direction. ull points are found on the a@is of the magnet at 14 cm from the centre of

the magnet. 'he earthBs magnetic field at the place is 0.3- = and the angle of dip is ero.*hat is the total magnetic field on the normal %isector of the magnet at the same distance as

the null−point i.e.# 14 cm from the centre of the magnet2 At n,!! p%ints# field due to a

magnet is e!ual and opposite to the horiontal component of earthBs magnetic field.

"arthBs magnetic field at the gi(en place# H $ 0.3- =

'he magnetic field at a distance d # on the a@is of the magnet is gi(en asI

*here#


$ >agnetic moment

'he magnetic field at the same distance d # on the e!uatorial line of the magnet is gi(en asI



'otal magnetic field#

ence# the magnetic field is 0.54 = in the direction of earthBs magnetic field.

Question 5.14:

f the %ar magnet in e@ercise 5.13 is turned around %& 10# where will the new null points %e

located2

'he magnetic field on the a@is of the magnet at a distance d 1 $ 14 cm# can %e written asI

*here#

$ >agnetic moment


H $ oriontal component of the magnetic field at d 1

f the %ar magnet is turned through 10°# then the neutral point will lie on the e!uatorial line.

ence# the magnetic field at a distance d )# on the e!uatorial line of the magnet can %e writtenasI

"!uating e!uations 1 and )# we getI



'he new null points will %e located 11.1 cm on the normal %isector.

Question 5.15:

A short %ar magnet of magnetic moment 5.)5 × 10−) V '−1 is placed with its a@is perpendicular

to the earthBs field direction. At what distance from the centre of the magnet# the resultant

field is inclined at 45 with earthBs field on

a its normal %isector and % its a@is. >agnitude of the earthBs field at the place is gi(en to

%e 0.4) =. gnore the length of the magnet in comparison to the distances in(ol(ed.

>agnetic moment of the %ar magnet# $ 5.)5 × 10−) V '−1

>agnitude of earthBs magnetic field at a place# H $ 0.4) = $ 0.4) × 10−4 '

(a) 'he magnetic field at a distance 9 from the centre of the magnet on the normal %isector isgi(en %& the relationI

*here#

$ /ermea%ilit& of free space $ 4^ × 10−+

'm A−1

*hen the resultant field is inclined at 45° with earthBs field# $ H



(b) 'he magnetic field at a distanced from the centre of the magnet on its a@is is gi(en asI

'he resultant field is inclined at 45° with earthBs field.

Question 5.16:


(a) *h& does a paramagnetic sample displa& greater magnetisation for the same

magnetising field when cooled2

(b) *h& is diamagnetism# in contrast# almost independent of temperature2



(c) f a toroid uses %ismuth for its core# will the field in the core %e slightl& greater or

slightl& less than when the core is empt&2

(d) s the permea%ilit& of a ferromagnetic material independent of the magnetic field2 f not#

is it more for lower or higher fields2

(e) >agnetic field lines are alwa&s nearl& normal to the surface of a ferromagnet at e(er&

point. 'his fact is analogous to the static electric field lines %eing normal to the surface of a

conductor at e(er& point. *h&2

(f ) *ould the ma@imum possi%le magnetisation of a paramagnetic sample %e of the same

order of magnitude as the magnetiation of a ferromagnet2

(a)Ewing to therandom thermal motion of molecules# the alignments of dipoles get disruptedat high temperatures. En cooling# this disruption is reduced. ence# a paramagnetic sample

displa&s greater magnetisation when cooled.

(b)'he induced dipole moment in a diamagnetic su%stance is alwa&s opposite to themagnetising field. ence# the internal motion of the atoms which is related to thetemperature does not affect the diamagnetism of a material.

(c)8ismuth is a diamagnetic su%stance. ence# a toroid with a %ismuth core has a magneticfield slightl& greater than a toroid whose core is empt&.

(d)'he permea%ilit& of ferromagnetic materials is not independent of the applied magneticfield. t is greater for a lower field and (ice (ersa.

(e)'he permea%ilit& of a ferromagnetic material is not less than one. t is alwa&s greater than

one. ence# magnetic field lines are alwa&s nearl& normal to the surface of such materials ate(er& point.

(f)'he ma@imum possi%le magnetisation of a paramagnetic sample can %e of the same orderof magnitude as the magnetisation of a ferromagnet. 'his re!uires high magnetising fields forsaturation.

Question 5.17:


(a) "@plain !ualitati(el& on the %asis of domain picture the irre(ersi%ilit& in the magnetisation

cur(e of a ferromagnet.

(b) 'he h&steresis loop of a soft iron piece has a much smaller area than that of a car%on

steel piece. f the material is to go through repeated c&cles of magnetisation# which piece will

dissipate greater heat energ&2



(c) A s&stem displa&ing a h&steresis loop such as a ferromagnet# is a de(ice for storing

memor&2B "@plain the meaning of this statement.

(d) *hat 7ind of ferromagnetic material is used for coating magnetic tapes in a cassette

pla&er# or for %uilding memor& storesB in a modern computer2

(e) A certain region of space is to %e shielded from magnetic fields.

Fuggest a method.

'he h&steresis cur(e ?H cur(e of a ferromagnetic material is shown in the following figure.

(a) t can %e o%ser(ed from the gi(en cur(e that magnetisation persists e(en when thee@ternal field is remo(ed. 'his reflects the irre(ersi%ilit& of a ferromagnet.

(b)'he dissipated heat energ& is directl& proportional to the area of a h&steresis loop. Acar%on steel piece has a greater h&steresis cur(e area. ence# it dissipates greater heatenerg&.

(c)'he (alue of magnetisation is memor& or record of h&steresis loop c&cles of magnetisation.'hese %its of information correspond to the c&cle of magnetisation. &steresis loops can %eused for storing information.

(d)Ceramic is used for coating magnetic tapes in cassette pla&ers and for %uilding memor&stores in modern computers.

(e)A certain region of space can %e shielded from magnetic fields if it is surrounded %& softiron rings. n such arrangements# the magnetic lines are drawn out of the region.

Question 5.18:

A long straight horiontal ca%le carries a current of ).5 A in the direction 10 south of west to

10° north of east. 'he magnetic meridian of the place happens to %e 10 west of the



geographic meridian. 'he earthBs magnetic field at the location is 0.33 =# and the angle of dip

is ero. <ocate the line of neutral points ignore the thic7ness of the ca%le. At ne,tr&! p%ints#

magnetic field due to a current?carr&ing ca%le is e!ual and opposite to the horiontal

component of earthBs magnetic field.

Current in the wire# 3 $ ).5 A

Angle of dip at the gi(en location on earth# $ 0°

"arthBs magnetic field# H $ 0.33 = $ 0.33 × 10−4 '

'he horiontal component of earthBs magnetic field is gi(en asI

H H $ H cos

'he magnetic field at the neutral point at a distance 9 from the ca%le is gi(en %& the relationI

*here#


ence# a set of neutral points parallel to and a%o(e the ca%le are located at a normal distanceof 1.51 cm.

Question 5.19:

A telephone ca%le at a place has four long straight horiontal wires carr&ing a current of 1.0 A

in the same direction east to west. 'he earthBs magnetic field at the place is 0.39 =# and the

angle of dip is 35. 'he magnetic declination is nearl& ero. *hat are the resultant magnetic

fields at points 4.0 cm %elow the ca%le2

um%er of horiontal wires in the telephone ca%le# n $ 4



Current in each wire# 3 $ 1.0 A

"arthBs magnetic field at a location# H $ 0.39 = $ 0.39 × 10−4 '

Angle of dip at the location# $ 35°

Angle of declination# θ ∼ 0°

F!$ a %!i&" ' c be!* "+e cabe#

istance# r $ 4 cm $ 0.04 m

'he horiontal component of earthBs magnetic field can %e written asI

H ' $ H cos −

*here#

$ >agnetic field at 4 cm due to current 3 in the four wires

$ /ermea%ilit& of free space $ 4^ × 10−+ 'm A−1

$ 0.) × 10−4

' $ 0.) =

∴ H ' $ 0.39 cos 35° − 0.)

$ 0.39 × 0.19 − 0.) 0.1) =

'he (ertical component of earthBs magnetic field is gi(en asI

H " $ H sin

$ 0.39 sin 35° $ 0.)) =

'he angle made %& the field with its horiontal component is gi(en asI

'he resultant field at the point is gi(en asI



F!$ a %!i&" ' c ab!ve "+e cabe#

oriontal component of earthBs magnetic fieldI

H ' $ H cos H

$ 0.39 cos 35° H 0.) $ 0.5) =

Mertical component of earthBs magnetic fieldI

H " $ H sin

$ 0.39 sin 35° $ 0.)) =

Angle# θ $ )).9°

And resultant fieldI

Question 5.20:

A compass needle free to turn in a horiontal plane is placed at the centre of circular coil of 30

turns and radius 1) cm. 'he coil is in a (ertical plane ma7ing an angle of 45 with the

magnetic meridian. *hen the current in the coil is 0.35 A# the needle points west to east.

(a) etermine the horiontal component of the earthBs magnetic field at the location.

(b) 'he current in the coil is re(ersed# and the coil is rotated a%out its (ertical a@is %& an

angle of 90 in the anticloc7wise sense loo7ing from a%o(e. /redict the direction of the needle.

'a7e the magnetic declination at the places to %e ero.

um%er of turns in the circular coil# B $ 30

,adius of the circular coil# r $ 1) cm $ 0.1) m

Current in the coil# 3 $ 0.35 A



Angle of dip# $ 45°

(a) 'he magnetic field due to current 3 # at a distance r # is gi(en asI

*here#

$ /ermea%ilit& of free space $ 4^ × 10−+ 'm A−1

$ 5.49 × 10−5 '

'he compass needle points from *est to "ast. ence# the horiontal component of earthBsmagnetic field is gi(en asI

H $ sin

$ 5.49 × 10−5 sin 45° $ 3. × 10−5 ' $ 0.3 =

(b) *hen the current in the coil is re(ersed and the coil is rotated a%out its (ertical a@is %& anangle of 90 # the needle will re(erse its original direction. n this case# the needle will pointfrom "ast to *est.

Question 5.21:

A magnetic dipole is under the influence of two magnetic fields. 'he angle %etween the field

directions is -0# and one of the fields has a magnitude of 1.) × 10−) '. f the dipole comes to

sta%le e!uili%rium at an angle of 15 with this field# what is the magnitude of the other field2

>agnitude of one of the magnetic fields# 1 $ 1.) × 10−) '

>agnitude of the other magnetic field $ )

Angle %etween the two fields# θ $ -0°

At sta%le e!uili%rium# the angle %etween the dipole and field 1# θ1 $ 15°

Angle %etween the dipole and field )# θ) $ θ − θ1 $ -0° − 15° $ 45°

At rotational e!uili%rium# the tor!ues %etween %oth the fields must %alance each other.

∴'or!ue due to field 1 $ 'or!ue due to field )

1 sinθ1 $ ) sinθ)



*here#

$ >agnetic moment of the dipole

ence# the magnitude of the other magnetic field is 4.39 × 10−3 '.

Question 5.22:

A monoenergetic 1 7eM electron %eam initiall& in the horiontal direction is su%Gected to a

horiontal magnetic field of 0.04 = normal to the initial direction. "stimate the up or down

deflection of the %eam o(er a distance of 30 cm me$ 9.11 × 10−19 C. NN!"e# C&t& in this

e@ercise are so chosen that the answer will gi(e &ou an idea of the effect of earthBs magnetic

field on the motion of the electron %eam from the electron gun to the screen in a 'M set.O

"nerg& of an electron %eam# E $ 1 7eM $ 1 × 103 eM


E $ 1 × 103 × 1.- × 10−19 V

>agnetic field# $ 0.04 =

>ass of an electron# me $ 9.11 × 10−19

7g

istance up to which the electron %eam tra(els# d $ 30 cm $ 0.3 m

*e can write the 7inetic energ& of the electron %eam asI

'he electron %eam deflects along a circular path of radius# r .

'he force due to the magnetic field %alances the centripetal force of the path.



<et the up and down deflection of the electron %eam %e

*here#

θ $ Angle of declination

'herefore# the up and down deflection of the %eam is 3.9 mm.

Question 5.23:

A sample of paramagnetic salt contains ).0 × 10)4 atomic dipoles each of dipole moment 1.5 ×

10−)3 V '−1. 'he sample is placed under a homogeneous magnetic field of 0.-4 '# and cooled to

a temperature of 4.) Z. 'he degree of magnetic saturation achie(ed is e!ual to 15_. *hat is

the total dipole moment of the sample for a magnetic field of 0.9 ' and a temperature of ).

Z2 Assume CurieBs law

um%er of atomic dipoles# n $ ).0 × 10)4

ipole moment of each atomic dipole# $ 1.5 × 10−)3 V '−1

*hen the magnetic field# 1 $ 0.-4 '

'he sample is cooled to a temperature# 4 1 $ 4.)°Z

'otal dipole moment of the atomic dipole# tot $ n ×

$ ) × 10)4 × 1.5 × 10−)3



$ 30 V '−1

>agnetic saturation is achie(ed at 15_.

ence# effecti(e dipole moment#

*hen the magnetic field# ) $ 0.9 '

'emperature# 4 ) $ ).°Z

ts total dipole moment $ )

According to CurieBs law# we ha(e the ratio of two magnetic dipoles asI

'herefore# is the total dipole moment of the sample for a magnetic field of 0.9 'and a temperature of ). Z.

Question 5.24:

A ,owland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core

of relati(e permea%ilit& 00. *hat is the magnetic field in the core for a magnetising current

of 1.) A2

>ean radius of a ,owland ring# r $ 15 cm $ 0.15 m

um%er of turns on a ferromagnetic core# B $ 3500

,elati(e permea%ilit& of the core material#

>agnetising current# 3 $ 1.) A

'he magnetic field is gi(en %& the relationI

*here#



:0 $ /ermea%ilit& of free space $ 4^ × 10−+ 'm A−1

'herefore# the magnetic field in the core is 4.4 '.

CHAPTR -, ELECTROAGNETIC INDCTION

Question 6.1:

/redict the direction of induced current in the situations descri%ed %& the following igs.-.1a to f .

(a)

(b)

(c)



(d)

(e)

(f)

'he direction of the induced current in a closed loop is gi(en %& <enBs law. 'he gi(en pairs offigures show the direction of the induced current when the orth pole of a %ar magnet ismo(ed towards and awa& from a closed loop respecti(el&.

Psing <enBs rule# the direction of the induced current in the gi(en situations can %e predictedas followsI

(a) 'he direction of the induced current is along /$%//

(b) 'he direction of the induced current is along %$/%.



(c) 'he direction of the induced current is along yzxy .

(d) 'he direction of the induced current is along zyxz .

(e) 'he direction of the induced current is along xryx .

(f) o current is induced since the field lines are l&ing in the plane of the closed loop.

Question 6.2:

A 1.0 m long metallic rod is rotated with an angular fre!uenc& of 400 rad s−1 a%out an a@is

normal to the rod passing through its one end. 'he other end of the rod is in contact with a

circular metallic ring. A constant and uniform magnetic field of 0.5 ' parallel to the a@is e@ists

e(er&where. Calculate the emf de(eloped %etween the centre and the ring.

<ength of the rod# ! $ 1 m

Angular fre!uenc&# $ 400 rad;s


Ene end of the rod has ero linear (elocit&# while the other end has a linear (elocit& of !/

A(erage linear (elocit& of the rod#

"mf de(eloped %etween the centre and the ring#

ence# the emf de(eloped %etween the centre and the ring is 100 M.

Question 6.3:

A long solenoid with 15 turns per cm has a small loop of area ).0 cm) placed inside the

solenoid normal to its a@is. f the current carried %& the solenoid changes steadil& from ).0 Ato 4.0 A in 0.1 s# what is the induced emf in the loop while the current is changing2

um%er of turns on the solenoid $ 15 turns;cm $ 1500 turns;m

um%er of turns per unit length# n $ 1500 turns

'he solenoid has a small loop of area# A $ ).0 cm) $ ) × 10−4 m)



Current carried %& the solenoid changes from ) A to 4 A.

Change in current in the solenoid# di $ 4 − ) $ ) A

Change in time# dt $ 0.1 s

nduced em) in the solenoid is gi(en %& arada&Bs law asI

*here#

$ nduced flu@ through the small loop

$ A ... ii

$ >agnetic field

$

:0 $ /ermea%ilit& of free space

= 4^×10−+ ;m

ence# e!uation i reduces toI

ence# the induced (oltage in the loop is

Question 6.4:

A rectangular wire loop of sides cm and ) cm with a small cut is mo(ing out of a region of

uniform magnetic field of magnitude 0.3 ' directed normal to the loop. *hat is the emf

de(eloped across the cut if the (elocit& of the loop is 1 cm s−1 in a direction normal to the a

longer side# % shorter side of the loop2 or how long does the induced (oltage last in each

case2



<ength of the rectangular wire# ! $ cm $ 0.0 m

*idth of the rectangular wire# 5 $ ) cm $ 0.0) m

ence# area of the rectangular loop#

A $ !5

$ 0.0 × 0.0)

$ 1- × 10−4 m)


Melocit& of the loop# " $ 1 cm;s $ 0.01 m;s

(a) "mf de(eloped in the loop is gi(en asI

e $ !"

$ 0.3 × 0.0 × 0.01 $ ).4 × 10−4 M

ence# the induced (oltage is ).4 × 10−4 M which lasts for ) s.

(b) "mf de(eloped# e $ 5"

$ 0.3 × 0.0) × 0.01 $ 0.- × 10−4 M

ence# the induced (oltage is 0.- × 10−4 M which lasts for s.

Question 6.6:

A circular coil of radius .0 cm and )0 turns is rotated a%out its (ertical diameter with an

angular speed of 50 rad s−1 in a uniform horiontal magnetic field of magnitude 3.0×10−) '.

E%tain the ma@imum and a(erage emf induced in the coil. f the coil forms a closed loop of

resistance 10# calculate the ma@imum (alue of current in the coil. Calculate the a(erage

power loss due to Voule heating. *here does this power come from2



>a@ induced em) $ 0.-03 M

A(erage induced em) $ 0 M

>a@ current in the coil $ 0.0-03 A

A(erage power loss $ 0.01 *

/ower comes from the e@ternal rotor

,adius of the circular coil# r $ cm $ 0.0 m

Area of the coil# A $ ^r ) $ ^ × 0.0) m)

um%er of turns on the coil# B $ )0

Angular speed# $ 50 rad;s

>agnetic field strength# $ 3 × 10−)

'

,esistance of the loop# 9 $ 10

>a@imum induced em) is gi(en asI

e $ B A

$ )0 × 50 × ^ × 0.0) × 3 × 10−)

$ 0.-03 M

'he ma@imum em) induced in the coil is 0.-03 M.

E(er a full c&cle# the a(erage em) induced in the coil is ero.

>a@imum current is gi(en asI

A(erage power loss due to Goule heatingI



'he current induced in the coil produces a tor!ue opposing the rotation of the coil. 'he rotor isan e@ternal agent. t must suppl& a tor!ue to counter this tor!ue in order to 7eep the coilrotating uniforml&. ence# dissipated power comes from the e@ternal rotor.

Question 6.7:

A horiontal straight wire 10 m long e@tending from east to west is falling with a speed of 5.0

m s−1# at right angles to the horiontal component of the earthBs magnetic field# 0.30 × 10−4

*% m−).

(a) *hat is the instantaneous (alue of the emf induced in the wire2

(b) *hat is the direction of the emf2

(c) *hich end of the wire is at the higher electrical potential2

<ength of the wire# ! $ 10 m

alling speed of the wire# " $ 5.0 m;s

>agnetic field strength# $ 0.3 × 10−4 *% m−)

(a) "mf induced in the wire#

e $ !"

(b) Psing lemingBs right hand rule# it can %e inferred that the direction of the induced emf isfrom *est to "ast.

(c) 'he eastern end of the wire is at a higher potential.

Question 6.8:

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. f an a(erage emf of )00 M induced# gi(e

an estimate of the self?inductance of the circuit.

nitial current# 3 1 $ 5.0 A

inal current# 3 ) $ 0.0 A

Change in current#

'ime ta7en for the change# t $ 0.1 s



A(erage emf# e $ )00 M

or self?inductance #I of the coil# we ha(e the relation for a(erage emf asI

e $ #

ence# the self induction of the coil is 4 .

Question 6.9:

A pair of adGacent coils has a mutual inductance of 1.5 . f the current in one coil changes

from 0 to )0 A in 0.5 s# what is the change of flu@ lin7age with the other coil2

>utual inductance of a pair of coils# J $ 1.5

nitial current# 3 1 $ 0 A

inal current 3 ) $ )0 A

Change in current#

'ime ta7en for the change# t $ 0.5 s

nduced emf#

*here is the change in the flu@ lin7age with the coil.

"mf is related with mutual inductance asI

"!uating e!uations 1 and )# we get



ence# the change in the flu@ lin7age is 30 *%.

Question 6.10:

A Get plane is tra(elling towards west at a speed of 100 7m;h. *hat is the (oltage difference

de(eloped %etween the ends of the wing ha(ing a span of )5 m# if the "arthBs magnetic field at

the location has a magnitude of 5 × 10−4 ' and the dip angle is 30°.

Fpeed of the Get plane# " $ 100 7m;h $ 500 m;s

*ing spanof Get plane# ! $ )5 m

"arthBs magnetic field strength# $ 5.0 × 10−4 '

Angle of dip#

Mertical component of "arthBs magnetic field#

7 $ sin

$ 5 × 10−4 sin 30°

$ ).5 × 10−4 '

Moltage difference %etween the ends of the wing can %e calculated asI

e $ 7 × ! × "

$ ).5 × 10−4 × )5 × 500

$ 3.1)5 M

ence# the (oltage difference de(eloped %etween the ends of the wings is

3.1)5 M.

Question 6.11:

Fuppose the loop in "@ercise -.4 is stationar& %ut the current feeding the electromagnet that

produces the magnetic field is graduall& reduced so that the field decreases from its initial



(alue of 0.3 ' at the rate of 0.0) ' s−1. f the cut is Goined and the loop has a resistance of 1.-

how much power is dissipated %& the loop as heat2 *hat is the source of this power2

Fides of the rectangular loop are cm and ) cm.

ence# area of the rectangular wire loop#

A $ length × width

$ × ) $ 1- cm)

$ 1- × 10−4 m)

nitial (alue of the magnetic field#

,ate of decrease of the magnetic field#

Em) de(eloped in the loop is gi(en asI

*here#

$ Change in flu@ through the loop area

$ A

,esistance of the loop# 9 $ 1.-

'he current induced in the loop is gi(en asI

/ower dissipated in the loop in the form of heat is gi(en asI



'he source of this heat loss is an e@ternal agent# which is responsi%le for changing themagnetic field with time.

Question 6.12:

A s!uare loop of side 1) cm with its sides parallel to Q and 6 a@es is mo(ed with a (elocit& of

cm s−1 in the positi(e x-direction in an en(ironment containing a magnetic field in the positi(e

z ?direction. 'he field is neither uniform in space nor constant in time. t has a gradient of 10−3

' cm−1 along the negati(e x-direction that is it increases %& 10− 3 ' cm−1 as one mo(es in the

negati(e x ?direction# and it is decreasing in time at the rate of 10−3 ' s−1. etermine the

direction and magnitude of the induced current in the loop if its resistance is 4.50 m.

Fide of the s!uare loop# s $ 1) cm $ 0.1) m

Area of the s!uare loop# A $ 0.1) × 0.1) $ 0.0144 m)

Melocit& of the loop# " $ cm;s $ 0.0 m;s

=radient of the magnetic field along negati(e x ?direction#

And# rate of decrease of the magnetic field#

,esistance of the loop#

,ate of change of the magnetic flu@ due to the motion of the loop in a non?uniform magneticfield is gi(en asI

,ate of change of the flu@ due to e@plicit time (ariation in field is gi(en asI



Fince the rate of change of the flu@ is the induced em) # the total induced em) in the loop can%e calculated asI

∴nduced current#

ence# the direction of the induced current is such that there is an increase in the flu@ throughthe loop along positi(e ?direction.

Question 6.13:

t is desired to measure the magnitude of field %etween the poles of a powerful loud spea7er

magnet. A small flat search coil of area ) cm) with )5 closel& wound turns# is positioned

normal to the field direction# and then !uic7l& snatched out of the field region. "!ui(alentl&#

one can gi(e it a !uic7 90° turn to %ring its plane parallel to the field direction. 'he total

charge flown in the coil measured %& a %allistic gal(anometer connected to coil is +.5 mC.

'he com%ined resistance of the coil and the gal(anometer is 0.50 . "stimate the field

strength of magnet.

Area of the small flat search coil# A $ ) cm) $ ) × 10−4 m)

um%er of turns on the coil# B $ )5

'otal charge flowing in the coil# Q $ +.5 mC $ +.5 × 10−3 C

'otal resistance of the coil and gal(anometer# 9 $ 0.50

nduced current in the coil#



nduced emf is gi(en asI

*here#

$ Charge in flu@

Com%ining e!uations 1 and )# we get

nitial flu@ through the coil# $ A

*here#

$ >agnetic field strength

inal flu@ through the coil#

ntegrating e!uation 3 on %oth sides# we ha(e

8ut total charge#

ence# the field strength of the magnet is 0.+5 '.



Question 6.14:

igure -.)0 shows a metal rod /S resting on the smooth rails A8 and positioned %etween the

poles of a permanent magnet. 'he rails# the rod# and the magnetic field are in three mutual

perpendicular directions. A gal(anometer = connects the rails through a switch Z. <ength of

the rod $ 15 cm# $ 0.50 '# resistance of the closed loop containing the rod $ 9.0 m.

Assume the field to %e uniform.

(a) Fuppose Z is open and the rod is mo(ed with a speed of 1) cm s −1 in the direction shown.

=i(e the polarit& and magnitude of the induced emf.

(b) s there an e@cess charge %uilt up at the ends of the rods when

Z is open2 *hat if Z is closed2

(c) *ith Z open and the rod mo(ing uniforml&# there is n% net )%r+e on the electrons in the

rod /S e(en though the& do e@perience magnetic force due to the motion of the rod. "@plain.

(d) *hat is the retarding force on the rod when Z is closed2

(e) ow much power is re!uired %& an e@ternal agent to 7eep the rod mo(ing at the same

speed $1) cm s−1 when Z is closed2 ow much power is re!uired when Z is open2

(f) ow much power is dissipated as heat in the closed circuit2

*hat is the source of this power2

(g) *hat is the induced emf in the mo(ing rod if the magnetic field is parallel to the rails

instead of %eing perpendicular2

<ength of the rod# ! $ 15 cm $ 0.15 m


,esistance of the closed loop# 9 $ 9 m $ 9 × 10−3

(a) nduced emf $ 9 mML polarit& of the induced emf is such that end 6 shows positi(e whileend Q shows negati(e ends.



Fpeed of the rod# " $ 1) cm;s $ 0.1) m;s

nduced emf is gi(en asI

e $ "!

= 0.5 × 0.1) × 0.15

$ 9 × 10−3 (

$ 9 mM

'he polarit& of the induced emf is such that end 6 shows positi(e while end Q shows negati(eends.

(b) 6esL when 7e& Z is closed# e@cess charge is maintained %& the continuous flow of current.

*hen 7e& Z is open# there is e@cess charge %uilt up at %oth ends of the rods.

*hen 7e& Z is closed# e@cess charge is maintained %& the continuous flow of current.

(c) >agnetic force is cancelled %& the electric force set?up due to the e@cess charge ofopposite nature at %oth ends of the rod.

'here is no net force on the electrons in rod /S when 7e& Z is open and the rod is mo(inguniforml&. 'his is %ecause magnetic force is cancelled %& the electric force set?up due to thee@cess charge of opposite nature at %oth ends of the rods.

(d) ,etarding force e@erted on the rod# F $ 3!

*here#

3 $ Current flowing through the rod

(e) 9 m*L no power is e@pended when 7e& Z is open.

Fpeed of the rod# " $ 1) cm;s $ 0.1) m;s

ence# power is gi(en asI



*hen 7e& Z is open# no power is e@pended.

(f) 9 m*L power is pro(ided %& an e@ternal agent.

/ower dissipated as heat $ 3 ) 9

$ 1) × 9 × 10−3

$ 9 m*

'he source of this power is an e@ternal agent.

(g) Rero

n this case# no emf is induced in the coil %ecause the motion of the rod does not cut acrossthe field lines.

Question 6.15:

An air?cored solenoid with length 30 cm# area of cross?section )5 cm) and num%er of turns

500# carries a current of ).5 A. 'he current is suddenl& switched off in a %rief time of 10−3 s.

ow much is the a(erage %ac7 emf induced across the ends of the open switch in the circuit2

gnore the (ariation in magnetic field near the ends of the solenoid.

<ength of the solenoid# ! $ 30 cm $ 0.3 m

Area of cross?section# A $ )5 cm) $ )5 × 10−4 m)

um%er of turns on the solenoid# B $ 500

Current in the solenoid# 3 $ ).5 A

Current flows for time# t $ 10−3 s

A(erage %ac7 emf#

*here#

$ Change in flu@

$ BA D )

*here#

$ >agnetic field strength



*here#

$ /ermea%ilit& of free space $ 4^ × 10−+ ' m A−1

Psing e!uations ) and 3 in e!uation 1# we get

ence# the a(erage %ac7 emf induced in the solenoid is -.5 M.

Question 6.16:

(a) E%tain an e@pression for the mutual inductance %etween a long straight wire and a s!uare

loop of side & as shown in ig. -.)1.

(b) ow assume that the straight wire carries a current of 50 A and the loop is mo(ed to the

right with a constant (elocit&# " $ 10 m;s.

Calculate the induced emf in the loop at the instant when x $ 0.) m.

'a7e & $ 0.1 m and assume that the loop has a large resistance.

(a) 'a7e a small element dy in the loop at a distance y from the long straight wire as shownin the gi(en figure.



>agnetic flu@ associated with element

*here#

dA $ Area of element dy $ & dy

$ >agnetic field at distance y

3 $ Current in the wire

$ /ermea%ilit& of free space $ 4^ × 10−+ ' m A−1

y tends from x to .

(b) "mf induced in the loop# e $ &"

=i(en#

3 $ 50 A



x $ 0.) m

& $ 0.1 m

" $ 10 m;s

Question 6.17:

A line charge X per unit length is lodged uniforml& onto the rim of a wheel of mass and

radius 9. 'he wheel has light non?conducting spo7es and is free to rotate without friction

a%out its a@is ig. -.)). A uniform magnetic field e@tends o(er a circular region within the

rim. t is gi(en %&#

$ − 80 0 r &L & 9

$ 0 otherwise

*hat is the angular (elocit& of the wheel after the field is suddenl& switched off2

<ine charge per unit length

*here#

r $ istance of the point within the wheel

>ass of the wheel $

,adius of the wheel $ 9



>agnetic field#

At distance r #themagnetic force is %alanced %& the centripetal force i.e.#

∴Angular (elocit&#

CHAPTER -ALTERNATING CRRENT

Question 7.1:

A 100 resistor is connected to a ))0 M# 50 ac suppl&.

(a) *hat is the rms (alue of current in the circuit2

(b) *hat is the net power consumed o(er a full c&cle2

,esistance of the resistor# 9 $ 100

Fuppl& (oltage# 7 $ ))0 M

re!uenc&# ν 50

(a) 'he rms (alue of current in the circuit is gi(en asI



ence# the rms (alue of current in the circuit is 15.9) A.

Question 7.4:

A -0 : capacitor is connected to a 110 M# -0 ac suppl&. etermine the rms (alue of the

current in the circuit.

Question 7.5:

n "@ercises +.3 and +.4# what is the net power a%sor%ed %& each circuit o(er a complete

c&cle. "@plain &our answer.

n the inducti(e circuit#

,ms (alue of current# 3 $ 15.9) A

,ms (alue of (oltage# 7 $ ))0 M

ence# the net power a%sor%ed can %e o%tained %& the relation#

6 $ 73 cos Φ

*here#

K $ /hase difference %etween 7 and 3

or a pure inducti(e circuit# the phase difference %etween alternating (oltage and current is90° i.e.# Φ$ 90°.

ence# 6 $ 0 i.e.# the net power is ero.

n the capaciti(e circuit#

,ms (alue of current# 3 $ ).49 A

,ms (alue of (oltage# 7 $ 110 M

ence# the net power a%sor%ed can (e o%tained asI

6 $ 73 Cos Φ

or a pure capaciti(e circuit# the phase difference %etween alternating (oltage and current is90° i.e.# Φ$ 90°.



ence# 6 $ 0 i.e.# the net power is ero.

Question 7.6:

E%tain the resonant fre!uenc& r of a series #$9 circuit with # $ ).0 # $ $ 3) : and 9 $ 10

. *hat is the Q?(alue of this circuit2

nductance# # $ ).0

Capacitance# $ $ 3) : $ 3) × 10−-

,esistance# 9 $ 10

,esonant fre!uenc& is gi(en %& the relation#

ow# Q?(alue of the circuit is gi(en asI

ence# the S?Malue of this circuit is )5.

Question 7.7:

A charged 30 : capacitor is connected to a )+ m inductor. *hat is the angular fre!uenc& of

free oscillations of the circuit2

Capacitance# $ $ 30: $ 30×10−-

nductance# # $ )+ m $ )+ × 10−3

Angular fre!uenc& is gi(en asI



At resonance#

Current in the circuit can %e calculated asI

ence# the a(erage power transferred to the circuit in one complete c&cle$ 73

$ )00 × 10 $ )000 *.

Question 7.10:

A radio can tune o(er the fre!uenc& range of a portion of >* %roadcast %andI 00 7 to

1)00 7. f its #$ circuit has an effecti(e inductance of )00 :# what must %e the range of

its (aria%le capacitor2

NHint: or tuning# the natural fre!uenc& i.e.# the fre!uenc& of free oscillations of the #$ circuit

should %e e!ual to the fre!uenc& of the radiowa(e.O

'he range of fre!uenc& ν of a radio is 00 7 to 1)00 7.

<ower tuning fre!uenc&# ν1 $ 00 7 $ 00 × 103

Ppper tuning fre!uenc&# ν) $ 1)00 7 $ 1)00× 103

"ffecti(e inductance of circuit # $ )00 : $ )00 × 10−-

Capacitance of (aria%le capacitor for ν1 is gi(en asI

$ 1

*here#

1 $ Angular fre!uenc& for capacitor $ 1



(c) etermine the rms potential drops across the three elements of the circuit. Fhow that the

potential drop across the #$ com%ination is ero at the resonating fre!uenc&.

nductance of the inductor# # $ 5.0

Capacitance of the capacitor# $ $ 0 : $ 0 × 10−-


/otential of the (aria%le (oltage source# 7 $ )30 M

(a) ,esonance angular fre!uenc& is gi(en asI

ence# the circuit will come in resonance for a source fre!uenc& of 50 rad;s.

(b) mpedance of the circuit is gi(en %& the relation#

At resonance#

Amplitude of the current at the resonating fre!uenc& is gi(en asI

*here#

7 0 $ /ea7 (oltage



ence# at resonance# the impedance of the circuit is 40 and the amplitude of the current is.13 A.

(c) ,ms potential drop across the inductor#

7 #rms $ 3 × 9#

*here#

3 $ rms current

/otential drop across the capacitor#

/otential drop across the resistor#

7 9rms $ 39

$ × 40 $ )30 M

/otential drop across the #$ com%ination#

At resonance#

∴7 #$ $ 0

ence# it is pro(ed that the potential drop across the #$ com%ination is ero at resonatingfre!uenc&.

Question 7.12:



An #$ circuit contains a )0 m inductor and a 50 : capacitor with an initial charge of 10 mC.

'he resistance of the circuit is negligi%le. <et the instant the circuit is closed %e t $ 0.

(a) *hat is the total energ& stored initiall&2 s it conser(ed during #$ oscillations2

(b) *hat is the natural fre!uenc& of the circuit2

(c) At what time is the energ& stored

i completel& electrical i.e.# stored in the capacitor2 ii completel& magnetic i.e.# stored in

the inductor2

(d) At what times is the total energ& shared e!uall& %etween the inductor and the capacitor2

(e) f a resistor is inserted in the circuit# how much energ& is e(entuall& dissipated as heat2

nductance of the inductor# # $ )0 m $ )0 × 10−3


nitial charge on the capacitor# Q $ 10 mC $ 10 × 10−3 C

(a) 'otal energ& stored initiall& in the circuit is gi(en asI

ence# the total energ& stored in the #$ circuit will %e conser(ed %ecause there is no resistorconnected in the circuit.

(b)atural fre!uenc& of the circuit is gi(en %& the relation#

atural angular fre!uenc&#



ence# the natural fre!uenc& of the circuit is 103 rad;s.

(c) (i) or time period 4 # total charge on the capacitor at time t #

or energ& stored is electrical# we can write QB $ Q.

ence# it can %e inferred that the energ& stored in the capacitor is completel& electrical at

time# t $

(ii) >agnetic energ& is the ma@imum when electrical energ&# Qj is e!ual to 0.

ence# it can %e inferred that the energ& stored in the capacitor is completel& magnetic at

time#

(d) Q1 $ Charge on the capacitor when total energ& is e!uall& shared %etween the capacitorand the inductor at time t .

*hen total energ& is e!uall& shared %etween the inductor and capacitor# the energ& stored in

the capacitor $ ma@imum energ&.



ence# total energ& is e!uall& shared %etween the inductor and the capacit& at time#

(e) f a resistor is inserted in the circuit# then total initial energ& is dissipated as heat energ&in the circuit. 'he resistance damps out the #$ oscillation.

Question 7.13:

A coil of inductance 0.50 and resistance 100 is connected to a )40 M# 50 ac suppl&.

(a) *hat is the ma@imum current in the coil2

(b) *hat is the time lag %etween the (oltage ma@imum and the current ma@imum2



/otential of the suppl& (oltage# 7 $ )40 M

re!uenc& of the suppl&# ν $ 50

(a) /ea7 (oltage is gi(en asI

Angular fre!uenc& of the suppl&#



$ ) ^ν

$ )^ × 50 $ 100 ^ rad;s

>a@imum current in the circuit is gi(en asI

(b) "!uation for (oltage is gi(en asI

7 $ 7 0 cos t

"!uation for current is gi(en asI

3 $ 3 0 cos t − Φ

*here#

K $ /hase difference %etween (oltage and current

At time# t $ 0.

7 $ 7 0(oltage is ma@imum

ort − Φ $ 0 i.e.# at time #

3 $ 3 0 current is ma@imum

ence# the time lag %etween ma@imum (oltage and ma@imum current is .

ow# phase angle Φis gi(en %& the relation#



ence# the time lag %etween ma@imum (oltage and ma@imum current is 3.) ms.

Question 7.14:

E%tain the answers a to % in "@ercise +.13 if the circuit is connected to a high fre!uenc&

suppl& )40 M# 10 7. ence# e@plain the statement that at (er& high fre!uenc&# an inductor

in a circuit nearl& amounts to an open circuit. ow does an inductor %eha(e in a dc circuit

after the stead& state2



/otential of the suppl& (oltages# 7 $ )40 M

re!uenc& of the suppl&#ν 10 7 $ 104

Angular fre!uenc&# $ )^ν $ )^ × 104 rad;s

(a) /ea7 (oltage#

>a@imum current#

(b) or phase differenceΦ# we ha(e the relationI



t can %e o%ser(ed that 3 0 is (er& small in this case. ence# at high fre!uencies# the inductoramounts to an open circuit.

n a dc circuit# after a stead& state is achie(ed# $ 0. ence# inductor < %eha(es li7e a pure

conducting o%Gect.

Question 7.15:

A 100 : capacitor in series with a 40 resistance is connected to a 110 M# -0 suppl&.

(a) *hat is the ma@imum current in the circuit2

(b) *hat is the time lag %etween the current ma@imum and the (oltage ma@imum2




(a) re!uenc& of oscillations# ν $ -0

Angular fre!uenc&#

or a 9$ circuit# we ha(e the relation for impedance asI

/ea7 (oltage# 7 0 $

>a@imum current is gi(en asI



(b) n a capacitor circuit# the (oltage lags %ehind the current %& a phase angle of Φ. 'his angleis gi(en %& the relationI

ence# the time lag %etween ma@imum current and ma@imum (oltage is 1.55 ms.

Question 7.16:

E%tain the answers to a and % in "@ercise +.15 if the circuit is connected to a 110 M# 1)

7 suppl&2 ence# e@plain the statement that a capacitor is a conductor at (er& high

fre!uencies. Compare this %eha(iour with that of a capacitor in a dc circuit after the stead&

state.






re!uenc& of the suppl&# ν $ 1) 7 $ 1) × 103

Angular re!uenc&# $ ) ^ν $ ) × ^ × 1) × 10303

$ )4^ × 103 rad;s

/ea7 (oltage#

>a@imum current#

or an 9$ circuit# the (oltage lags %ehind the current %& a phase angle of Φ gi(en asI



ence# Φ tends to %ecome ero at high fre!uencies. At a high fre!uenc&# capacitor C acts as aconductor.

n a dc circuit# after the stead& state is achie(ed# $ 0. ence# capacitor C amounts to anopen circuit.

Question 7.17:

Zeeping the source fre!uenc& e!ual to the resonating fre!uenc& of the series #$9 circuit# if the

three elements# ## $ and 9 are arranged in parallel# show that the total current in the parallel

#$9 circuit is minimum at this fre!uenc&. E%tain the current rms (alue in each %ranch of the

circuit for the elements and source specified in "@ercise +.11 for this fre!uenc&.

An inductor ## a capacitor $ # and a resistor 9 is connected in parallel with each other in acircuit where#

# $ 5.0

$ $ 0 : $ 0 × 10−-

9 $ 40

/otential of the (oltage source# 7 $ )30 M

mpedance R of the gi(en parallel #$9 circuit is gi(en asI

*here#

$ Angular fre!uenc&

At resonance#

ence# the magnitude of 8 is the ma@imum at 50 rad;s. As a result# the total current isminimum.

,ms current flowing through inductor < is gi(en asI



,ms current flowing through capacitor C is gi(en asI

,ms current flowing through resistor , is gi(en asI

Question 7.18:

A circuit containing a 0 m inductor and a -0 : capacitor in series is connected to a )30 M#

50 suppl&. 'he resistance of the circuit is negligi%le.

(a) E%tain the current amplitude and rms (alues.

(b) E%tain the rms (alues of potential drops across each element.

(c) *hat is the a(erage power transferred to the inductor2

(d) *hat is the a(erage power transferred to the capacitor2

(e) *hat is the total a(erage power a%sor%ed %& the circuit2 NA(erageB implies a(eraged o(er

one c&cleB.O

nductance# # $ 0 m $ 0 × 10−3

Capacitance# $ $ -0 : $ -0 × 10−-


re!uenc&# ν $ 50

Angular fre!uenc&# $ )^ν $ 100 ^ rad;s



/ea7 (oltage# 7 0 $

(a) >a@imum current is gi(en asI

'he negati(e sign appears %ecause

Amplitude of ma@imum current#

ence# rms (alue of current#

(b) /otential difference across the inductor#

7 #$ 3 × #

$ .)) × 100 ^ × 0 × 10−3

$ )0-.-1 M

/otential difference across the capacitor#

(c) A(erage power consumed %& the inductor is ero as actual (oltage leads the current %& .

(d) A(erage power consumed %& the capacitor is ero as (oltage lags current %& .



(e) 'he total power a%sor%ed a(eraged o(er one c&cle is ero.

Question 7.19:

Fuppose the circuit in "@ercise +.1 has a resistance of 15 . E%tain the a(erage power

transferred to each element of the circuit# and the total power a%sor%ed.

A(erage power transferred to the resistor $ +.44 *

A(erage power transferred to the capacitor $ 0 *

'otal power a%sor%ed %& the circuit $ +.44 *

nductance of inductor# # $ 0 m $ 0 × 10−3

Capacitance of capacitor# $ $ -0 K $ -0 × 10−-

,esistance of resistor# 9 $ 15

/otential of (oltage suppl&# 7 $ )30 M

re!uenc& of signal# ν $ 50

Angular fre!uenc& of signal# $ )^ν $ )^ × 50 $ 100^ rad;s

'he elements are connected in series to each other. ence# impedance of the circuit is gi(enasI

Current flowing in the circuit#

A(erage power transferred to resistance is gi(en asI

6 9$ 3 )9

$ +.)5) × 15 $ +.44 *

A(erage power transferred to capacitor# 6 $ $ A(erage power transferred to inductor# 6 # $ 0



'otal power a%sor%ed %& the circuitI

$ 6 9 < 6 $ < 6 #

$ +.44 H 0 H 0 $ +.44 *

ence# the total power a%sor%ed %& the circuit is +.44 *.

Question 7.20:

A series #$9 circuit with # $ 0.1) # $ $ 40 n# 9 $ )3 is connected to a )30 M (aria%le

fre!uenc& suppl&.

(a) *hat is the source fre!uenc& for which current amplitude is ma@imum. E%tain this

ma@imum (alue.

(b) *hat is the source fre!uenc& for which a(erage power a%sor%ed %& the circuit isma@imum. E%tain the (alue of this ma@imum power.

(c) or which fre!uencies of the source is the power transferred to the circuit half the power

at resonant fre!uenc&2 *hat is the current amplitude at these fre!uencies2

(d) *hat is the Q?factor of the gi(en circuit2

nductance# # $ 0.1)

Capacitance# $ $ 40 n $ 40 × 10−9

,esistance# 9 $ )3


/ea7 (oltage is gi(en asI

7 0 $ $ 3)5.)) M

(a) Current flowing in the circuit is gi(en %& the relation#

*here#

3 0 $ ma@imum at resonance

At resonance# we ha(e



'o impro(e the sharpness of the resonance %& reducing its full width at half ma@imumB %& a

factor of ) without changing # we need to reduce 9 to half i.e.#

,esistance $

Question 7.22:


(a) n an& ac circuit# is the applied instantaneous (oltage e!ual to the alge%raic sum of the

instantaneous (oltages across the series elements of the circuit2 s the same true for rms

(oltage2

(b) A capacitor is used in the primar& circuit of an induction coil.

(c) An applied (oltage signal consists of a superposition of a dc (oltage and an ac (oltage of

high fre!uenc&. 'he circuit consists of an inductor and a capacitor in series. Fhow that the dc

signal will appear across $ and the ac signal across #.

(d) A cho7e coil in series with a lamp is connected to a dc line. 'he lamp is seen to shine

%rightl&. nsertion of an iron core in the cho7e causes no change in the lampBs %rightness.

/redict the corresponding o%ser(ations if the connection is to an ac line.

(e) *h& is cho7e coil needed in the use of fluorescent tu%es with ac mains2 *h& can we not

use an ordinar& resistor instead of the cho7e coil2

(a) 6esL the statement is not true for rms (oltage

t is true that in an& ac circuit# the applied (oltage is e!ual to the a(erage sum of the

instantaneous (oltages across the series elements of the circuit. owe(er# this is not true forrms (oltage %ecause (oltages across different elements ma& not %e in phase.

(b) igh induced (oltage is used to charge the capacitor.

A capacitor is used in the primar& circuit of an induction coil. 'his is %ecause when the circuitis %ro7en# a high induced (oltage is used to charge the capacitor to a(oid spar7s.



(c) 'he dc signal will appear across capacitor $ %ecause for dc signals# the impedance of aninductor # is negligi%le while the impedance of a capacitor $ is (er& high almost infinite.ence# a dc signal appears across $ . or an ac signal of high fre!uenc&# the impedance of # ishigh and that of $ is (er& low. ence# an ac signal of high fre!uenc& appears across #.

(d) f an iron core is inserted in the cho7e coil which is in series with a lamp connected to theac line# then the lamp will glow diml&. 'his is %ecause the cho7e coil and the iron coreincrease the impedance of the circuit.

(e) A cho7e coil is needed in the use of fluorescent tu%es with ac mains %ecause it reduces the(oltage across the tu%e without wasting much power. An ordinar& resistor cannot %e usedinstead of a cho7e coil for this purpose %ecause it wastes power in the form of heat.

Question 7.23:

A power transmission line feeds input power at )300 M to a stepdown transformer with its

primar& windings ha(ing 4000 turns. *hat should %e the num%er of turns in the secondar& in

order to get output power at )30 M2

nput (oltage# 7 1 $ )300

um%er of turns in primar& coil# n1 $ 4000

Eutput (oltage# 7 ) $ )30 M

um%er of turns in secondar& coil $ n)

Moltage is related to the num%er of turns asI

ence# there are 400 turns in the second winding.

Question 7.24:

At a h&droelectric power plant# the water pressure head is at a height of 300 m and the water

flow a(aila%le is 100 m3 s−1. f the tur%ine generator efficienc& is -0_# estimate the electric

power a(aila%le from the plant g$ 9. m s−).

eight of water pressure head# ' $ 300 m

Molume of water flow per second# 7 $ 100 m3 ;s



"fficienc& of tur%ine generator# n $ -0_ $ 0.-


ensit& of water# ρ $ 103 7g;m3

"lectric power a(aila%le from the plant $ L × 'ρg7

$ 0.- × 300 × 103 × 9. × 100

$ 1+-.4 × 10- *

$ 1+-.4 >*

Question 7.25:

A small town with a demand of 00 7* of electric power at ))0 M is situated 15 7m awa& from

an electric plant generating power at 440 M. 'he resistance of the two wire line carr&ing poweris 0.5 per 7m. 'he town gets power from the line through a 4000?))0 M step?down

transformer at a su%?station in the town.

(a) "stimate the line power loss in the form of heat.

(b) ow much power must the plant suppl&# assuming there is negligi%le power loss due to

lea7age2

(c) Characterise the step up transformer at the plant.

'otal electric power re!uired# 6 $ 00 7* $ 00 × 103 *

Fuppl& (oltage# 7 $ ))0 M

Moltage at which electric plant is generating power# 7 k $ 440 M

istance %etween the town and power generating station# d $ 15 7m

,esistance of the two wire lines carr&ing power $ 0.5 ;7m

'otal resistance of the wires# 9 $ 15 H 150.5 $ 15

A step?down transformer of rating 4000 − ))0 M is used in the su%?station.

nput (oltage# 7 1 $ 4000 M

Eutput (oltage# 7 ) $ ))0 M

,ms current in the wire lines is gi(en asI



(a) <ine power loss $ 3 )9

$ )00) × 15

$ -00 × 103 *

$ -00 7*

(b) Assuming that the power loss is negligi%le due to the lea7age of the currentI

'otal power supplied %& the plant $ 00 7* H -00 7*

$ 1400 7*

(c) Moltage drop in the power line $ 39 $ )00 × 15 $ 3000 M

ence# total (oltage transmitted from the plant $ 3000 H 4000

$ +000 M

Also# the power generated is 440 M.

ence# the rating of the step?up transformer situated at the power plant is 440 M − +000 M.

Question 7.26:

o the same e@ercise as a%o(e with the replacement of the earlier transformer %& a 40#000?

))0 M step?down transformer eglect# as %efore# lea7age losses though this ma& not %e a

good assumption an& longer %ecause of the (er& high (oltage transmission in(ol(ed. ence#

e@plain wh& high (oltage transmission is preferred2

'he rating of a step?down transformer is 40000 M−))0 M.

nput (oltage# 7 1 $ 40000 M

Eutput (oltage# 7 ) $ ))0 M

'otal electric power re!uired# 6 $ 00 7* $ 00 × 103 *

Fource potential# 7 $ ))0 M

Moltage at which the electric plant generates power# 7 k $ 440 M



istance %etween the town and power generating station# d $ 15 7m

,esistance of the two wire lines carr&ing power $ 0.5 ;7m

'otal resistance of the wire lines# 9 $ 15 H 150.5 $ 15

6 $ M13

,ms current in the wire line is gi(en asI

(a) <ine power loss $ 3 )9

$ )0) × 15

$ - 7*

(b) Assuming that the power loss is negligi%le due to the lea7age of current.

ence# power supplied %& the plant $ 00 7* H -7* $ 0- 7*

(c) Moltage drop in the power line $ 39 $ )0 × 15 $ 300 M

ence# (oltage that is transmitted %& the power plant

$ 300 H 40000 $ 40300 M

'he power is %eing generated in the plant at 440 M.

ence# the rating of the step?up transformer needed at the plant is

440 M − 40300 M.

ence# power loss during transmission $

n the pre(ious e@ercise# the power loss due to the same reason is . Fincethe power loss is less for a high (oltage transmission# high (oltage transmissions are preferredfor this purpose.



CHAPTER- 3 ELECTROAGNETIC 4A4ES

Question 8.1:

igure .- shows a capacitor made of two circular plates each of radius 1) cm# and separated

%& 5.0 cm. 'he capacitor is %eing charged %& an e@ternal source not shown in the figure. 'he

charging current is constant and e!ual to 0.15 A.

(a) Calculate the capacitance and the rate of charge of potential difference %etween the

plates.

(b) E%tain the displacement current across the plates.

(c) s ZirchhoffBs first rule Gunction rule (alid at each plate of the capacitor2 "@plain.

,adius of each circular plate# r $ 1) cm $ 0.1) m

istance %etween the plates# d $ 5 cm $ 0.05 m

Charging current# 3 $ 0.15 A

/ermitti(it& of free space# $ .5 × 10−1) C) −1 m−)

(a) Capacitance %etween the two plates is gi(en %& the relation#

$

*here#




Charge on each plate# q $ $7

*here#

M $ /otential difference across the plates

ifferentiation on %oth sides with respect to time t gi(esI

'herefore# the change in potential difference %etween the plates is 1.+ ×109 M;s.

(b) 'he displacement current across the plates is the same as the conduction current. ence#the displacement current# i d is 0.15 A.

(c) 6es

ZirchhoffBs first rule is (alid at each plate of the capacitor pro(ided that we ta7e the sum ofconduction and displacement for current.

Question 8.2:

A parallel plate capacitor ig. .+ made of circular plates each of radius 9 $ -.0 cm has a

capacitance $ $ 100 p. 'he capacitor is connected to a )30 M ac suppl& with a angular

fre!uenc& of 300 rad s−1.

(a) *hat is the rms (alue of the conduction current2

(b) s the conduction current e!ual to the displacement current2

(c) etermine the amplitude of at a point 3.0 cm from the a@is %etween the plates.



,adius of each circular plate# 9 $ -.0 cm $ 0.0- m

Capacitance of a parallel plate capacitor# $ $ 100 p $ 100 × 10−1)


Angular fre!uenc&# $ 300 rad s−1

(a) ,ms (alue of conduction current# 3

*here#

2 $ $ Capaciti(e reactance

∴ 3 $ 7 × $

$ )30 × 300 × 100 × 10−1)

$ -.9 × 10−- A

$ -.9 :A

ence# the rms (alue of conduction current is -.9 :A.

(b) 6es# conduction current is e!ual to displacement current.

(c) >agnetic field is gi(en asI

*here#

:0 $ ree space permea%ilit&

3 0 $ >a@imum (alue of current $

r $ istance %etween the plates from the a@is $ 3.0 cm $ 0.03 m

∴



$ 1.-3 × 10−11 '

ence# the magnetic field at that point is 1.-3 × 10−11 '.

Question 8.3:

*hat ph&sical !uantit& is the same for Q?ra&s of wa(elength 10−10 m# red light of wa(elength

-00 Y and radiowa(es of wa(elength 500 m2

'he speed of light 3 × 10 m;s in a (acuum is the same for all wa(elengths. t isindependent of the wa(elength in the (acuum.

Question 8.4:

A plane electromagnetic wa(e tra(els in (acuum along ?direction. *hat can &ou sa& a%out the

directions of its electric and magnetic field (ectors2 f the fre!uenc& of the wa(e is 30 >#

what is its wa(elength2

'he electromagnetic wa(e tra(els in a (acuum along the ?direction. 'he electric field E andthe magnetic field H are in the x ?y plane. 'he& are mutuall& perpendicular.

re!uenc& of the wa(e# ν $ 30 > $ 30 × 10- s−1

Fpeed of light in a (acuum# + $ 3 × 10 m;s

*a(elength of a wa(e is gi(en asI

Question 8.5:

A radio can tune in to an& station in the +.5 > to 1) > %and. *hat is the corresponding

wa(elength %and2

A radio can tune to minimum fre!uenc&# ν 1 $ +.5 >$ +.5 × 10-

>a@imum fre!uenc&# ν ) $ 1) > $ 1) × 10-

Fpeed of light# + $ 3 × 10 m;s

Corresponding wa(elength for ν 1 can %e calculated asI



Corresponding wa(elength for ν ) can %e calculated asI

'hus# the wa(elength %and of the radio is 40 m to )5 m.

Question 8.6:

A charged particle oscillates a%out its mean e!uili%rium position with a fre!uenc& of 109 .

*hat is the fre!uenc& of the electromagnetic wa(es produced %& the oscillator2

'he fre!uenc& of an electromagnetic wa(e produced %& the oscillator is the same as that of acharged particle oscillating a%out its mean position i.e.# 109 .

Question 8.7:

'he amplitude of the magnetic field part of a harmonic electromagnetic wa(e in (acuum is 0

$ 510 n'. *hat is the amplitude of the electric field part of the wa(e2

Amplitude of magnetic field of an electromagnetic wa(e in a (acuum#

0 $ 510 n' $ 510 × 10−9 '

Fpeed of light in a (acuum# + $ 3 × 10 m;s

Amplitude of electric field of the electromagnetic wa(e is gi(en %& the relation#

E $ +0

$ 3 × 10 × 510 × 10−9 $ 153 ;C

'herefore# the electric field part of the wa(e is 153 ;C.

Question 8.8:

Fuppose that the electric field amplitude of an electromagnetic wa(e is E 0 $ 1)0 ;C and that

its fre!uenc& is ν $ 50.0 >. a etermine# 0 D # kD and X. % ind e@pressions for E and

.



"lectric field amplitude# E 0 $ 1)0 ;C

re!uenc& of source# ν $ 50.0 > $ 50 × 10-


(a) >agnitude of magnetic field strength is gi(en asI

Angular fre!uenc& of source is gi(en asI

$ )^ν

$ )^ × 50 × 10-

$ 3.14 × 10 rad;s

/ropagation constant is gi(en asI

*a(elength of wa(e is gi(en asI

(b) Fuppose the wa(e is propagating in the positi(e x direction. 'hen# the electric field (ectorwill %e in the positi(e y direction and the magnetic field (ector will %e in the positi(e z

direction. 'his is %ecause all three (ectors are mutuall& perpendicular.

"!uation of electric field (ector is gi(en asI



And# magnetic field (ector is gi(en asI

Question 8.9:

'he terminolog& of different parts of the electromagnetic spectrum is gi(en in the te@t. Pse

the formula E $ 'ν for energ& of a !uantum of radiationI photon and o%tain the photon

energ& in units of eM for different parts of the electromagnetic spectrum. n what wa& are the

different scales of photon energies that &ou o%tain related to the sources of electromagnetic

radiation2

"nerg& of a photon is gi(en asI

*here#

' $ /lanc7Bs constant $ -.- × 10−34 Vs

+ $ Fpeed of light $ 3 × 10 m;s

X $ *a(elength of radiation

'he gi(en ta%le lists the photon energies for different parts of an electromagnetic spectrum fordifferent λ.

X

m

103 1 10−3 10−- 10− 10−10 10−1)

E eM

1).3+5×

10−10

1).3+5 ×10−+

1).3+5 ×10−4

1).3+5 ×10−1

1).3+5 ×101

1).3+5 ×103

1).3+5× 105



'he photon energies for the different parts of the spectrum of a source indicate the spacing ofthe rele(ant energ& le(els of the source.

Question 8.10:

n a plane electromagnetic wa(e# the electric field oscillates sinusoidall& at a fre!uenc& of ).0

× 1010 and amplitude 4 M m−1.

(a) *hat is the wa(elength of the wa(e2

(b) *hat is the amplitude of the oscillating magnetic field2

(c) Fhow that the a(erage energ& densit& of the E field e!uals the a(erage energ& densit& of

the field. N+ $ 3 × 10 m s−1.O

re!uenc& of the electromagnetic wa(e# ν $ ).0 × 1010

"lectric field amplitude# E 0 $ 4 M m−1


(a) *a(elength of a wa(e is gi(en asI

(b) >agnetic field strength is gi(en asI

(c) "nerg& densit& of the electric field is gi(en asI

And# energ& densit& of the magnetic field is gi(en asI



*here#


:0 $ /ermea%ilit& of free space

*e ha(e the relation connecting E and asI

E $ + D 1

*here#

D )

/utting e!uation ) in e!uation 1# we get

F!uaring %oth sides# we get

Question 8.11:

Fuppose that the electric field part of an electromagnetic wa(e in (acuum is E $ 3.1 ;C

cos N1. rad;m y H 5.4 × 10- rad;st O .

(a) *hat is the direction of propagation2

(b) *hat is the wa(elength X2

(c) *hat is the fre!uenc& ν 2

(d) *hat is the amplitude of the magnetic field part of the wa(e2



(e) *rite an e@pression for the magnetic field part of the wa(e.

(a) rom the gi(en electric field (ector# it can %e inferred that the electric field is directedalong the negati(e x direction. ence# the direction of motion is along the negati(e y direction

i.e.# .

(b) t is gi(en that#

'he general e!uation for the electric field (ector in the positi(e x direction can %e written asI

En comparing e!uations 1 and )# we get

"lectric field amplitude# E 0 $ 3.1 ;C

Angular fre!uenc&# $ 5.4 × 10 rad;s

*a(e num%er# k $ 1. rad;m

*a(elength# $ 3.490 m

(c) re!uenc& of wa(e is gi(en asI

(d) >agnetic field strength is gi(en asI

*here#




(e) En o%ser(ing the gi(en (ector field# it can %e o%ser(ed that the magnetic field (ector isdirected along the negati(e direction. ence# the general e!uation for the magnetic field(ector is written asI

Question 8.12:

A%out 5_ of the power of a 100 * light %ul% is con(erted to (isi%le radiation. *hat is the

a(erage intensit& of (isi%le radiation

(a) at a distance of 1 m from the %ul%2

(b) at a distance of 10 m2

Assume that the radiation is emitted isotropicall& and neglect reflection.

/ower rating of %ul%# 6 $ 100 *

t is gi(en that a%out 5_ of its power is con(erted into (isi%le radiation.

/ower of (isi%le radiation#

ence# the power of (isi%le radiation is 5*.

(a) istance of a point from the %ul%# d $ 1 m

ence# intensit& of radiation at that point is gi(en asI

(b) istance of a point from the %ul%# d 1 $ 10 m

ence# intensit& of radiation at that point is gi(en asI



Question 8.13:

Pse the formula Xm 4 $ 0.)9 cm Z to o%tain the characteristic temperature ranges for different

parts of the electromagnetic spectrum. *hat do the num%ers that &ou o%tain tell &ou2

A %od& at a particular temperature produces a continous spectrum of wa(elengths. n case ofa %lac7 %od&# the wa(elength corresponding to ma@imum intensit& of radiation is gi(enaccording to /lanc7Bs law. t can %e gi(en %& the relation#

*here#

Xm $ ma@imum wa(elength

4 $ temperature

'hus# the temperature for different wa(elengths can %e o%tained asI

or λm $ 10

−4

cmL

or λm $ 5 ×10−5 cmL

or λm $ 10−- cmL and so on.

'he num%ers o%tained tell us that temperature ranges are re!uired for o%taining radiations indifferent parts of an electromagnetic spectrum. As the wa(elength decreases# thecorresponding temperature increases.

Question 8.14:

=i(en %elow are some famous num%ers associated with electromagnetic radiations in different

conte@ts in ph&sics. Ftate the part of the electromagnetic spectrum to which each %elongs.

(a) )1 cm wa(elength emitted %& atomic h&drogen in interstellar space.



(b) 105+ > fre!uenc& of radiation arising from two close energ& le(els in h&drogenL 7nown

as <am% shift.

(c) ).+ Z Ntemperature associated with the isotropic radiation filling all space?thought to %e a

relic of the %ig?%angB origin of the uni(erseO.

(d) 590 Y ? 59- Y Ndou%le lines of sodiumO

(e) 14.4 7eM Nenerg& of a particular transition in 5+e nucleus associated with a famous high

resolution spectroscopic method

>ss%auer spectroscop&O.

(a) ,adio wa(esL it %elongs to the short wa(elength end of the electromagnetic spectrum.

(b) ,adio wa(esL it %elongs to the short wa(elength end.

(c) 'emperature# 4 $ ).+ °Z

Xm is gi(en %& /lanc7Bs law asI

'his wa(elength corresponds to microwa(es.

(d) 'his is the &ellow light of the (isi%le spectrum.

(e) 'ransition energ& is gi(en %& the relation#

E $ 'ν

*here#

' $ /lanc7Bs constant = -.- × 10−34 Vs

ν $ re!uenc& of radiation

"nerg&# E $ 14.4 Z eM



'his corresponds to Q?ra&s.

Question 8.15:


(a) <ong distance radio %roadcasts use short?wa(e %ands. *h&2

(b) t is necessar& to use satellites for long distance 'M transmission. *h&2

(c) Eptical and radio telescopes are %uilt on the ground %ut Q?ra& astronom& is possi%le onl&

from satellites or%iting the earth. *h&2

(d) 'he small oone la&er on top of the stratosphere is crucial for human sur(i(al. *h&2

(e) f the earth did not ha(e an atmosphere# would its a(erage surface temperature %e higheror lower than what it is now2

(f) Fome scientists ha(e predicted that a glo%al nuclear war on the earth would %e followed %&

a se(ere nuclear winterB with a de(astating effect on life on earth. *hat might %e the %asis of

this prediction2

(a) <ong distance radio %roadcasts use shortwa(e %ands %ecause onl& these %ands can %erefracted %& the ionosphere.

(b) t is necessar& to use satellites for long distance 'M transmissions %ecause tele(ision

signals are of high fre!uencies and high energies. 'hus# these signals are not reflected %& theionosphere. ence# satellites are helpful in reflecting 'M signals. Also# the& help in longdistance 'M transmissions.

(c) *ith reference to Q?ra& astronom&# Q?ra&s are a%sor%ed %& the atmosphere. owe(er#(isi%le and radio wa(es can penetrate it. ence# optical and radio telescopes are %uilt on theground# while Q?ra& astronom& is possi%le onl& with the help of satellites or%iting the "arth.

(d) 'he small oone la&er on the top of the atmosphere is crucial for human sur(i(al %ecauseit a%sor%s harmful ultra(iolet radiations present in sunlight and pre(ents it from reaching the"arthBs surface.

(e) n thea%senceof an atmosphere# there would %e no greenhouse effect on the surface of

the "arth. As a result# the temperature of the "arth would decrease rapidl&# ma7ing it chill&and difficult for human sur(i(al.

(f) A glo%al nuclear war on the surface of the "arth would ha(e disastrous conse!uences./ost?nuclear war# the "arth will e@perience se(ere winter as the war will produce clouds ofsmo7e that would co(er ma@imum parts of the s7&# there%& pre(enting solar light formreaching the atmosphere. Also# it will lead to the depletion of the oone la&er.

CHAPTER 5 RA6 OPTICS AND OPTICAL INSTRENTS



Question 9.1:

A small candle# ).5 cm in sie is placed at )+ cm in front of a conca(e mirror of radius of

cur(ature 3- cm. At what distance from the mirror should a screen %e placed in order to

o%tain a sharp image2 escri%e the nature and sie of the image. f the candle is mo(ed closer

to the mirror# how would the screen ha(e to %e mo(ed2

Fie of the candle# ' $ ).5 cm

mage sie $ 'B

E%Gect distance# , $ −)+ cm

,adius of cur(ature of the conca(e mirror# 9 $ −3- cm

ocal length of the conca(e mirror#

mage distance $ "

'he image distance can %e o%tained using the mirror formulaI

'herefore# the screen should %e placed 54 cm awa& from the mirror to o%tain a sharp image.

'he magnification of the image is gi(en asI



'he height of the candleBs image is 5 cm. 'he negati(e sign indicates that the image isin(erted and (irtual.

f the candle is mo(ed closer to the mirror# then the screen will ha(e to %e mo(ed awa& fromthe mirror in order to o%tain the image.

Question 9.2:

A 4.5 cm needle is placed 1) cm awa& from a con(e@ mirror of focal length 15 cm. =i(e the

location of the image and the magnification. escri%e what happens as the needle is mo(ed

farther from the mirror.

eight of the needle# '1 $ 4.5 cm

E%Gect distance# , $ −1) cm

ocal length of the con(e@ mirror# ) $ 15 cm

mage distance $ "

'he (alue of " can %e o%tained using the mirror formulaI

ence# the image of the needle is -.+ cm awa& from the mirror. Also# it is on the other side ofthe mirror.

'he image sie is gi(en %& the magnification formulaI

ence# magnification of the image#



'he height of the image is ).5 cm. 'he positi(e sign indicates that the image is erect# (irtual#and diminished.

f the needle is mo(ed farther from the mirror# the image will also mo(e awa& from the mirror#and the sie of the image will reduce graduall&.

Question 9.3:

A tan7 is filled with water to a height of 1).5 cm. 'he apparent depth of a needle l&ing at the

%ottom of the tan7 is measured %& a microscope to %e 9.4 cm. *hat is the refracti(e inde@ of

water2 f water is replaced %& a li!uid of refracti(e inde@ 1.-3 up to the same height# %& what

distance would the microscope ha(e to %e mo(ed to focus on the needle again2

Actual depth of the needle in water# '1 $ 1).5 cm

Apparent depth of the needle in water# ') $ 9.4 cm

,efracti(e inde@ of water $ K

'he (alue of Kcan %e o%tained as followsI

ence# the refracti(e inde@ of water is a%out 1.33.

*ater is replaced %& a li!uid of refracti(e inde@#

'he actual depth of the needle remains the same# %ut its apparent depth changes. <et y %e thenew apparent depth of the needle. ence# we can write the relationI

ence# the new apparent depth of the needle is +.-+ cm. t is less than '). 'herefore# to focusthe needle again# the microscope should %e mo(ed up.

∴istance %& which the microscope should %e mo(ed up $ 9.4 − +.-+

$ 1.+3 cm



Question 9.4:

igures 9.34a and % show refraction of a ra& in air incident at -0° with the normal to a

glass?air and water?air interface# respecti(el&. /redict the angle of refraction in glass when theangle of incidence in water is 45 with the normal to a water?glass interface Nig. 9.34cO.

As per the gi(en figure# for the glass − air interfaceI

Angle of incidence# i $ -0°

Angle of refraction# r $ 35°

'he relati(e refracti(e inde@ of glass with respect to air is gi(en %& FnellBs law asI

As per the gi(en figure# for the air − water interfaceI

Angle of incidence# i $ -0°

Angle of refraction# r $ 4+°

'he relati(e refracti(e inde@ of water with respect to air is gi(en %& FnellBs law asI

Psing 1 and )# the relati(e refracti(e inde@ of glass with respect to water can %e o%tainedasI



'he following figure shows the situation in(ol(ing the glass − water interface.

Angle of incidence# i $ 45°

Angle of refraction $ r

rom FnellBs law# r can %e calculated asI

ence# the angle of refraction at the water − glass interface is 3.-°

Question 9.5:

A small %ul% is placed at the %ottom of a tan7 containing water to a depth of 0 cm. *hat is

the area of the surface of water through which light from the %ul% can emerge out2 ,efracti(e

inde@ of water is 1.33. Consider the %ul% to %e a point source.



Actual depth of the %ul% in water# d 1 $ 0 cm $ 0. m

,efracti(e inde@ of water#

'he gi(en situation is shown in the following figureI

*here#

i $ Angle of incidence

r $ Angle of refraction $ 90°

Fince the %ul% is a point source# the emergent light can %e considered as a circle of radius#

Psing FnellB law# we can write the relation for the refracti(e inde@ of water asI

Psing the gi(en figure# we ha(e the relationI

∴9 $ tan 4.+5° × 0. $ 0.91 m

∴Area of the surface of water $ ^9) $ ^ 0.91) $ ).-1 m)

ence# the area of the surface of water through which the light from the %ul% can emerge isappro@imatel& ).-1 m)



Question 9.6:

A prism is made of glass of un7nown refracti(e inde@. A parallel %eam of light is incident on a

face of the prism. 'he angle of minimum de(iation is measured to %e 40°. *hat is the

refracti(e inde@ of the material of the prism2 'he refracting angle of the prism is -0°. f the

prism is placed in water refracti(e inde@ 1.33# predict the new angle of minimum de(iation of

a parallel %eam of light.

Angle of minimum de(iation# $ 40°

Angle of the prism# A $ -0°

,efracti(e inde@ of water# J $ 1.33

,efracti(e inde@ of the material of the prism $

'he angle of de(iation is related to refracti(e inde@ asI

ence# the refracti(e inde@ of the material of the prism is 1.53).

Fince the prism is placed in water# let %e the new angle of minimum de(iation for the sameprism.

'he refracti(e inde@ of glass with respect to water is gi(en %& the relationI



ence# the image is formed 4 cm awa& from the lens# toward its right.

Question 9.9:

An o%Gect of sie 3.0 cm is placed 14 cm in front of a conca(e lens of focal length )1 cm.

escri%e the image produced %& the lens. *hat happens if the o%Gect is mo(ed further awa&

from the lens2

Fie of the o%Gect# '1 $ 3 cm

E%Gect distance# , $ −14 cm

ocal length of the conca(e lens# ) $ −)1 cm

mage distance $ "

According to the lens formula# we ha(e the relationI

ence# the image is formed on the other side of the lens# .4 cm awa& from it. 'he negati(esign shows that the image is erect and (irtual.

'he magnification of the image is gi(en asI

ence# the height of the image is 1. cm.




mage distance for the o%Gecti(e lens#

E%Gect distance for the o%Gecti(e lens $ ,1


>agnitude of the o%Gect distance# $ ).5 cm

'he magnif&ing power of a compound microscope is gi(en %& the relationI

ence# the magnif&ing power of the microscope is )0.

(b) 'he final image is formed at infinit&.

∴mage distance for the e&epiece#

E%Gect distance for the e&epiece $ ,)




mage distance for the o%Gecti(e lens#

E%Gect distance for the o%Gecti(e lens $ ,1


>agnitude of the o%Gect distance# $ ).59 cm

'he magnif&ing power of a compound microscope is gi(en %& the relationI

ence# the magnif&ing power of the microscope is 13.51.

Question 9.12:

A person with a normal near point )5 cm using a compound microscope with o%Gecti(e of

focal length .0 mm and an e&epiece of focal length ).5 cm can %ring an o%Gect placed at 9.0

mm from the o%Gecti(e in sharp focus. *hat is the separation %etween the two lenses2

Calculate the magnif&ing power of the microscope#

ocal length of the o%Gecti(e lens# ) o $ mm $ 0. cm

ocal length of the e&epiece# ) e $ ).5 cm



E%Gect distance for the o%Gecti(e lens# ,o $ −9.0 mm $ −0.9 cm

<east distance of distant (ision# d $ )5 cm

mage distance for the e&epiece# " e $ −d $ −)5 cm

E%Gect distance for the e&epiece $

Psing the lens formula# we can o%tain the (alue of asI

*e can also o%tain the (alue of the image distance for the o%Gecti(e lens using the lensformula.

'he distance %etween the o%Gecti(e lens and the e&epiece

'he magnif&ing power of the microscope is calculated asI



ence# the magnif&ing power of the microscope is .

Question 9.13:

A small telescope has an o%Gecti(e lens of focal length 144 cm and an e&epiece of focal length

-.0 cm. *hat is the magnif&ing power of the telescope2 *hat is the separation %etween the

o%Gecti(e and the e&epiece2

ocal length of the o%Gecti(e lens# ) o $ 144 cm

ocal length of the e&epiece# ) e $ -.0 cm

'he magnif&ing power of the telescope is gi(en asI

'he separation %etween the o%Gecti(e lens and the e&epiece is calculated asI

ence# the magnif&ing power of the telescope is )4 and the separation %etween the o%Gecti(elens and the e&epiece is 150 cm.

Question 9.14:

(a) A giant refracting telescope at an o%ser(ator& has an o%Gecti(e lens of focal length 15 m.

f an e&epiece of focal length 1.0 cm is used# what is the angular magnification of the

telescope2

(b) f this telescope is used to (iew the moon# what is the diameter of the image of the moonformed %& the o%Gecti(e lens2 'he diameter of the moon is 3.4 × 10- m# and the radius of

lunar or%it is 3. × 10 m.



ocal length of the o%Gecti(e lens# ) o $ 15 m $ 15 × 10) cm

ocal length of the e&epiece# ) e $ 1.0 cm

(a) 'he angular magnification of a telescope is gi(en asI

ence# the angular magnification of the gi(en refracting telescope is 1500.

(b) iameter of the moon# d $ 3.4 × 10- m

,adius of the lunar or%it# r 0 $ 3. × 10 m

<et %e the diameter of the image of the moon formed %& the o%Gecti(e lens.

'he angle su%tended %& the diameter of the moon is e!ual to the angle su%tended %& theimage.

ence# the diameter of the moonBs image formed %& the o%Gecti(e lens is 13.+4 cm

Question 9.15:

Pse the mirror e!uation to deduce thatI

(a) an o%Gect placed %etween ) and )) of a conca(e mirror produces a real image %e&ond ))/

(b) a con(e@ mirror alwa&s produces a (irtual image independent of the location of the o%Gect.



(c) the (irtual image produced %& a con(e@ mirror is alwa&s diminished in sie and is located

%etween the focus and the pole.

(d) an o%Gect placed %etween the pole and focus of a conca(e mirror produces a (irtual and

enlarged image.

NB%te: 'his e@ercise helps &ou deduce alge%raicall& properties of

images that one o%tains from e@plicit ra& diagrams.O

(a) or a conca(e mirror# the focal length ) is negati(e.

∴) 0

*hen the o%Gect is placed on the left side of the mirror# the o%Gect distance , is negati(e.

∴, 0

or image distance " # we can write the lens formula asI

'he o%Gect lies %etween ) and )) .

Psing e!uation 1# we getI

∴ is negati(e# i.e.# " is negati(e.



'herefore# the image lies %e&ond )) .

(b) or a con(e@ mirror# the focal length ) is positi(e.

∴ ) [ 0


∴ , 0

or image distance " # we ha(e the mirror formulaI

'hus# the image is formed on the %ac7 side of the mirror.

ence# a con(e@ mirror alwa&s produces a (irtual image# regardless of the o%Gect distance.

(c) or a con(e@ mirror# the focal length ) is positi(e.

∴) [ 0

*hen the o%Gect is placed on the left side of the mirror# the o%Gect distance , is negati(e#

∴, 0




ence# the image formed is diminished and is located %etween the focus ) and the pole.

(d) or a conca(e mirror# the focal length ) is negati(e.

∴) 0


∴, 0

t is placed %etween the focus ) and the pole.


'he image is formed on the right side of the mirror. ence# it is a (irtual image.

or , 0 and " [ 0# we can writeI



>agnification# m [ 1

ence# the formed image is enlarged.

Question 9.16:

A small pin fi@ed on a ta%le top is (iewed from a%o(e from a distance of 50 cm. 8& what

distance would the pin appear to %e raised if it is (iewed from the same point through a 15 cm

thic7 glass sla% held parallel to the ta%le2 ,efracti(e inde@ of glass $ 1.5. oes the answerdepend on the location of the sla%2

Actual depth of the pin# d $ 15 cm

Apparent dept of the pin $

,efracti(e inde@ of glass#

,atio of actual depth to the apparent depth is e!ual to the refracti(e inde@ of glass# i.e.

'he distance at which the pin appears to %e raised $

or a small angle of incidence# this distance does not depend upon the location of the sla%.

Question 9.17:

(a) igure 9.35 shows a cross?section of a light pipeB made of a glass fi%re of refracti(e inde@

1.-. 'he outer co(ering of the pipe is made of a material of refracti(e inde@ 1.44. *hat is the



range of the angles of the incident ra&s with the a@is of the pipe for which total reflections

inside the pipe ta7e place# as shown in the figure.

(b) *hat is the answer if there is no outer co(ering of the pipe2

(a) ,efracti(e inde@ of the glass fi%re#

,efracti(e inde@ of the outer co(ering of the pipe# $ 1.44

Angle of incidence $ i

Angle of refraction $ r

Angle of incidence at the interface $ i B

'he refracti(e inde@ K of the inner core − outer core interface is gi(en asI

or the critical angle# total internal reflection ', ta7es place onl& when # i.e.# i [ 59°

>a@imum angle of reflection#

<et# %e the ma@imum angle of incidence.

'he refracti(e inde@ at the air − glass interface#

*e ha(e the relation for the ma@imum angles of incidence and reflection asI



'hus# all the ra&s incident at angles l&ing in the range 0 i -0° will suffer total internalreflection.

(b) f the outer co(ering of the pipe is not present# thenI

,efracti(e inde@ of the outer pipe#

or the angle of incidence i $ 90°# we can write FnellBs law at the air − pipe interface asI

.

Question 9.18:


(a) 6ou ha(e learnt that plane and con(e@ mirrors produce (irtual images of o%Gects. Can the&produce real images under some circumstances2 "@plain.

(b) A (irtual image# we alwa&s sa&# cannot %e caught on a screen.

6et when we seeB a (irtual image# we are o%(iousl& %ringing it on to the screenB i.e.# the

retina of our e&e. s there a contradiction2



(c) A di(er under water# loo7s o%li!uel& at a fisherman standing on the %an7 of a la7e. *ould

the fisherman loo7 taller or shorter to the di(er than what he actuall& is2

(d) oes the apparent depth of a tan7 of water change if (iewed o%li!uel&2 f so# does the

apparent depth increase or decrease2

(e) 'he refracti(e inde@ of diamond is much greater than that of ordinar& glass. s this fact of

some use to a diamond cutter2

(a) 6es

/lane and con(e@ mirrors can produce real images as well. f the o%Gect is (irtual# i.e.# if thelight ra&s con(erging at a point %ehind a plane mirror or a con(e@ mirror are reflected to apoint on a screen placed in front of the mirror# then a real image will %e formed.

(b) o

A (irtual image is formed when light ra&s di(erge. 'he con(e@ lens of the e&e causes thesedi(ergent ra&s to con(erge at the retina. n this case# the (irtual image ser(es as an o%Gect forthe lens to produce a real image.

(c) 'he di(er is in the water and the fisherman is on land i.e.# in air. *ater is a densermedium than air. t is gi(en that the di(er is (iewing the fisherman. 'his indicates that thelight ra&s are tra(elling from a denser medium to a rarer medium. ence# the refracted ra&swill mo(e awa& from the normal. As a result# the fisherman will appear to %e taller.

(d) 6esL ecrease

'he apparent depth of a tan7 of water changes when (iewed o%li!uel&. 'his is %ecause light%ends on tra(elling from one medium to another. 'he apparent depth of the tan7 when (iewedo%li!uel& is less than the near?normal (iewing.

(e) 6es

'he refracti(e inde@ of diamond ).4) is more than that of ordinar& glass 1.5. 'he criticalangle for diamond is less than that for glass. A diamond cutter uses a large angle of incidenceto ensure that the light entering the diamond is totall& reflected from its faces. 'his is thereason for the spar7ling effect of a diamond.

Question 9.19:

'he image of a small electric %ul% fi@ed on the wall of a room is to %e o%tained on the opposite

wall 3 m awa& %& means of a large con(e@ lens. *hat is the ma@imum possi%le focal length of

the lens re!uired for the purpose2



istance %etween the o%Gect and the image# d $ 3 m

>a@imum focal length of the con(e@ lens $

or real images# the ma@imum focal length is gi(en asI

ence# for the re!uired purpose# the ma@imum possi%le focal length of the con(e@ lens is 0.+5m.

Question 9.20:

A screen is placed 90 cm from an o%Gect. 'he image of the o%Gect on the screen is formed %& a

con(e@ lens at two different locations separated %& )0 cm. etermine the focal length of the

lens.

istance %etween the image screen and the o%Gect# C $ 90 cm

istance %etween two locations of the con(e@ lens# d $ )0 cm

ocal length of the lens $ )

ocal length is related to d and C asI

'herefore# the focal length of the con(e@ lens is )1.39 cm

Question 9.21:

(a) etermine the effecti(e focal lengthB of the com%ination of the two lenses in "@ercise

9.10# if the& are placed .0 cm apart with their principal a@es coincident. oes the answer

depend on which side of the com%ination a %eam of parallel light is incident2 s the notion of

effecti(e focal length of this s&stem useful at all2



(b) An o%Gect 1.5 cm in sie is placed on the side of the con(e@ lens in the arrangement a

a%o(e. 'he distance %etween the o%Gect and the con(e@ lens is 40 cm. etermine the

magnification produced %& the two?lens s&stem# and the sie of the image.

ocal length of the con(e@ lens# ) 1 $ 30 cm

ocal length of the conca(e lens# ) ) $ −)0 cm

istance %etween the two lenses# d $ .0 cm

(a) 4+e& "+e %a$ae bea !f ig+" i7 i&cide&" !& "+e c!&ve8 e&7 fi$7"#

According to the lens formula# we ha(eI

*here#

$ E%Gect distance $

" 1 $ mage distance

'he image will act as a (irtual o%Gect for the conca(e lens.

Appl&ing lens formula to the conca(e lens# we ha(eI

*here#

$ E%Gect distance

$ 30 − d $ 30 − $ )) cm

$ mage distance



'he parallel incident %eam appears to di(erge from a point that is

from the centre of the com%ination of the two lenses.

(ii) 4+e& "+e %a$ae bea !f ig+" i7 i&cide&"9 f$! "+e ef"9 !& "+e c!&cave e&7fi$7"#


*here#

$ E%Gect distance $ −

$ mage distance

'he image will act as a real o%Gect for the con(e@ lens.

Appl&ing lens formula to the con(e@ lens# we ha(eI

*here#

$ E%Gect distance

$ −)0 H d $ −)0 H $ −) cm

$ mage distance



*here#

$ E%Gect distance

$ H1)0 − $ 11) cm.

$ mage distance

>agnification#

ence# the magnification due to the conca(e lens is .

'he magnification produced %& the com%ination of the two lenses is calculated asI

'he magnification of the com%ination is gi(en asI

*here#

'1 $ E%Gect sie $ 1.5 cm

') $ Fie of the image

ence# the height of the image is 0.9 cm.



According to FnellBs law# we ha(e the relationI

ence# the angle of incidence is )9.+5°.

Question 9.23:

6ou are gi(en prisms made of crown glass and flint glass with a wide (ariet& of angles.

Fuggest a com%ination of prisms which will

(a) de(iate a pencil of white light without much dispersion#

(b) disperse and displace a pencil of white light without much de(iation.

(a)/lace the two prisms %eside each other. >a7e sure that their %ases are on the oppositesides of the incident white light# with their faces touching each other. *hen the white light isincident on the first prism# it will get dispersed. *hen this dispersed light is incident on thesecond prism# it will recom%ine and white light will emerge from the com%ination of the twoprisms.

(b)'a7e the s&stem of the two prisms as suggested in answer (a). AdGust increase the angleof the flint?glass?prism so that the de(iations due to the com%ination of the prisms %ecome

e!ual. 'his com%ination will disperse the pencil of white light without much de(iation.

Question 9.24:

or a normal e&e# the far point is at infinit& and the near point of distinct (ision is a%out )5cm

in front of the e&e. 'he cornea of the e&e pro(ides a con(erging power of a%out 40 dioptres#

and the least con(erging power of the e&e?lens %ehind the cornea is a%out )0 dioptres. rom

this rough data estimate the range of accommodation i.e.# the range of con(erging power of

the e&e?lens of a normal e&e.

<east distance of distinct (ision# d $ )5 cm

ar point of a normal e&e#

Con(erging power of the cornea#

<east con(erging power of the e&e?lens#



'o see the o%Gects at infinit&# the e&e uses its least con(erging power.

/ower of the e&e?lens# 6 $ 6 c H 6 e $ 40 H )0 $ -0

/ower of the e&e?lens is gi(en asI

'o focus an o%Gect at the near point# o%Gect distance , $ −d $ −)5 cm

ocal length of the e&e?lens $ istance %etween the cornea and the retina

$ mage distance

ence# image distance#

According to the lens formula# we can writeI

*here#

$ ocal length

∴/ower of the e&e?lens $ -4 − 40 $ )4

ence# the range of accommodation of the e&e?lens is from )0 to )4 .



Question 9.25:

oes short?sightedness m&opia or long?sightedness h&permetropia impl& necessaril& that

the e&e has partiall& lost its a%ilit& of accommodation2 f not# what might cause these defects

of (ision2

A m&opic or h&permetropic person can also possess the normal a%ilit& of accommodation ofthe e&e?lens. >&opia occurs when the e&e?%alls get elongated from front to %ac7.&permetropia occurs when the e&e?%alls get shortened. *hen the e&e?lens loses its a%ilit& ofaccommodation# the defect is called pres%&opia.

Question 9.26:

A m&opic person has %een using spectacles of power −1.0 dioptre for distant (ision. uring old

age he also needs to use separate reading glass of power H ).0 dioptres. "@plain what ma&

ha(e happened.

'he power of the spectacles used %& the m&opic person# 6 $ −1.0

ocal length of the spectacles#

ence# the far point of the person is 100 cm. e might ha(e a normal near point of )5 cm.*hen he uses the spectacles# the o%Gects placed at infinit& produce (irtual images at 100 cm.e uses the a%ilit& of accommodation of the e&e?lens to see the o%Gects placed %etween 100cm and )5 cm.

uring old age# the person uses reading glasses of power#

'he a%ilit& of accommodation is lost in old age. 'his defect is called pres%&opia. As a result# heis una%le to see clearl& the o%Gects placed at )5 cm.

Question 9.27:

A person loo7ing at a person wearing a shirt with a pattern comprising (ertical and horiontal

lines is a%le to see the (ertical lines more distinctl& than the horiontal ones. *hat is this

defect due to2 ow is such a defect of (ision corrected2

n the gi(en case# the person is a%le to see (ertical lines more distinctl& than horiontal lines.

'his means that the refracting s&stem cornea and e&e?lens of the e&e is not wor7ing in thesame wa& in different planes. 'his defect is called astigmatism. 'he personBs e&e has enoughcur(ature in the (ertical plane. owe(er# the cur(ature in the horiontal plane is insufficient.ence# sharp images of the (ertical lines are formed on the retina# %ut horiontal lines appear%lurred. 'his defect can %e corrected %& using c&lindrical lenses.



Question 9.28:

A man with normal near point )5 cm reads a %oo7 with small print using a magnif&ing glassI

a thin con(e@ lens of focal length 5 cm.

(a) *hat is the closest and the farthest distance at which he should 7eep the lens from the

page so that he can read the %oo7 when (iewing through the magnif&ing glass2

(b) *hat is the ma@imum and the minimum angular magnification magnif&ing power

possi%le using the a%o(e simple microscope2

(a) ocal length of the magnif&ing glass# ) $ 5 cm

<east distance of distance (ision# d = )5 cm

Closest o%Gect distance $ ,

mage distance# " $ −d $ −)5 cm


ence# the closest distance at which the person can read the %oo7 is 4.1-+ cm.

or the o%Gect at the farthest distant ,B# the image distance




ence# the farthest distance at which the person can read the %oo7 is5 cm.

(b) >a@imum angular magnification is gi(en %& the relationI

>inimum angular magnification is gi(en %& the relationI

Question 9.29:

A card sheet di(ided into s!uares each of sie 1 mm) is %eing (iewed at a distance of 9 cm

through a magnif&ing glass a con(erging lens of focal length 9 cm held close to the e&e.

(a) *hat is the magnification produced %& the lens2 ow much is the area of each s!uare in

the (irtual image2

(b) *hat is the angular magnification magnif&ing power of the lens2

(c) s the magnification in a e!ual to the magnif&ing power in %2

"@plain.

(a) Area of each s!uare# A $ 1 mm)

E%Gect distance# , $ −9 cm

ocal length of a con(erging lens# ) $ 10 cm

or image distance " # the lens formula can %e written asI



>agnification#

∴Area of each s!uare in the (irtual image $ 10) A

$ 10) × 1 $ 100 mm)

$ 1 cm)

(b) >agnif&ing power of the lens

(c) 'he magnification in (a) is not the same as the magnif&ing power in (b).

'he magnification magnitude is and the magnif&ing power is .

'he two !uantities will %e e!ual when the image is formed at the near point )5 cm.

Question 9.30:

(a) At what distance should the lens %e held from the figure in

"@ercise 9.)9 in order to (iew the s!uares distinctl& with the ma@imum possi%le magnif&ing

power2

(b) *hat is the magnification in this case2

(c) s the magnification e!ual to the magnif&ing power in this case2

"@plain.



(a) 'he ma@imum possi%le magnification is o%tained when the image is formed at the nearpoint d $ )5 cm.

mage distance# " $ −d $ −)5 cm

ocal length# ) $ 10 cm

E%Gect distance $ ,


ence# to (iew the s!uares distinctl&# the lens should %e 7ept +.14 cm awa& from them.

(b) >agnification $

(c) >agnif&ing power $

Fince the image is formed at the near point )5 cm# the magnif&ing power is e!ual to themagnitude of magnification.

Question 9.31:

*hat should %e the distance %etween the o%Gect in "@ercise 9.30 and the magnif&ing glass if

the (irtual image of each s!uare in the figure is to ha(e an area of -.)5 mm). *ould &ou %e

a%le to see the s!uares distinctl& with &our e&es (er& close to the magnifier2

NB%te: "@ercises 9.)9 to 9.31 will help &ou clearl& understand the difference %etween

magnification in a%solute sie and the angular magnification or magnif&ing power of an

instrument.O

Area of the (irtual image of each s!uare# A $ -.)5 mm)



Area of each s!uare# A0 $ 1 mm)

ence# the linear magnification of the o%Gect can %e calculated asI

ocal length of the magnif&ing glass# ) $ 10 cm


'he (irtual image is formed at a distance of 15 cm# which is less than the near point i.e.# )5cm of a normal e&e. ence# it cannot %e seen %& the e&es distinctl&.

Question 9.32:


(a) 'he angle su%tended at the e&e %& an o%Gect is e!ual to the angle su%tended at the e&e %&

the (irtual image produced %& a magnif&ing glass. n what sense then does a magnif&ing glass

pro(ide angular magnification2

(b) n (iewing through a magnif&ing glass# one usuall& positions oneBs e&es (er& close to the

lens. oes angular magnification change if the e&e is mo(ed %ac72



(c) >agnif&ing power of a simple microscope is in(ersel& proportional to the focal length of

the lens. *hat then stops us from using a con(e@ lens of smaller and smaller focal length and

achie(ing greater and greater magnif&ing power2

(d) *h& must %oth the o%Gecti(e and the e&epiece of a compound microscope ha(e short focal

lengths2

(e) *hen (iewing through a compound microscope# our e&es should %e positioned not on the

e&epiece %ut a short distance awa& from it for %est (iewing. *h&2 ow much should %e that

short distance %etween the e&e and e&epiece2

(a)'hough the image sie is %igger than the o%Gect# the angular sie of the image is e!ual tothe angular sie of the o%Gect. A magnif&ing glass helps one see the o%Gects placed closer thanthe least distance of distinct (ision i.e.# )5 cm. A closer o%Gect causes a larger angular sie. Amagnif&ing glass pro(ides angular magnification. *ithout magnification# the o%Gect cannot %eplaced closer to the e&e. *ith magnification# the o%Gect can %e placed much closer to the e&e.

(b) 6es# the angular magnification changes. *hen the distance %etween the e&e and amagnif&ing glass is increased# the angular magnification decreases a little. 'his is %ecause theangle su%tended at the e&e is slightl& less than the angle su%tended at the lens. magedistance does not ha(e an& effect on angular magnification.

(c) 'he focal length of a con(e@ lens cannot %e decreased %& a greater amount. 'his is%ecause ma7ing lenses ha(ing (er& small focal lengths is not eas&. Fpherical and chromatica%errations are produced %& a con(e@ lens ha(ing a (er& small focal length.

(d) 'he angular magnification produced %& the e&epiece of a compound microscope is

*here#

) e $ ocal length of the e&epiece

t can %e inferred that if ) e is small# then angular magnification of the e&epiece will %e large.

'he angular magnification of the o%Gecti(e lens of a compound microscope is gi(en as

*here#

$ E%Gect distance for the o%Gecti(e lens

$ ocal length of the o%Gecti(e



'he magnification is large when [ . n the case of a microscope# the o%Gect is 7ept close

to the o%Gecti(e lens. ence# the o%Gect distance is (er& little. Fince is small# will %e e(en

smaller. 'herefore# and are %oth small in the gi(en condition.

(e)*hen we place our e&es too close to the e&epiece of a compound microscope# we areuna%le to collect much refracted light. As a result# the field of (iew decreases su%stantiall&.ence# the clarit& of the image gets %lurred.

'he %est position of the e&e for (iewing through a compound microscope is at the e&e?ringattached to the e&epiece. 'he precise location of the e&e depends on the separation %etweenthe o%Gecti(e lens and the e&epiece.

Question 9.33:

An angular magnification magnif&ing power of 30Q is desired using an o%Gecti(e of focal

length 1.)5 cm and an e&epiece of focal length 5 cm. ow will &ou set up the compound

microscope2

ocal length of the o%Gecti(e lens# $ 1.)5 cm

ocal length of the e&epiece# ) e $ 5 cm


Angular magnification of the compound microscope $ 30Q

'otal magnif&ing power of the compound microscope# m $ 30

'he angular magnification of the e&epiece is gi(en %& the relationI

'he angular magnification of the o%Gecti(e lens mo is related to me asI

$ m



Appl&ing the lens formula for the o%Gecti(e lensI

'he o%Gect should %e placed 1.5 cm awa& from the o%Gecti(e lens to o%tain the desiredmagnification.

Appl&ing the lens formula for the e&epieceI

*here#

$ mage distance for the e&epiece $ −d $ −)5 cm

$ E%Gect distance for the e&epiece



Feparation %etween the o%Gecti(e lens and the e&epiece

'herefore# the separation %etween the o%Gecti(e lens and the e&epiece should %e 11.-+ cm.

Question 9.34:

A small telescope has an o%Gecti(e lens of focal length 140 cm and an e&epiece of focal length

5.0 cm. *hat is the magnif&ing power of the telescope for (iewing distant o%Gects when

(a) the telescope is in normal adGustment i.e.# when the final image

is at infinit&2

(b) the final image is formed at the least distance of distinct (ision

)5 cm2

ocal length of the o%Gecti(e lens# $ 140 cm



(a) *hen the telescope is in normal adGustment# its magnif&ing power is gi(en asI

(b) *hen the final image is formed at d #the magnif&ing power of the telescope is gi(en asI



Question 9.35:

a or the telescope descri%ed in "@ercise 9.34 a# what is the separation %etween the

o%Gecti(e lens and the e&epiece2

% f this telescope is used to (iew a 100 m tall tower 3 7m awa&# what is the height of the

image of the tower formed %& the o%Gecti(e lens2

c *hat is the height of the final image of the tower if it is formed at )5 cm2

ocal length of the o%Gecti(e lens# ) o $ 140 cm


(a) n normal adGustment# the separation %etween the o%Gecti(e lens and the e&epiece

(b) eight of the tower# '1 $ 100 m

istance of the tower o%Gect from the telescope# , $ 3 7m $ 3000 m

'he angle su%tended %& the tower at the telescope is gi(en asI

'he angle su%tended %& the image produced %& the o%Gecti(e lens is gi(en asI

*here#

') $ eight of the image of the tower formed %& the o%Gecti(e lens



ence# focal length of the secondar& mirror#

'he image of an o%Gect placed at infinit&# formed %& the o%Gecti(e mirror# will act as a (irtualo%Gect for the secondar& mirror.

ence# the (irtual o%Gect distance for the secondar& mirror#

Appl&ing the mirror formula for the secondar& mirror# we can calculate image distance " asI

ence# the final image will %e formed 315 mm awa& from the secondar& mirror.

Question 9.37:

<ight incident normall& on a plane mirror attached to a gal(anometer coil retraces %ac7wards

as shown in ig. 9.3-. A current in the coil produces a deflection of 3.5° of the mirror. *hat is

the displacement of the reflected spot of light on a screen placed 1.5 m awa&2

Angle of deflection# θ $ 3.5°

istance of the screen from the mirror# C $ 1.5 m

'he reflected ra&s get deflected %& an amount twice the angle of deflection i.e.# )θ$ +.0°

'he displacement d of the reflected spot of light on the screen is gi(en asI



ence# the displacement of the reflected spot of light is 1.4 cm.

Question 9.38:

igure 9.3+ shows an e!uicon(e@ lens of refracti(e inde@ 1.50 in contact with a li!uid la&er

on top of a plane mirror. A small needle with its tip on the principal a@is is mo(ed along the

a@is until its in(erted image is found at the position of the needle. 'he distance of the needle

from the lens is measured to %e 45.0 cm. 'he li!uid is remo(ed and the e@periment is

repeated. 'he new distance is measured to %e 30.0 cm. *hat is the refracti(e inde@ of the

li!uid2

ocal length of the con(e@ lens# ) 1 $ 30 cm

'he li!uid acts as a mirror. ocal length of the li!uid $ ) )

ocal length of the s&stem con(e@ lens H li!uid# ) $ 45 cm

or a pair of optical s&stems placed in contact# the e!ui(alent focal length is gi(en asI

<et the refracti(e inde@ of the lens %e and the radius of cur(ature of one surface %e 9.ence# the radius of cur(ature of the other surface is −9.



9 can %e o%tained using the relationI

<et %e the refracti(e inde@ of the li!uid.

• ,adius of cur(ature of the li!uid on the side of the plane mirror $

,adius of cur(ature of the li!uid on the side of the lens# 9 $ −30 cm

'he (alue of can %e calculated using the relationI

ence# the refracti(e inde@ of the li!uid is 1.33.

CHAPTER- 10 WAVE OPTICS

Question 10.1:

>onochromatic light of wa(elength 59 nm is incident from air on a water surface. *hat are

the wa(elength# fre!uenc& and speed of a reflected# and % refracted light2 ,efracti(e inde@of water is 1.33.

*a(elength of incident monochromatic light#

X $ 59 nm $ 59 × 10−9 m

Fpeed of light in air# + $ 3 × 10 m;s



,efracti(e inde@ of water# K $ 1.33

(a) 'he ra& will reflect %ac7 in the same medium as that of incident ra&. ence# thewa(elength# speed# and fre!uenc& of the reflected ra& will %e the same as that of the incidentra&.

re!uenc& of light is gi(en %& the relation#

ence# the speed# fre!uenc&# and wa(elength of the reflected light are 3 × 10 m;s# 5.09×1014 # and 59 nm respecti(el&.

(b) re!uenc& of light does not depend on the propert& of the medium in which it is tra(elling.ence# the fre!uenc& of the refracted ra& in water will %e e!ual to the fre!uenc& of theincident or reflected light in air.

,efracted fre!uenc&# ν $ 5.09 ×1014

Fpeed of light in water is related to the refracti(e inde@ of water asI

*a(elength of light in water is gi(en %& the relation#

ence# the speed# fre!uenc&# and wa(elength of refracted light are ).)- ×10 m;s# 444.01nm#and 5.09 × 1014 respecti(el&.

Question 10.2:

*hat is the shape of the wa(efront in each of the following casesI



(a) <ight di(erging from a point source.

(b) <ight emerging out of a con(e@ lens when a point source is placed at its focus.

(c) 'he portion of the wa(efront of light from a distant star intercepted %& the "arth.

(a) 'he shape of the wa(efront in case of a light di(erging from a point source is spherical.'he wa(efront emanating from a point source is shown in the gi(en figure.

(b) 'he shape of the wa(efront in case of a light emerging out of a con(e@ lens when a pointsource is placed at its focus is a parallel grid. 'his is shown in the gi(en figure.

(c) 'he portion of the wa(efront of light from a distant star intercepted %& the "arth is aplane.

Question 10.3:

(a) 'he refracti(e inde@ of glass is 1.5. *hat is the speed of light in glass2 Fpeed of light in

(acuum is 3.0 × 10 m s−1

(b) s the speed of light in glass independent of the colour of light2 f not# which of the two

colours red and (iolet tra(els slower in a glass prism2

(a) ,efracti(e inde@ of glass# K $ 1.5

Fpeed of light# c $ 3 × 10 m;s



Fpeed of light in glass is gi(en %& the relation#

ence# the speed of light in glass is ) × 10 m;s.

(b) 'he speed of light in glass is not independent of the colour of light.

'he refracti(e inde@ of a (iolet component of white light is greater than the refracti(e inde@ ofa red component. ence# the speed of (iolet light is less than the speed of red light in glass.ence# (iolet light tra(els slower than red light in a glass prism.

Question 10.4:

n a 6oungBs dou%le?slit e@periment# the slits are separated %& 0.) mm and the screen is

placed 1.4 m awa&. 'he distance %etween the central %right fringe and the fourth %right fringe

is measured to %e 1.) cm. etermine the wa(elength of light used in the e@periment.

istance %etween the slits# d $ 0.) mm $ 0.) × 10−3 m

istance %etween the slits and the screen# C $ 1.4 m

istance %etween the central fringe and the fourth n $ 4 fringe#

, $ 1.) cm $ 1.) × 10−) m

n case of a constructi(e interference# we ha(e the relation for the distance %etween the twofringes asI

*here#

n $ Erder of fringes $ 4

X $ *a(elength of light used

∴



ence# the wa(elength of the light is -00 nm

Question 10.5:

n 6oungBs dou%le?slit e@periment using monochromatic light of wa(elengthX# the intensit& of

light at a point on the screen where path difference is X# is units. *hat is the intensit& of

light at a point where path difference is X ;32

<et 3 1 and 3 ) %e the intensit& of the two light wa(es. 'heir resultant intensities can %e o%tainedasI

*here#

$ /hase difference %etween the two wa(es

or monochromatic light wa(es#

/hase difference $

Fince path difference $ X#

/hase difference#

=i(en#

3 B $



*hen path difference #

/hase difference#

ence# resultant intensit&#

Psing e!uation 1# we can writeI

ence# the intensit& of light at a point where the path difference is is units.

Question 10.6:

A %eam of light consisting of two wa(elengths# -50 nm and 5)0 nm# is used to o%taininterference fringes in a 6oungBs dou%le?slit e@periment.

(a) ind the distance of the third %right fringe on the screen from the central ma@imum for

wa(elength -50 nm.

(b) *hat is the least distance from the central ma@imum where the %right fringes due to %oth

the wa(elengths coincide2

*a(elength of the light %eam#

*a(elength of another light %eam#

istance of the slits from the screen $ C

istance %etween the two slits $ d



(a) istance of the nth %right fringe on the screen from the central ma@imum is gi(en %& therelation#

(b) <et the nth %right fringe due to wa(elength and n − 1th %right fringe due to wa(elength

coincide on the screen. *e can e!uate the conditions for %right fringes asI

ence# the least distance from the central ma@imum can %e o%tained %& the relationI

oteI 'he (alue of d and C are not gi(en in the !uestion.

Question 10.7:

n a dou%le?slit e@periment the angular width of a fringe is found to %e 0.)° on a screen placed

1 m awa&. 'he wa(elength of light used is -00 nm. *hat will %e the angular width of the

fringe if the entire e@perimental apparatus is immersed in water2 'a7e refracti(e inde@ of

water to %e 4;3.

istance of the screen from the slits# C $ 1 m

*a(elength of light used#

Angular width of the fringe in

Angular width of the fringe in water $



,efracti(e inde@ of water#

,efracti(e inde@ is related to angular width asI

'herefore# the angular width of the fringe in water will reduce to 0.15°.

Question 10.8:

*hat is the 8rewster angle for air to glass transition2 ,efracti(e inde@ of glass $ 1.5.

,efracti(e inde@ of glass#

8rewster angle $ θ

8rewster angle is related to refracti(e inde@ asI

'herefore# the 8rewster angle for air to glass transition is 5-.31°.

Question 10.9:

<ight of wa(elength 5000 Y falls on a plane reflecting surface. *hat are the wa(elength andfre!uenc& of the reflected light2 or what angle of incidence is the reflected ra& normal to theincident ra&2 *a(elength of incident light# X $ 5000 Y $ 5000 × 10−10 m

Fpeed of light# + $ 3 × 10 m

re!uenc& of incident light is gi(en %& the relation#

'he wa(elength and fre!uenc& of incident light is the same as that of reflected ra&. ence# thewa(elength of reflected light is 5000 Y and its fre!uenc& is - × 1014 .



*hen reflected ra& is normal to incident ra&# the sum of the angle of incidence# and angle

of reflection# is 90°.

According to the law of reflection# the angle of incidence is alwa&s e!ual to the angle ofreflection. ence# we can write the sum asI

'herefore# the angle of incidence for the gi(en condition is 45°.

Question 10.10:

"stimate the distance for which ra& optics is good appro@imation for an aperture of 4 mm and

wa(elength 400 nm.

resnelBs distance 8 F is the distance for which the ra& optics is a good appro@imation. t isgi(en %& the relation#

*here#

Aperture width# & $ 4 mm $ 4 ×10−3 m

*a(elength of light# λ $ 400 nm $ 400 × 10−9 m

'herefore# the distance for which the ra& optics is a good appro@imation is 40 m.

Question 10.11:

'he -5-3 Y line emitted %& h&drogen in a star is found to %e red shifted %& 15 Y. "stimate

the speed with which the star is receding from the "arth.

*a(elength of line emitted %& h&drogen#

X $ -5-3 Y

$ -5-3 × 10−10 m.



FtarBs red?shift#

Fpeed of light#

<et the (elocit& of the star receding awa& from the "arth %e " .

'he red shift is related with (elocit& asI

'herefore# the speed with which the star is receding awa& from the "arth is -.+ × 105 m;s.

Question 10.12:

"@plain how Corpuscular theor& predicts the speed of light in a medium# sa&# water# to %e

greater than the speed of light in (acuum. s the prediction confirmed %& e@perimental

determination of the speed of light in water2 f not# which alternati(e picture of light is

consistent with e@periment2

oL *a(e theor&

ewtonBs corpuscular theor& of light states that when light corpuscles stri7e the interface oftwo media from a rarer air to a denser water medium# the particles e@perience forces ofattraction normal to the surface. ence# the normal component of (elocit& increases while thecomponent along the surface remains unchanged.

ence# we can write the e@pressionI

D i

*here#

i $ Angle of incidence

r $ Angle of reflection

+ $ Melocit& of light in air

" $ Melocit& of light in water



*e ha(e the relation for relati(e refracti(e inde@ of water with respect to air asI

ence# e!uation i reduces to

8ut# [ 1

ence# it can %e inferred from e!uation ii that " [ + . 'his is not possi%le since this predictionis opposite to the e@perimental results of + [ " .

'he wa(e picture of light is consistent with the e@perimental results.

Question 10.13:

6ou ha(e learnt in the te@t how u&gensB principle leads to the laws of reflection and

refraction. Pse the same principle to deduce directl& that a point o%Gect placed in front of a

plane mirror produces a (irtual image whose distance from the mirror is e!ual to the o%Gect

distance from the mirror.

<et an o%Gect at E %e placed in front of a plane mirror >EB at a distance r as shown in thegi(en figure.

A circle is drawn from the centre E such that it Gust touches the plane mirror at point EB.According to u&gensB /rinciple# Q6 is the wa(efront of incident light.

f the mirror is a%sent# then a similar wa(efront QB6B as Q6 would form %ehind EB at distancer as shown in the gi(en figure.



can %e considered as a (irtual reflected ra& for the plane mirror. ence# a point o%Gectplaced in front of the plane mirror produces a (irtual image whose distance from the mirror ise!ual to the o%Gect distance r .

Question 10.14:

<et us list some of the factors# which could possi%l& influence the speed of wa(e propagationI

(i) ature of the source.

(ii) irection of propagation.

(iii) >otion of the source and;or o%ser(er.

(iv) *a(e length.

(v) ntensit& of the wa(e.

En which of these factors# if an&# does

(a) 'he speed of light in (acuum#

(b) 'he speed of light in a medium sa&# glass or water# depend2

(a) 'hespeed of light in a (acuum i.e.# 3 × 10 m;s appro@imatel& is a uni(ersal constant. tis not affected %& the motion of the source# the o%ser(er# or %oth. ence# the gi(en factor doesnot affect the speed of light in a (acuum.

(b) Eut of the listed factors# the speed of light in a medium depends on the wa(elength oflight in that medium.

Question 10.15:

or sound wa(es# the oppler formula for fre!uenc& shift differs slightl& %etween the two

situationsI i source at restL o%ser(er mo(ing# and ii source mo(ingL o%ser(er at rest. 'he

e@act oppler formulas for the case of light wa(es in (acuum are# howe(er# strictl& identical



for these situations. "@plain wh& this should %e so. *ould &ou e@pect the formulas to %e

strictl& identical for the two situations in case of light tra(elling in a medium2

o

Found wa(es can propagate onl& through a medium. 'he two gi(en situations are notscientificall& identical %ecause the motion of an o%ser(er relati(e to a medium is different inthe two situations. ence# the oppler formulas for the two situations cannot %e the same.

n case of light wa(es# sound can tra(el in a (acuum. n a (acuum# the a%o(e two cases areidentical %ecause the speed of light is independent of the motion of the o%ser(er and themotion of the source. *hen light tra(els in a medium# the a%o(e two cases are not identical%ecause the speed of light depends on the wa(elength of the medium.

Question 10.16:

n dou%le?slit e@periment using light of wa(elength -00 nm# the angular width of a fringe

formed on a distant screen is 0.1. *hat is the spacing %etween the two slits2

*a(elength of light used# λ $ -000 nm $ -00 × 10−9 m

Angular width of fringe#

Angular width of a fringe is related to slit spacing d asI

'herefore# the spacing %etween the slits is .

Question 10.17:


(a) n a single slit diffraction e@periment# the width of the slit is made dou%le the original

width. ow does this affect the sie and intensit& of the central diffraction %and2



(b) n what wa& is diffraction from each slit related to the interference pattern in a dou%le?slit

e@periment2

(c) *hen a tin& circular o%stacle is placed in the path of light from a distant source# a %right

spot is seen at the centre of the shadow of the o%stacle. "@plain wh&2

(d) 'wo students are separated %& a + m partition wall in a room 10 m high. f %oth light and

sound wa(es can %end around o%stacles# how is it that the students are una%le to see each

other e(en though the& can con(erse easil&.

(e) ,a& optics is %ased on the assumption that light tra(els in a straight line. iffraction

effects o%ser(ed when light propagates through small apertures;slits or around small

o%stacles dispro(e this assumption. 6et the ra& optics assumption is so commonl& used in

understanding location and se(eral other properties of images in optical instruments. *hat is

the Gustification2

(a) n a single slit diffraction e@periment# if the width of the slit is made dou%le the originalwidth# then the sie of the central diffraction %and reduces to half and the intensit& of thecentral diffraction %and increases up to four times.

(b) 'he interference pattern in a dou%le?slit e@periment is modulated %& diffraction from eachslit. 'he pattern is the result of the interference of the diffracted wa(e from each slit.

(c) *hen a tin& circular o%stacle is placed in the path of light from a distant source# a %right

spot is seen at the centre of the shadow of the o%stacle. 'his is %ecause light wa(es arediffracted from the edge of the circular o%stacle# which interferes constructi(el& at the centreof the shadow. 'his constructi(e interference produces a %right spot.

(d) 8ending of wa(es %& o%stacles %& a large angle is possi%le when the sie of the o%stacle iscompara%le to the wa(elength of the wa(es.

En the one hand# the wa(elength of the light wa(es is too small in comparison to the sie ofthe o%stacle. 'hus# the diffraction angle will %e (er& small. ence# the students are una%le tosee each other. En the other hand# the sie of the wall is compara%le to the wa(elength of thesound wa(es. 'hus# the %ending of the wa(es ta7es place at a large angle. ence# thestudents are a%le to hear each other.

(e) 'he Gustification is that in ordinar& optical instruments# the sie of the aperture in(ol(ed ismuch larger than the wa(elength of the light used.

Question 10.18:



'wo towers on top of two hills are 40 7m apart. 'he line Goining them passes 50 m a%o(e a hill

halfwa& %etween the towers. *hat is the longest wa(elength of radio wa(es# which can %e

sent %etween the towers without apprecia%le diffraction effects2

istance %etween the towers# d $ 40 7m

eight of the line Goining the hills# d $ 50 m.

'hus# the radial spread of the radio wa(es should not e@ceed 50 7m.

Fince the hill is located halfwa& %etween the towers# resnelBs distance can %e o%tained asI

8 6 $ )0 7m $ ) × 104 m

Aperture can %e ta7en asI

& $ d $ 50 m

resnelBs distance is gi(en %& the relation#

*here#

X $ *a(elength of radio wa(es

'herefore# the wa(elength of the radio wa(es is 1).5 cm.

Question 10.19:

A parallel %eam of light of wa(elength 500 nm falls on a narrow slit and the resulting

diffraction pattern is o%ser(ed on a screen 1 m awa&. t is o%ser(ed that the first minimum is

at a distance of ).5 mm from the centre of the screen. ind the width of the slit.

*a(elength of light %eam# λ $ 500 nm $ 500 × 10−9 m

istance of the screen from the slit# C $ 1 m

or first minima# n $ 1

istance %etween the slits $ d



*idth of each slit#

Angle of diffraction is gi(en %& the relation#

ow# each of these infinitesimall& small slit sends ero intensit& in directionθ. ence# thecom%ination of these slits will gi(e ero intensit&.

CA/'",?11 PA< A'P," E ,AA'E A >A''",

Question 11.1:

ind the

(a) ma@imum fre!uenc&# and

(b) minimum wa(elength of Q?ra&s produced %& 30 7M electrons.

/otential of the electrons# 7 $ 30 7M $ 3 × 104 M

ence# energ& of the electrons# E $ 3 × 104 eM

*here#

e $ Charge on an electron $ 1.- × 10−19 C

(a)>a@imum fre!uenc& produced %& the Q?ra&s $ ν

'he energ& of the electrons is gi(en %& the relationI

E $ 'ν

*here#

' $ /lanc7Bs constant $ -.-)- × 10−34 Vs



Question 11.3:

'he photoelectric cut?off (oltage in a certain e@periment is 1.5 M. *hat is the ma@imum 7inetic

energ& of photoelectrons emitted2

/hotoelectric cut?off (oltage# 7 0 $ 1.5 M

'he ma@imum 7inetic energ& of the emitted photoelectrons is gi(en asI

*here#


'herefore# the ma@imum 7inetic energ& of the photoelectrons emitted in the gi(en e@perimentis ).4 × 10−19 V.

Question 11.4:

>onochromatic light of wa(elength -3). nm is produced %& a helium?neon laser. 'he power

emitted is 9.4) m*.

(a) ind the energ& and momentum of each photon in the light %eam#

(b) ow man& photons per second# on the a(erage# arri(e at a target irradiated %& this %eam2

Assume the %eam to ha(e uniform cross?section which is less than the target area# and

(c) ow fast does a h&drogen atom ha(e to tra(el in order to ha(e the same momentum as

that of the photon2

*a(elength of the monochromatic light# λ $ -3). nm $ -3). × 10−9 m

/ower emitted %& the laser# 6 $ 9.4) m* $ 9.4) × 10−3 *

/lanc7Bs constant# ' $ -.-)- × 10−34 Vs


>ass of a h&drogen atom# m $ 1.-- × 10−)+ 7g

(a)'he energ& of each photon is gi(en asI



'he momentum of each photon is gi(en asI

(b)um%er of photons arri(ing per second# at a target irradiated %& the %eam $ n

Assume that the %eam has a uniform cross?section that is less than the target area.

ence# the e!uation for power can %e written asI

(c)>omentum of the h&drogen atom is the same as the momentum of the photon#

>omentum is gi(en asI

*here#

" $ Fpeed of the h&drogen atom



Question 11.6:

n an e@periment on photoelectric effect# the slope of the cut?off (oltage (ersus fre!uenc& of

incident light is found to %e 4.1) × 10−15 M s. Calculate the (alue of /lanc7Bs constant.

'he slope of the cut?off (oltage 7 (ersus fre!uenc& ν of an incident light is gi(en asI

*here#


' $ /lanc7Bs constant

'herefore# the (alue of /lanc7Bs constant is

Question 11.7:

A 100 * sodium lamp radiates energ& uniforml& in all directions. 'he lamp is located at the

centre of a large sphere that a%sor%s all the sodium light which is incident on it. 'he

wa(elength of the sodium light is 59 nm. a *hat is the energ& per photon associated with

the sodium light2 % At what rate are the photons deli(ered to the sphere2

/ower of the sodium lamp# 6 $ 100 *

*a(elength of the emitted sodium light# λ $ 59 nm $ 59 × 10−9 m





(a)'he energ& per photon associated with the sodium light is gi(en asI

(b)um%er of photons deli(ered to the sphere $ n

'he e!uation for power can %e written asI

'herefore# e(er& second# photons are deli(ered to the sphere.

Question 11.8:

'he threshold fre!uenc& for a certain metal is 3.3 × 1014 . f light of fre!uenc& .) × 1014

is incident on the metal# predict the cutoff (oltage for the photoelectric emission.

'hreshold fre!uenc& of the metal#

re!uenc& of light incident on the metal#

Charge on an electron# e $ 1.- × 10−19

C


Cut?off (oltage for the photoelectric emission from the metal $

'he e!uation for the cut?off energ& is gi(en asI



'herefore# the cut?off (oltage for the photoelectric emission is

Question 11.9:

'he wor7 function for a certain metal is 4.) eM. *ill this metal gi(e photoelectric emission for

incident radiation of wa(elength 330 nm2

o

*or7 function of the metal#



*a(elength of the incident radiation# λ $ 330 nm $ 330 × 10−9 m


'he energ& of the incident photon is gi(en asI

t can %e o%ser(ed that the energ& of the incident radiation is less than the wor7 function ofthe metal. ence# no photoelectric emission will ta7e place.



Question 11.10:

<ight of fre!uenc& +.)1 × 1014 is incident on a metal surface. "lectrons with a ma@imum

speed of -.0 × 105 m;s are eGected from the surface. *hat is the threshold fre!uenc& for

photoemission of electrons2

re!uenc& of the incident photon#

>a@imum speed of the electrons# " $ -.0 × 105 m;s


>ass of an electron# m $ 9.1 × 10−31 7g

or threshold fre!uenc& ν 0# the relation for 7inetic energ& is written asI

'herefore# the threshold fre!uenc& for the photoemission of electrons is

Question 11.11:

<ight of wa(elength 4 nm is produced %& an argon laser which is used in the photoelectric

effect. *hen light from this spectral line is incident on the emitter# the stopping cut?off

potential of photoelectrons is 0.3 M. ind the wor7 function of the material from which the

emitter is made.

*a(elength of light produced %& the argon laser# X $ 4 nm

$ 4 × 10−9 m

Ftopping potential of the photoelectrons# 7 0 $ 0.3 M

1eM $ 1.- × 10−19 V



∴ 7 0 $

/lanc7Bs constant# ' $ -.- × 10−34 Vs



rom "insteinBs photoelectric effect# we ha(e the relation in(ol(ing the wor7 function Φ0 of thematerial of the emitter asI

'herefore# the material with which the emitter is made has the wor7 function of ).1- eM.

*a(elength of light produced %& the argon laser# X $ 4 nm

$ 4 × 10−9 m

Ftopping potential of the photoelectrons# 7 0 $ 0.3 M

1eM $ 1.- × 10−19 V

∴ 7 0 $




rom "insteinBs photoelectric effect# we ha(e the relation in(ol(ing the wor7 function Φ0 of thematerial of the emitter asI



'herefore# the material with which the emitter is made has the wor7 function of ).1- eM.

Question 11.12:

Calculate the

(a) momentum# and

(b) de 8roglie wa(elength of the electrons accelerated through a potential difference of 5- M.

/otential difference# 7 $ 5- M


>ass of an electron# m $ 9.1 × 10−31

7g


(a) At e!uili%rium# the 7inetic energ& of each electron is e!ual to the accelerating potential#i.e.# we can write the relation for (elocit& " of each electron asI



'he momentum of each accelerated electron is gi(en asI

p $ m"

$ 9.1 × 10−31 × 4.44 × 10-

$ 4.04 × 10−)4 7g m s−1

'herefore# the momentum of each electron is 4.04 × 10−)4 7g m s−1.

(b) e 8roglie wa(elength of an electron accelerating through a potential 7 # is gi(en %& therelationI

'herefore# the de 8roglie wa(elength of each electron is 0.1-39 nm.

Question 11.13:

*hat is the

(a) momentum#

(b) speed# and

(c) de 8roglie wa(elength of an electron with 7inetic energ& of 1)0 eM.

Zinetic energ& of the electron# E k $ 1)0 eM


>ass of an electron# m $ 9.1 × 10−31 7g


(a) or the electron# we can write the relation for 7inetic energ& asI

*here#



" $ Fpeed of the electron

>omentum of the electron# p $ m"

$ 9.1 × 10−31 × -.49- × 10-

$ 5.91 × 10

−)4

7g m s

−1

'herefore# the momentum of the electron is 5.91 × 10−)4 7g m s−1.

(b) Fpeed of the electron# " $ -.49- × 10- m;s

(c) e 8roglie wa(elength of an electron ha(ing a momentum p# is gi(en asI

'herefore# the de 8roglie wa(elength of the electron is 0.11) nm.

Question 11.14:

'he wa(elength of light from the spectral emission line of sodium is 59 nm. ind the 7inetic

energ& at which

(a) an electron# and

(b) a neutron# would ha(e the same de 8roglie wa(elength.



*a(elength of light of a sodium line# λ $ 59 nm $ 59 × 10−9 m

>ass of an electron# me$ 9.1 × 10−31 7g

>ass of a neutron# mn$ 1.-- × 10−)+ 7g


(a) or the 7inetic energ& # of an electron accelerating with a (elocit& " # we ha(e therelationI

*e ha(e the relation for de 8roglie wa(elength asI

Fu%stituting e!uation ) in e!uation 1# we get the relationI

ence# the 7inetic energ& of the electron is -.9 × 10−)5 V or 4.31 :eM.

(b) Psing e!uation 3# we can write the relation for the 7inetic energ& of the neutron asI



(c)>ass of the dust particle# m $ 1 × 10−9 7g

Fpeed of the dust particle# " $ ).) m;s

e 8roglie wa(elength of the dust particle is gi(en %& the relationI

Question 11.16:

An electron and a photon each ha(e a wa(elength of 1.00 nm. ind

(a) their momenta#

(b) the energ& of the photon# and

(c) the 7inetic energ& of electron.

*a(elength of an electron

$ 1 × 10−9 m

/lanc7Bs constant# ' $ -.-3 × 10−34 Vs

(a) 'he momentum of an elementar& particle is gi(en %& de 8roglie relationI

t is clear that momentum depends onl& on the wa(elength of the particle. Fince thewa(elengths of an electron and a photon are e!ual# %oth ha(e an e!ual momentum.



(b) 'he energ& of a photon is gi(en %& the relationI

*here#


'herefore# the energ& of the photon is 1.)43 7eM.

(c) 'he 7inetic energ& of an electron ha(ing momentum p#is gi(en %& the relationI

*here#

m $ >ass of the electron $ 9.1 × 10−31 7g

p $ -.-3 × 10−)5 7g m s−1

ence# the 7inetic energ& of the electron is 1.51 eM.

Question 11.17:

(a) or what 7inetic energ& of a neutron will the associated de 8roglie wa(elength %e 1.40 ×

10−10 m2

(b) Also find the de 8roglie wa(elength of a neutron# in thermal e!uili%rium with matter#

ha(ing an a(erage 7inetic energ& of 3;) k4 at 300 Z.

(a) e 8roglie wa(elength of the neutron# λ $ 1.40 × 10−10 m



'herefore# the de 8roglie wa(elength of the neutron is 0.14- nm.

Question 11.18:

Fhow that the wa(elength of electromagnetic radiation is e!ual to the de 8roglie wa(elength of

its !uantum photon.

'he momentum of a photon ha(ing energ& 'ν is gi(en asI

*here#

X $ *a(elength of the electromagnetic radiation

+ $ Fpeed of light


e 8roglie wa(elength of the photon is gi(en asI



*here#

m $ >ass of the photon

" $ Melocit& of the photon

ence# it can %e inferred from e!uations i and ii that the wa(elength of theelectromagnetic radiation is e!ual to the de 8roglie wa(elength of the photon.

Question 11.19:

*hat is the de 8roglie wa(elength of a nitrogen molecule in air at 300 Z2 Assume that the

molecule is mo(ing with the root?mean s!uare speed of molecules at this temperature.

Atomic mass of nitrogen $ 14.00+- u

'emperature of the nitrogen molecule# 4 $ 300 Z

Atomic mass of nitrogen $ 14.00+- u

ence# mass of the nitrogen molecule# m $ ) × 14.00+- $ ).015) u

8ut 1 u $ 1.-- × 10−)+ 7g

∴m $ ).015) ×1.-- × 10−)+ 7g

/lanc7Bs constant# ' $ -.-3 × 10−34 Vs

8oltmann constant# k $ 1.3 × 10−)3 V Z−1

*e ha(e the e@pression that relates mean 7inetic energ& of the nitrogen molecule

with the root mean s!uare speed asI



ence# the de 8roglie wa(elength of the nitrogen molecule is gi(en asI

'herefore# the de 8roglie wa(elength of the nitrogen molecule is 0.0) nm.

Question 11.20:

(a) "stimate the speed with which electrons emitted from a heated emitter of an e(acuated

tu%e impinge on the collector maintained at a potential difference of 500 M with respect to the

emitter. gnore the small initial speeds of the electrons. 'he spe+i)i+ +'&rge of the electron#i.e.# its eMm is gi(en to %e 1.+- × 1011 C 7g−1.

(b) Pse the same formula &ou emplo& in a to o%tain electron speed for an collector potential

of 10 >M. o &ou see what is wrong2 n what wa& is the formula to %e modified2

(a)/otential difference across the e(acuated tu%e# 7 $ 500 M

Fpecific charge of an electron# eMm $ 1.+- × 1011 C 7g−1

'he speed of each emitted electron is gi(en %& the relation for 7inetic energ& asI



'herefore# the speed of each emitted electron is

(b)/otential of the anode# 7 $ 10 >M $ 10 × 10- M

'he speed of each electron is gi(en asI

'his result is wrong %ecause nothing can mo(e faster than light. n the a%o(e formula# thee@pression m" ) ;) for energ& can onl& %e used in the non?relati(istic limit# i.e.# for " + .

or (er& high speed pro%lems# relati(istic e!uations must %e considered for sol(ing them. nthe relati(istic limit# the total energ& is gi(en asI

E $ m+ )

*here#

m $ ,elati(istic mass

m0 $ >ass of the particle at rest

Zinetic energ& is gi(en asI

$ m+ ) − m0+ )



ence# e!uation 1 reduces toI

'herefore# the radius of the circular path is )).+ cm.

(b) "nerg& of the electron %eam# E $ )0 >eM

'he energ& of the electron is gi(en asI

'his result is incorrect %ecause nothing can mo(e faster than light. n the a%o(e formula# thee@pression m" ) ;) for energ& can onl& %e used in the non?relati(istic limit# i.e.# for " +

*hen (er& high speeds are concerned# the relati(istic domain comes into consideration.

n the relati(istic domain# mass is gi(en asI

*here#

$ >ass of the particle at rest



ence# the radius of the circular path is gi(en asI

Question 11.22:

An electron gun with its collector at a potential of 100 M fires out electrons in a spherical %ul%

containing h&drogen gas at low pressure ∼10−) mm of g. A magnetic field of ).3 × 10−4 '

cur(es the path of the electrons in a circular or%it of radius 1).0 cm. 'he path can %e (iewed

%ecause the gas ions in the path focus the %eam %& attracting electrons# and emitting light %&

electron captureL this method is 7nown as the fine %eam tu%eB method. etermine e ;m fromthe data.

/otential of an anode# 7 $ 100 M

>agnetic field e@perienced %& the electrons# $ ).3 × 10−4 '

,adius of the circular or%it r $ 1).0 cm $ 1).0 × 10−) m

>ass of each electron $ m

Charge on each electron $ e

Melocit& of each electron $ "

'he energ& of each electron is e!ual to its 7inetic energ&# i.e.#

t is the magnetic field# due to its %ending nature# that pro(ides the centripetal force

for the %eam. ence# we can writeI

Centripetal force $ >agnetic force



/utting the (alue of " in e!uation 1# we getI

'herefore# the specific charge ratio eMm is

Question 11.23:

(a) An Q?ra& tu%e produces a continuous spectrum of radiation with its short wa(elength end

at 0.45 Y. *hat is the ma@imum energ& of a photon in the radiation2

b) rom &our answer to a# guess what order of accelerating (oltage for electrons is

re!uired in such a tu%e2

(a) *a(elength produced %& an Q?ra& tu%e#



'he ma@imum energ& of a photon is gi(en asI

'herefore# the ma@imum energ& of an Q?ra& photon is )+.- 7eM.



(b) Accelerating (oltage pro(ides energ& to the electrons for producing Q?ra&s. 'o get an Q?ra&of )+.- 7eM# the incident electrons must possess at least )+.- 7eM of 7inetic electric energ&.ence# an accelerating (oltage of the order of 30 7eM is re!uired for producing Q?ra&s.

Question 11.24:

n an accelerator e@periment on high?energ& collisions of electrons with positrons# a certain

e(ent is interpreted as annihilation of an electron?positron pair of total energ& 10.) 8eM into

two γ ?ra&s of e!ual energ&. *hat is the wa(elength associated with each γ ?ra&2 18eM $ 109

eM

'otal energ& of two γ ?ra&sI

E $ 10. ) 8eM

$ 10.) × 109 eM

$ 10.) × 109 × 1.- × 10−10 V

ence# the energ& of each γ ?ra&I

/lanc7Bs constant#

Fpeed of light#

"nerg& is related to wa(elength asI

'herefore# the wa(elength associated with each γ ?ra& is

Question 11.25:



"stimating the following two num%ers should %e interesting. 'he first num%er will tell &ou wh&

radio engineers do not need to worr& much a%out photons` 'he second num%er tells &ou wh&

our e&e can ne(er count photonsB# e(en in %arel& detecta%le light.

(a) 'he num%er of photons emitted per second %& a >edium wa(e transmitter of 10 7*

power# emitting radiowa(es of wa(elength 500 m.

(b) 'he num%er of photons entering the pupil of our e&e per second corresponding to theminimum intensit& of white light that we humans can percei(e ∼10−10 * m−). 'a7e the areaof the pupil to %e a%out 0.4 cm)# and the a(erage fre!uenc& of white light to %e a%out - × 1014

. (a) /ower of the medium wa(e transmitter# 6 $ 10 7* $ 104 * $ 104 V;s

ence# energ& emitted %& the transmitter per second# E $ 104

*a(elength of the radio wa(e# λ $ 500 m

'he energ& of the wa(e is gi(en asI

*here#



<et n %e the num%er of photons emitted %& the transmitter.

∴nE 1 $ E

'he energ& E 1 of a radio photon is (er& less# %ut the num%er of photons n emitted persecond in a radio wa(e is (er& large.



Question 11.26:

Pltra(iolet light of wa(elength ))+1 Y from a 100 * mercur& source irradiates a photo?cell

made of mol&%denum metal. f the stopping potential is −1.3 M# estimate the wor7 function of

the metal. ow would the photo?cell respond to a high intensit& ∼105 * m−) red light of

wa(elength -3) Y produced %& a e?e laser2

*a(elength of ultra(iolet light# λ $ ))+1 Y $ ))+1 × 10−10 m

Ftopping potential of the metal# 7 0 $ 1.3 M

/lanc7Bs constant# ' $ -.- × 10−34 V


*or7 function of the metal $

re!uenc& of light $ ν

*e ha(e the photo?energ& relation from the photoelectric effect asI

$ 'ν − e7 0

<et ν 0 %e the threshold fre!uenc& of the metal.

∴ $ 'ν 0

*a(elength of red light# $ -3) × 10−10 m



∴re!uenc& of red light#

Fince ν 0[ ν r # the photocell will not respond to the red light produced %& the laser.

Question 11.27:

>onochromatic radiation of wa(elength -40.) nm 1nm $ 10−9 m from a neon lamp irradiates

photosensiti(e material made of caesium on tungsten. 'he stopping (oltage is measured to %e

0.54 M. 'he source is replaced %& an iron source and its 4)+.) nm line irradiates the same

photo?cell. /redict the new stopping (oltage.

*a(elength of the monochromatic radiation# λ $ -40.) nm

$ -40.) × 10−9 m

Ftopping potential of the neon lamp# 7 0 $ 0.54 M



<et %e the wor7 function and ν %e the fre!uenc& of emitted light.

*e ha(e the photo?energ& relation from the photoelectric effect asI

e7 0 $ 'ν −

*a(elength of the radiation emitted from an iron source# λk $ 4)+.) nm

= 4)+.) × 10−9 m



<et %e the new stopping potential. ence# photo?energ& is gi(en asI

ence# the new stopping potential is 1.50 eM.

Question 11.28:

A mercur& lamp is a con(enient source for stud&ing fre!uenc& dependence of photoelectric

emission# since it gi(es a num%er of spectral lines ranging from the PM to the red end of the

(isi%le spectrum. n our e@periment with ru%idium photo?cell# the following lines from a

mercur& source were usedI

λ1 $ 3-50 Y# λ)$ 404+ Y# λ3$ 435 Y# λ4$ 54-1 Y# λ5$ -90+ Y#

'he stopping (oltages# respecti(el&# were measured to %eI

7 01 $ 1.) M# 7 0) $ 0.95 M# 7 03 $ 0.+4 M# 7 04 $ 0.1- M# 7 05 $ 0 M

etermine the (alue of /lanc7Bs constant '# the threshold fre!uenc& and wor7 function for the

material.

NB%te: 6ou will notice that to get ' from the data# &ou will need to 7now e which &ou can ta7e

to %e 1.- × 10−19 C. "@periments of this 7ind on a# <i# Z# etc. were performed %& >illi7an#

who# using his own (alue of e from the oil?drop e@periment confirmed "insteinBs

photoelectric e!uation and at the same time ga(e an independent estimate of the (alue of '.O

"insteinBs photoelectric e!uation is gi(en asI

e7 0 $ 'ν −



*here#

7 0 $ Ftopping potential


e $ Charge on an electron

ν $ re!uenc& of radiation

$ *or7 function of a material

t can %e concluded from e!uation 1 that potential 7 0 is directl& proportional to fre!uenc& ν .

re!uenc& is also gi(en %& the relationI

'his relation can %e used to o%tain the fre!uencies of the (arious lines of the gi(enwa(elengths.

'he gi(en !uantities can %e listed in ta%ular form asI

F$e/:e&c; < 1=1' H> .)19 +.41) -.4 5.493 4.343

S"!%%i&g %!"e&"ia V = 1.) 0.95 0.+4 0.1- 0

'he following figure shows a graph %etween ν and 7 0.



t can %e o%ser(ed that the o%tained cur(e is a straight line. t intersects the ν ?a@is at 5 × 1014

# which is the threshold fre!uenc& ν 0 of the material. /oint corresponds to a fre!uenc&less than the threshold fre!uenc&. ence# there is no photoelectric emission for the λ5 line# andtherefore# no stopping (oltage is re!uired to stop the current.

Flope of the straight line $

rom e!uation 1# the slope can %e written asI

'he wor7 function of the metal is gi(en asI

$ ' ν0

$ -.5+3 × 10−34 × 5 × 1014

$ 3.)- × 10−19 V

$ ).054 eM

Question 11.29:

'he wor7 function for the following metals is gi(enI



aI ).+5 eML ZI ).30 eML >oI 4.1+ eML iI 5.15 eM. *hich of these metals will not gi(e

photoelectric emission for a radiation of wa(elength 3300 Y from a e?Cd laser placed 1 m

awa& from the photocell2 *hat happens if the laser is %rought nearer and placed 50 cm awa&2

>o and i will not show photoelectric emission in %oth cases

*a(elength for a radiation# λ $ 3300 Y $ 3300 × 10−10 m



'he energ& of incident radiation is gi(en asI

t can %e o%ser(ed that the energ& of the incident radiation is greater than the wor7 functionof a and Z onl&. t is less for >o and i. ence# >o and i will not show photoelectricemission.

f the source of light is %rought near the photocells and placed 50 cm awa& from them# thenthe intensit& of radiation will increase. 'his does not affect the energ& of the radiation. ence#the result will %e the same as %efore. owe(er# the photoelectrons emitted from a and Z willincrease in proportion to intensit&.

Question 11.30:

<ight of intensit& 10−5 * m−) falls on a sodium photo?cell of surface area ) cm). Assuming that

the top 5 la&ers of sodium a%sor% the incident energ&# estimate time re!uired for photoelectric

emission in the wa(e?picture of radiation. 'he wor7 function for the metal is gi(en to %e a%out

) eM. *hat is the implication of &our answer2

ntensit& of incident light# 3 $ 10−5 * m−)

Furface area of a sodium photocell# A $ ) cm) $ ) × 10−4 m)

ncident power of the light# 6 = 3 × A

$ 10−5 × ) × 10−4

$ ) × 10−9 *



*or7 function of the metal# $ ) eM

$ ) × 1.- × 10−19

$ 3.) × 10−19 V

um%er of la&ers of sodium that a%sor%s the incident energ&# n $ 5

*e 7now that the effecti(e atomic area of a sodium atom# Ae is 10−)0 m).

ence# the num%er of conduction electrons in n la&ers is gi(en asI

'he incident power is uniforml& a%sor%ed %& all the electrons continuousl&. ence# the amountof energ& a%sor%ed per second per electron isI

'ime re!uired for photoelectric emissionI

'he time re!uired for the photoelectric emission is nearl& half a &ear# which is not practical.ence# the wa(e picture is in disagreement with the gi(en e@periment.

Question 11.31:

Cr&stal diffraction e@periments can %e performed using Q?ra&s# or electrons accelerated

through appropriate (oltage. *hich pro%e has greater energ&2 or !uantitati(e comparison#

ta7e the wa(elength of the pro%e e!ual to 1 Y# which is of the order of inter?atomic spacing in

the lattice me$ 9.11 × 10−31 7g.

An Q?ra& pro%e has a greater energ& than an electron pro%e for the same wa(elength.



*a(elength of light emitted from the pro%e# λ $ 1 Y $ 10−10 m

>ass of an electron# me $ 9.11 × 10−31 7g



'he 7inetic energ& of the electron is gi(en asI

*here#

" $ Melocit& of the electron

me" $ >omentum p of the electron

According to the de 8roglie principle# the de 8roglie wa(elength is gi(en asI

"nerg& of a photon#

ence# a photon has a greater energ& than an electron for the same wa(elength.

Question 11.32:



(a) E%tain the de 8roglie wa(elength of a neutron of 7inetic energ& 150 eM. As &ou ha(e seen

in "@ercise 11.31# an electron %eam of this energ& is suita%le for cr&stal diffraction

e@periments. *ould a neutron %eam of the same energ& %e e!uall& suita%le2 "@plain. mn$

1.-+5 × 10−)+ 7g

(b) E%tain the de 8roglie wa(elength associated with thermal neutrons at room temperature

)+ C. ence e@plain wh& a fast neutron %eam needs to %e thermalised with the

en(ironment %efore it can %e used for neutron diffraction e@periments.

(a) e 8roglie wa(elength $ L neutron is not suita%le for the diffractione@periment

Zinetic energ& of the neutron# $ 150 eM

$ 150 × 1.- × 10−19

$ ).4 × 10−1+ V

>ass of a neutron# mn $ 1.-+5 × 10−)+ 7g

'he 7inetic energ& of the neutron is gi(en %& the relationI

*here#

" $ Melocit& of the neutron

mn" $ >omentum of the neutron

e?8roglie wa(elength of the neutron is gi(en asI



t is gi(en in the pre(ious pro%lem that the inter?atomic spacing of a cr&stal is a%out 1 Y# i.e.#10−10 m. ence# the inter?atomic spacing is a%out a hundred times greater. ence# a neutron%eam of energ&150 eM is not suita%le for diffraction e@periments.

(b) e 8roglie wa(elength $

,oom temperature# 4 $ )+°C $ )+ H )+3 $ 300 Z

'he a(erage 7inetic energ& of the neutron is gi(en asI

*here#

k $ 8oltmann constant $ 1.3 × 10−)3 V mol−1 Z−1

'he wa(elength of the neutron is gi(en asI

'his wa(elength is compara%le to the inter?atomic spacing of a cr&stal. ence# the high?energ&

neutron %eam should first %e thermalised# %efore using it for diffraction.

Question 11.33:

An electron microscope uses electrons accelerated %& a (oltage of 50 7M. etermine the de

8roglie wa(elength associated with the electrons. f other factors such as numerical aperture#

etc. are ta7en to %e roughl& the same# how does the resol(ing power of an electron

microscope compare with that of an optical microscope which uses &ellow light2

"lectrons are accelerated %& a (oltage# 7 $ 50 7M $ 50 × 103 M



*a(elength of &ellow light $ 5.9 × 10−+ m

'he 7inetic energ& of the electron is gi(en asI



E $ e7

$ 1.- × 10−19 × 50 × 103

$ × 10−15 V

e 8roglie wa(elength is gi(en %& the relationI

'his wa(elength is nearl& 105 times less than the wa(elength of &ellow light.

'he resol(ing power of a microscope is in(ersel& proportional to the wa(elength of light used.'hus# the resol(ing power of an electron microscope is nearl& 105 times that of an opticalmicroscope.

Question 11.34:

'he wa(elength of a pro%e is roughl& a measure of the sie of a structure that it can pro%e in

some detail. 'he !uar7 structure of protons and neutrons appears at the minute length?scale

of 10−15 m or less. 'his structure was first pro%ed in earl& 19+0Bs using high energ& electron

%eams produced %& a linear accelerator at Ftanford# PFA. =uess what might ha(e %een the

order of energ& of these electron %eams. ,est mass energ& of electron $ 0.511 >eM.

*a(elength of a proton or a neutron# λ 10−15 m

,est mass energ& of an electronI

m0+ ) $ 0.511 >eM

$ 0.511 × 10- × 1.- × 10−19

$ 0.1+- × 10−13 V

/lanc7Bs constant# ' $ -.- × 10

−34

Vs


'he momentum of a proton or a neutron is gi(en asI



'he relati(istic relation for energ& E is gi(en asI

'hus# the electron energ& emitted from the accelerator at Ftanford# PFA might %e of the orderof 1.)4 8eM.

Question 11.35:

ind the t&pical de 8roglie wa(elength associated with a e atom in helium gas at room

temperature )+ C and 1 atm pressureL and compare it with the mean separation %etween

two atoms under these conditions.

e 8roglie wa(elength associated with e atom $

,oom temperature# 4 $ )+°C $ )+ H )+3 $ 300 Z

Atmospheric pressure# 6 $ 1 atm $ 1.01 × 105 /a

Atomic weight of a e atom $ 4

A(ogadroBs num%er# A $ -.0)3 × 10)3

8oltmann constant# k $ 1.3 × 10−)3 V mol−1 Z−1

A(erage energ& of a gas at temperature 4 #is gi(en asI



e 8roglie wa(elength is gi(en %& the relationI

*here#

m $ >ass of a e atom

*e ha(e the ideal gas formulaI

67 = 94

67 = kB4

*here#

7 $ Molume of the gas

B $ um%er of moles of the gas

>ean separation %etween two atoms of the gas is gi(en %& the relationI



ence# the mean separation %etween the atoms is much greater than the de 8rogliewa(elength.

Question 11.36:

Compute the t&pical de 8roglie wa(elength of an electron in a metal at )+ C and compare it

with the mean separation %etween two electrons in a metal which is gi(en to %e a%out ) ×

10−10 m.

NB%te: "@ercises 11.35 and 11.3- re(eal that while the wa(e?pac7ets associated with gaseous

molecules under ordinar& conditions are non?o(erlapping# the electron wa(e?pac7ets in a

metal strongl& o(erlap with one another. 'his suggests that whereas molecules in an ordinar&

gas can %e distinguished apart# electrons in a metal cannot %e distinguished apart from one

another. 'his indistinguishi%ilit& has man& fundamental implications which &ou will e@plore in

more ad(anced /h&sics courses.O

'emperature# 4 $ )+°C $ )+ H )+3 $ 300 Z

>ean separation %etween two electrons# r $ ) × 10−10 m

e 8roglie wa(elength of an electron is gi(en asI

*here#


m $ >ass of an electron $ 9.11 × 10−31 7g

k $ 8oltmann constant $ 1.3 × 10−)3 V mol−1 Z−1



ence# the de 8roglie wa(elength is much greater than the gi(en inter?electron separation.

Question 11.37:


(a) Suar7s inside protons and neutrons are thought to carr& fractional charges NH);3e L

−1;3eO. *h& do the& not show up in >illi7anBs oil?drop e@periment2

(b) *hat is so special a%out the com%ination eMm2 *h& do we not simpl& tal7 of e and m

separatel&2

(c) *h& should gases %e insulators at ordinar& pressures and start conducting at (er& low

pressures2

(d) "(er& metal has a definite wor7 function. *h& do all photoelectrons not come out with the

same energ& if incident radiation is monochromatic2 *h& is there an energ& distri%ution of

photoelectrons2

(e) 'he energ& and momentum of an electron are related to the fre!uenc& and wa(elength of

the associated matter wa(e %& the relationsI

E = 'ν # p $

8ut while the (alue of λ is ph&sicall& significant# the (alue of ν and therefore# the (alue of the

phase speed ν λ has no ph&sical significance. *h&2

(a) Suar7s inside protons and neutrons carr& fractional charges. 'his is %ecause nuclear forceincreases e@tremel& if the& are pulled apart. 'herefore# fractional charges ma& e@ist in natureLo%ser(a%le charges are still the integral multiple of an electrical charge.

(b) 'he %asic relations for electric field and magnetic field are

.



'hese relations include e electric charge# " (elocit&# m mass# 7 potential# r radius#and magnetic field. 'hese relations gi(e the (alue of (elocit& of an electron as

and

t can %e o%ser(ed from these relations that the d&namics of an electron is determined not %&e and m separatel&# %ut %& the ratio e;m.

(c) At atmospheric pressure# the ions of gases ha(e no chance of reaching their respecti(eelectrons %ecause of collision and recom%ination with other gas molecules. ence# gases areinsulators at atmospheric pressure. At low pressures# ions ha(e a chance of reaching theirrespecti(e electrodes and constitute a current. ence# the& conduct electricit& at thesepressures.

(d) 'he wor7 function of a metal is the minimum energ& re!uired for a conduction electron toget out of the metal surface. All the electrons in an atom do not ha(e the same energ& le(el.*hen a ra& ha(ing some photon energ& is incident on a metal surface# the electrons come outfrom different le(els with different energies. ence# these emitted electrons show differentenerg& distri%utions.

(e) 'he a%solute (alue of energ& of a particle is ar%itrar& within the additi(e constant. ence#wa(elength λ is significant# %ut the fre!uenc& ν associated with an electron has no directph&sical significance.

'herefore# the product ν λphase speedhas no ph&sical significance.

=roup speed is gi(en asI

'his !uantit& has a ph&sical meaning.



C+a%"e$ 12 ? A"!7

Choose the correct alternati(e from the clues gi(en at the end of the each statementI

(a) 'he sie of the atom in 'homsonBs model is .......... the atomic sie in ,utherfordBs model.

much greater than;no different from;much less than.

(b) n the ground state of .......... electrons are in sta%le e!uili%rium# while in ..........

electrons alwa&s e@perience a net force.

'homsonBs model; ,utherfordBs model.

(c) A +!&ssi+&! atom %ased on .......... is doomed to collapse.


(d) An atom has a nearl& continuous mass distri%ution in a .......... %ut has a highl& non?

uniform mass distri%ution in ..........


(e) 'he positi(el& charged part of the atom possesses most of the mass in ..........

,utherfordBs model;%oth the models.

a) 'he sies of the atoms ta7en in 'homsonBs model and ,utherfordBs model ha(e the sameorder of magnitude.

(b) n the ground state of 'homsonBs model# the electrons are in sta%le e!uili%rium. owe(er#in ,utherfordBs model# the electrons alwa&s e@perience a net force.

(c) A +!&ssi+&! atom %ased on ,utherfordBs model is doomed to collapse.

(d) An atom has a nearl& continuous mass distri%ution in 'homsonBs model# %ut has a highl&

non?uniform mass distri%ution in ,utherfordBs model.

(e) 'he positi(el& charged part of the atom possesses most of the mass in %oth the models.

Question 12.2:



Fuppose &ou are gi(en a chance to repeat the alpha?particle scattering e@periment using a

thin sheet of solid h&drogen in place of the gold foil. &drogen is a solid at temperatures

%elow 14 Z. *hat results do &ou e@pect2

n the alpha?particle scattering e@periment# if a thin sheet of solid h&drogen is used in place of

a gold foil# then the scattering angle would not %e large enough. 'his is %ecause the mass ofh&drogen 1.-+ × 10−)+ 7g is less than the mass of incident α−particles -.-4 × 10−)+ 7g.'hus# the mass of the scattering particle is more than the target nucleus h&drogen. As aresult# the α−particles would not %ounce %ac7 if solid h&drogen is used in the α?particlescattering e@periment.

Question 12.3:

*hat is the shortest wa(elength present in the /aschen series of spectral lines2

,&d%ergBs formula is gi(en asI

*here#



Nn1 and n) are integers

'he shortest wa(elength present in the /aschen series of the spectral lines is gi(en for (aluesn1 $ 3 and n) $ .

Question 12.4:

A difference of ).3 eM separates two energ& le(els in an atom. *hat is the fre!uenc& of

radiation emitted when the atom ma7es a transition from the upper le(el to the lower le(el2

Feparation of two energ& le(els in an atom#



E $ ).3 eM

$ ).3 × 1.- × 10−19

$ 3.- × 10−19 V

<et ν %e the fre!uenc& of radiation emitted when the atom transits from the upper le(el to thelower le(el.

*e ha(e the relation for energ& asI

E $ '"

*here#


ence# the fre!uenc& of the radiation is 5.- × 1014 .

Question 12.5:

'he ground state energ& of h&drogen atom is −13.- eM. *hat are the 7inetic and potentialenergies of the electron in this state2

=round state energ& of h&drogen atom# E $ − 13.- eM

'his is the total energ& of a h&drogen atom. Zinetic energ& is e!ual to the negati(e of the totalenerg&.

Zinetic energ& $ − E $ − − 13.- $ 13.- eM

/otential energ& is e!ual to the negati(e of two times of 7inetic energ&.

/otential energ& = − ) × 13.- $ − )+ .) eM

Question 12.6:

A h&drogen atom initiall& in the ground le(el a%sor%s a photon# which e@cites it to the n $ 4

le(el. etermine the wa(elength and fre!uenc& of the photon.



or ground le(el# n1 $ 1

<et E 1 %e the energ& of this le(el. t is 7nown that E 1 is related with n1 asI

'he atom is e@cited to a higher le(el# n) $ 4.

<et E ) %e the energ& of this le(el.

'he amount of energ& a%sor%ed %& the photon is gi(en asI

E $ E ) − E 1

or a photon of wa(elength λ# the e@pression of energ& is written asI

*here#





And# fre!uenc& of a photon is gi(en %& the relation#

ence# the wa(elength of the photon is 9+ nm while the fre!uenc& is 3.1 × 1015 .

Question 12.7:

a Psing the 8ohrBs model calculate the speed of the electron in a h&drogen atom in the n $

1# )# and 3 le(els. % Calculate the or%ital period in each of these le(els.

(a) <et ν 1 %e the or%ital speed of the electron in a h&drogen atom in the ground state le(el# n1 $ 1. or charge eI of an electron# ν 1 is gi(en %& the relation#

*here#

e $ 1.- × 10−19 C

∈0 $ /ermitti(it& of free space $ .5 × 10 −1) −1 C) m−)

' $ /lanc7Bs constant $ -.-) × 10−34 Vs

or le(el n) $ )# we can write the relation for the corresponding or%ital speed asI



And# for n3 $ 3# we can write the relation for the corresponding or%ital speed asI

ence# the speed of the electron in a h&drogen atom in n $ 1# n$)# and n$3 is ).1 × 10- m;s# 1.09 × 10- m;s# +.)+ × 105 m;s respecti(el&.

(b) <et 4 1 %e the or%ital period of the electron when it is in le(el n1 $ 1.

Er%ital period is related to or%ital speed asI

*here#

r 1 $ ,adius of the or%it



∈0 $ /ermitti(it& of free space $ .5 × 10−1) −1 C) m−)

m $ >ass of an electron $ 9.1 × 10−31 7g



or le(el n) $ )# we can write the period asI

*here#

r ) $ ,adius of the electron in n) $ )

And# for le(el n3 $ 3# we can write the period asI

*here#

r 3 $ ,adius of the electron in n3 $ 3



ence# the or%ital period in each of these le(els is 1.5) × 10−1- s# 1.)) × 10−15 s# and 4.1) ×10−15 s respecti(el&

Question 12.8:

'he radius of the innermost electron or%it of a h&drogen atom is 5.3 ×10−11 m. *hat are the

radii of the n $ ) and n $3 or%its2

'he radius of the innermost or%it of a h&drogen atom# r 1 $ 5.3 × 10−11 m.

<et r ) %e the radius of the or%it at n $ ). t is related to the radius of the innermost or%it asI

or n $ 3# we can write the corresponding electron radius asI

ence# the radii of an electron for n $ ) and n $ 3 or%its are ).1) × 10−10 m and 4.++ × 10−10 m respecti(el&.

Question 12.9:

A 1).5 eM electron %eam is used to %om%ard gaseous h&drogen at room temperature. *hat

series of wa(elengths will %e emitted2

t is gi(en that the energ& of the electron %eam used to %om%ard gaseous h&drogen at room

temperature is 1).5 eM. Also# the energ& of the gaseous h&drogen in its ground state at roomtemperature is −13.- eM.

*hen gaseous h&drogen is %om%arded with an electron %eam# the energ& of the gaseoush&drogen %ecomes −13.- H 1).5 eM i.e.# −1.1 eM.

Er%ital energ& is related to or%it le(el nI asI



or n $ 3#

'his energ& is appro@imatel& e!ual to the energ& of gaseous h&drogen. t can %e concludedthat the electron has Gumped from n $ 1 to n $ 3 le(el.

uring its de?e@citation# the electrons can Gump from n $ 3 to n $ 1 directl&# which forms aline of the <&man series of the h&drogen spectrum.

*e ha(e the relation for wa(e num%er for <&man series asI

*here#

9& $ ,&d%erg constant $ 1.09+ × 10+ m−1

X= *a(elength of radiation emitted %& the transition of the electron

or n $ 3# we can o%tain λasI

f the electron Gumps from n $ ) to n $ 1# then the wa(elength of the radiation is gi(en asI

f the transition ta7es place from n $ 3 to n $ )# then the wa(elength of the radiation is gi(enasI



'his radiation corresponds to the 8almer series of the h&drogen spectrum.

ence# in <&man series# two wa(elengths i.e.# 10).5 nm and 1)1.5 nm are emitted. And in the8almer series# one wa(elength i.e.# -5-.33 nm is emitted.

Question 12.10:

n accordance with the 8ohrBs model# find the !uantum num%er that characterises the earthBs

re(olution around the sun in an or%it of radius 1.5 × 1011

m with or%ital speed 3 × 104

m;s.>ass of earth $ -.0 × 10)4 7g.

,adius of the or%it of the "arth around the Fun# r $ 1.5 × 1011 m

Er%ital speed of the "arth# ν $ 3 × 104 m;s

>ass of the "arth# m $ -.0 × 10)4 7g

According to 8ohrBs model# angular momentum is !uantied and gi(en asI

*here#


n $ Suantum num%er

ence# the !uanta num%er that characteries the "arthB re(olution is ).- × 10+4.

Question 12.11:



Answer the following !uestions# which help &ou understand the difference %etween 'homsonBs

model and ,utherfordBs model %etter.

(a) s the a(erage angle of deflection of α ?particles %& a thin gold foil predicted %& 'homsonBs

model much less# a%out the same# or much greater than that predicted %& ,utherfordBs model2

(b) s the pro%a%ilit& of %ac7ward scattering i.e.# scattering of α ?particles at angles greater

than 90° predicted %& 'homsonBs model much less# a%out the same# or much greater than

that predicted %& ,utherfordBs model2

(c) Zeeping other factors fi@ed# it is found e@perimentall& that for small thic7ness t # the

num%er of α ?particles scattered at moderate angles is proportional to t . *hat clue does this

linear dependence on t pro(ide2

(d) n which model is it completel& wrong to ignore multiple scattering for the calculation ofa(erage angle of scattering of α ?particles %& a thin foil2

(a) a%out the same

'he a(erage angle of deflection of α ?particles %& a thin gold foil predicted %& 'homsonBs modelis a%out the same sie as predicted %& ,utherfordBs model. 'his is %ecause the a(erage anglewas ta7en in %oth models.

(b) much less

'he pro%a%ilit& of scattering of α ?particles at angles greater than 90° predicted %& 'homsonBs

model is much less than that predicted %& ,utherfordBs model.

(c) Fcattering is mainl& due to single collisions. 'he chances of a single collision increaseslinearl& with the num%er of target atoms. Fince the num%er of target atoms increase with anincrease in thic7ness# the collision pro%a%ilit& depends linearl& on the thic7ness of the target.

(d) 'homsonBs model

t is wrong to ignore multiple scattering in 'homsonBs model for the calculation of a(erageangle of scattering of α −particles %& a thin foil. 'his is %ecause a single collision causes (er&little deflection in this model. ence# the o%ser(ed a(erage scattering angle can %e e@plainedonl& %& considering multiple scattering.

Question 12.12:

'he gra(itational attraction %etween electron and proton in a h&drogen atom is wea7er than

the coulom% attraction %& a factor of a%out 10−40. An alternati(e wa& of loo7ing at this fact is

to estimate the radius of the first 8ohr or%it of a h&drogen atom if the electron and proton

were %ound %& gra(itational attraction. 6ou will find the answer interesting.



,adius of the first 8ohr or%it is gi(en %& the relation#

*here#


' $ /lanc7Bs constant $ -.-3 × 10−34 Vs

me $ >ass of an electron $ 9.1 × 10−31 7g

e $ Charge of an electron $ 1.9 × 10−19 C

mp $ >ass of a proton $ 1.-+ × 10−)+ 7g

r $ istance %etween the electron and the proton

Coulom% attraction %etween an electron and a proton is gi(en asI

=ra(itational force of attraction %etween an electron and a proton is gi(en asI

*here#

= $ =ra(itational constant $ -.-+ × 10−11 m) ;7g)

f the electrostatic Coulom% force and the gra(itational force %etween an electron and aproton are e!ual# then we can writeI

∴F = $ F C

/utting the (alue of e!uation 4 in e!uation 1# we getI



t is 7nown that the uni(erse is 15- %illion light &ears wide or 1.5 × 10)+ m wide. ence# wecan conclude that the radius of the first 8ohr or%it is much greater than the estimated sie ofthe whole uni(erse.

Question 12.13:

E%tain an e@pression for the fre!uenc& of radiation emitted when a h&drogen atom de?e@cites

from le(el n to le(el n−1. or large n# show that this fre!uenc& e!uals the classicalfre!uenc& of re(olution of the electron in the or%it.

t is gi(en that a h&drogen atom de?e@cites from an upper le(el nI to a lower le(el n−1.

*e ha(e the relation for energ& E 1 of radiation at le(el n asI

ow# the relation for energ& E ) of radiation at le(el n − 1 is gi(enasI

"nerg& E released as a result of de?e@citationI

E $ E )−E 1



'ν $ E ) − E 1 D iii

*here#

ν $ re!uenc& of radiation emitted

/utting (alues from e!uations i and ii in e!uation iii# we getI

or large nD we can write

Classical relation of fre!uenc& of re(olution of an electron is gi(en asI

*here#

Melocit& of the electron in the nth or%it is gi(en asI

" $

And# radius of the nth or%it is gi(en asI

r $

/utting the (alues of e!uations (i and (ii in e!uation (# we getI



ence# the fre!uenc& of radiation emitted %& the h&drogen atom is e!ual to its classical or%italfre!uenc&.

Question 12.14:

Classicall&# an electron can %e in an& or%it around the nucleus of an atom. 'hen what

determines the t&pical atomic sie2 *h& is an atom not# sa&# thousand times %igger than its

t&pical sie2 'he !uestion had greatl& puled 8ohr %efore he arri(ed at his famous model of

the atom that &ou ha(e learnt in the te@t. 'o simulate what he might well ha(e done %efore his

disco(er&# let us pla& as follows with the %asic constants of nature and see if we can get a

!uantit& with the dimensions of length that is roughl& e!ual to the 7nown sie of an atom

10−10 m.

(a) Construct a !uantit& with the dimensions of length from the fundamental constants e# me#

and + . etermine its numerical (alue.

(b) 6ou will find that the length o%tained in a is man& orders of magnitude smaller than the

atomic dimensions. urther# it in(ol(es + . 8ut energies of atoms are mostl& in non?relati(istic

domain where + is not e@pected to pla& an& role. 'his is what ma& ha(e suggested 8ohr to

discard + and loo7 for something elseB to get the right atomic sie. ow# the /lanc7Bs constant

' had alread& made its appearance elsewhere. 8ohrBs great insight la& in recognising that '#me# and e will &ield the right atomic sie. Construct a !uantit& with the dimension of length

from '# me# and e and confirm that its numerical (alue has indeed the correct order of

magnitude.

(a) Charge on an electron# e $ 1.- × 10−19 C


Fpeed of light# + $ 3 ×10 m;s

<et us ta7e a !uantit& in(ol(ing the gi(en !uantities as

*here#




And#

'he numerical (alue of the ta7en !uantit& will %eI

ence# the numerical (alue of the ta7en !uantit& is much smaller than the t&pical sie of anatom.

(b) Charge on an electron# e $ 1.- × 10−19 C


/lanc7Bs constant# ' $ -.-3 ×10−34 Vs

<et us ta7e a !uantit& in(ol(ing the gi(en !uantities as

*here#


And#

'he numerical (alue of the ta7en !uantit& will %eI



ence# the (alue of the !uantit& ta7en is of the order of the atomic sie

Question 12.15:

'he total energ& of an electron in the first e@cited state of the h&drogen atom is a%out −3.4

eM.

(a) *hat is the 7inetic energ& of the electron in this state2

(b) *hat is the potential energ& of the electron in this state2

(c) *hich of the answers a%o(e would change if the choice of the ero of potential energ& is

changed2

(a) 'otal energ& of the electron# E $ −3.4 eM

Zinetic energ& of the electron is e!ual to the negati(e of the total energ&.

$ −E

$ − − 3.4 $ H3.4 eM

ence# the 7inetic energ& of the electron in the gi(en state is H3.4 eM.

(b) /otential energ& O of the electron is e!ual to the negati(e of twice of its 7inetic energ&.

O $ −)

$ − ) × 3.4 $ − -. eM

ence# the potential energ& of the electron in the gi(en state is − -. eM.

(c) 'he potential energ& of a s&stem depends on the reference point ta7en. ere# thepotential energ& of the reference point is ta7en as ero. f the reference point is changed# thenthe (alue of the potential energ& of the s&stem also changes. Fince total energ& is the sum of7inetic and potential energies# total energ& of the s&stem will also change.

Question 12.16:

f 8ohrBs !uantisation postulate angular momentum $ n'M )^ is a %asic law of nature# it

should %e e!uall& (alid for the case of planetar& motion also. *h& then do we ne(er spea7 of

!uantisation of or%its of planets around the sun2

*e ne(er spea7 of !uantiation of or%its of planets around the Fun %ecause the angularmomentum associated with planetar& motion is largel& relati(e to the (alue of /lanc7Bsconstant N'I. 'he angular momentum of the "arth in its or%it is of the order of 10+0'. 'hisleads to a (er& high (alue of !uantum le(els n of the order of 10+0. or large (alues of n#



successi(e energies and angular momenta are relati(el& (er& small. ence# the !uantumle(els for planetar& motion are considered continuous.

Question 12.17:

E%tain the first 8ohrBs radius and the ground state energ& of a m,%ni+ 'ydr%gen &t%m Ni.e.# an

atom in which a negati(el& charged muon :− of mass a%out )0+me or%its around a protonO.

>ass of a negati(el& charged muon#

According to 8ohrBs model#

8ohr radius#

And# energ& of a ground state e!e+tr%ni+ 'ydr%gen &t%m#

*e ha(e the (alue of the first 8ohr or%it#

<et r K %e the radius of m,%ni+ 'ydr%gen &t%m.

At e!uili%rium# we can write the relation asI

ence# the (alue of the first 8ohr radius of a m,%ni+ 'ydr%gen &t%m is

).5- × 10−13 m.

*e ha(e#

E e$ − 13.- eM

'a7e the ratio of these energies asI



ence# the ground state energ& of a m,%ni+ 'ydr%gen &t%m is −).1 7eM..

CA/'",?13 PC<"

Question 13.1:

a 'wo sta%le isotopes of lithium and ha(e respecti(e a%undances of +.5_ and

9).5_. 'hese isotopes ha(e masses -.0151) u and +.01-00 u# respecti(el&. ind the atomicmass of lithium.

% 8oron has two sta%le isotopes# and . 'heir respecti(e masses are 10.01)94 u and

11.00931 u# and the atomic mass of %oron is 10.11 u. ind the a%undances of and .

(a) >ass of lithium isotope # m1 $ -.0151) u

>ass of lithium isotope # m) $ +.01-00 u

A%undance of # L1$ +.5_

A%undance of # L)$ 9).5_

'he atomic mass of lithium atom is gi(en asI

(b) >ass of %oron isotope # m1 $ 10.01)94 u



>ass of %oron isotope # m) $ 11.00931 u

A%undance of # L1 $ x _

A%undance of # L)$ 100 − x _

Atomic mass of %oron# m $ 10.11 u

'he atomic mass of %oron atom is gi(en asI

And 100 − x $ 0.11_

ence# the a%undance of is 19.9_ and that of is 0.11_.

Question 13.2:

'he three sta%le isotopes of neonI and ha(e respecti(e a%undances of

90.51_# 0.)+_ and 9.))_. 'he atomic masses of the three isotopes are 19.99 u# )0.99 u

and )1.99 u# respecti(el&. E%tain the a(erage atomic mass of neon.

Atomic mass of # m1$ 19.99 u

A%undance of # L1 $ 90.51_



Atomic mass of # m) $ )0.99 u

A%undance of # L) $ 0.)+_

Atomic mass of # m3 $ )1.99 u

A%undance of # L3 $ 9.))_

'he a(erage atomic mass of neon is gi(en asI

Question 13.3:

E%tain the %inding energ& in >eM of a nitrogen nucleus # gi(en $14.0030+ u

Atomic mass of nitrogen # m $ 14.0030+ u

A nucleus of nitrogen contains + protons and + neutrons.

ence# the mass defect of this nucleus# ]m $ +mH H +mn − m

*here#

>ass of a proton# mH $ 1.00+)5 u

>ass of a neutron# mn$ 1.00--5 u

∴]m $ + × 1.00+)5 H + × 1.00--5 − 14.0030+

$ +.054++5 H +.0-055 − 14.0030+

$ 0.11)3- u

8ut 1 u $ 931.5 >eM;c)



∴]m $ 0.11)3- × 931.5 >eM;+ )

ence# the %inding energ& of the nucleus is gi(en asI

E 5 $ ]m+ )

*here#

+ $ Fpeed of light

∴E 5 $ 0.11)3- × 931.5

$ 104.--334 >eM

ence# the %inding energ& of a nitrogen nucleus is 104.--334 >eM.

Question 13.4:

E%tain the %inding energ& of the nuclei and in units of >eM from the following dataI

$ 55.934939 u $ )0.903 u

Atomic mass of # m1 $ 55.934939 u

nucleus has )- protons and 5- − )- $ 30 neutrons

ence# the mass defect of the nucleus# ]m $ )- × mH H 30 × mn − m1

*here#


>ass of a neutron# mn $ 1.00--5 u

∴]m $ )- × 1.00+)5 H 30 × 1.00--5 − 55.934939

$ )-.)0345 H 30.)5995 − 55.934939

$ 0.5)4-1 u

8ut 1 u $ 931.5 >eM;c)

∴]m $ 0.5)4-1 × 931.5 >eM;+ )

'he %inding energ& of this nucleus is gi(en asI



E 51 $ ]m+ )

*here#

+ $ Fpeed of light

∴E 51 $ 0.5)4-1 × 931.5

$ 49).)- >eM

A(erage %inding energ& per nucleon

Atomic mass of # m) $ )0.903 u

nucleus has 3 protons and )09 − 3 1)- neutrons.

ence# the mass defect of this nucleus is gi(en asI

]mk $ 3 × mH H 1)- × mn − m)

*here#



∴]mk $ 3 × 1.00+)5 H 1)- × 1.00--5 − )0.903

$ 3.-494+5 H 1)+.091+90 − )0.903

$ 1.+-0++ u

8ut 1 u $ 931.5 >eM;+ )

∴]mk $ 1.+-0++ × 931.5 >eM;+ )

ence# the %inding energ& of this nucleus is gi(en asI

E 5) $ ]mk+ )

$ 1.+-0++ × 931.5

$ 1-40.)- >eM



A(erage %indingenerg& per nucleon $

Question 13.5:

A gi(en coin has a mass of 3.0 g. Calculate the nuclear energ& that would %e re!uired to

separate all the neutrons and protons from each other. or simplicit& assume that the coin is

entirel& made of atoms of mass -).9)9-0 u.

>ass of a copper coin# mB $ 3 g

Atomic mass of atom# m $ -).9)9-0 u

'he total num%er of atoms in the coin

*here#

A $ A(ogadroBs num%er $ -.0)3 × 10)3 atoms ;g

>ass num%er $ -3 g

nucleus has )9 protons and -3 − )9 34 neutrons

∴>ass defect of this nucleus# ]mk $ )9 × mH H 34 × mn − m

*here#



∴]mk $ )9 × 1.00+)5 H 34 × 1.00--5 − -).9)9-

$ 0.591935 u

>ass defect of all the atoms present in the coin# ]m $ 0.591935 × ).- × 10))

$ 1.-9+--95 × 10)) u

8ut 1 u $ 931.5 >eM;+ )



∴]m $ 1.-9+--95 × 10)) × 931.5 >eM;+ )

ence# the %inding energ& of the nuclei of the coin is gi(en asI

E 5$ ]m+ )

$ 1.-9+--95 × 10)) × 931.5

$ 1.51 × 10)5 >eM

8ut 1 >eM $ 1.- × 10−13 V

E 5 = 1.51 × 10)5 × 1.- × 10−13

$ ).5)9- × 101) V

'his much energ& is re!uired to separate all the neutrons and protons from the gi(en coin.

Question 13.6:

*rite nuclear reaction e!uations for

(i) P?deca& of (ii) P?deca& of

(iii) ;?deca& of (iv) −?deca& of

(v) <?deca& of (vi) <?deca& of

(vii) "lectron capture of

q is a nucleus of helium and is an electron e− for ; and eH for <. n e(er& P?deca&# there is a loss of ) protons and 4 neutrons. n e(er& <?deca&# there is a loss of 1proton and a neutrino is emitted from the nucleus. n e(er& ;?deca&# there is a gain of 1proton and an antineutrino is emitted from the nucleus.

or the gi(en cases# the (arious nuclear reactions can %e written asI



Question 13.7:

A radioacti(e isotope has a half?life of 4 &ears. ow long will it ta7e the acti(it& to reduce to a

3.1)5_# % 1_ of its original (alue2

alf?life of the radioacti(e isotope $ 4 &ears

Eriginal amount of the radioacti(e isotope $ B 0

(a) After deca&# the amount of the radioacti(e isotope $ B

t is gi(en that onl& 3.1)5_ of B 0 remains after deca&. ence# we can writeI

*here#

X $ eca& constant

t $ 'ime



ence# the isotope will ta7e a%out 54 &ears to reduce to 3.1)5_ of its original (alue.

(b) After deca&# the amount of the radioacti(e isotope $ B

t is gi(en that onl& 1_ of B 0 remains after deca&. ence# we can writeI

Fince# λ $ 0.-93;4

ence# the isotope will ta7e a%out -.-454 &ears to reduce to 1_ of its original (alue.



Question 13.8:

'he normal acti(it& of li(ing car%on?containing matter is found to %e a%out 15 deca&s per

minute for e(er& gram of car%on. 'his acti(it& arises from the small proportion of radioacti(e

present with the sta%le car%on isotope . *hen the organism is dead# its interaction

with the atmosphere which maintains the a%o(e e!uili%rium acti(it& ceases and its acti(it&

%egins to drop. rom the 7nown half?life 5+30 &ears of # and the measured acti(it&# the

age of the specimen can %e appro@imatel& estimated. 'his is the principle of dating used

in archaeolog&. Fuppose a specimen from >ohenGodaro gi(es an acti(it& of 9 deca&s per

minute per gram of car%on. "stimate the appro@imate age of the ndus?Malle& ci(ilisation.

eca& rate of li(ing car%on?containing matter# 9 $ 15 deca&;min

<et %e the num%er of radioacti(e atoms present in a normal car%on? containing matter.

alf life of # $ 5+30 &ears

'he deca& rate of the specimen o%tained from the >ohenGodaro siteI

9k $ 9 deca&s;min

<et k %e the num%er of radioacti(e atoms present in the specimen during the >ohenGodaroperiod.

'herefore# we can relate the deca& constant# λand time# t asI



ence# the appro@imate age of the ndus?Malle& ci(ilisation is 4))3.5 &ears.

Question 13.9:

E%tain the amount of necessar& to pro(ide a radioacti(e source of .0 mCi strength. 'he

half?life of is 5.3 &ears.

'he strength of the radioacti(e source is gi(en asI

*here#

B $ ,e!uired num%er of atoms

alf?life of # $ 5.3 &ears

$ 5.3 × 3-5 × )4 × -0 × -0

$ 1.-+ × 10 s

or deca& constant λ# we ha(e the rate of deca& asI

*here# λ



or I

>ass of -.0)3 × 10)3 A(ogadroBs num%er atoms $ -0 g

∴>ass of atoms

ence# the amount of necessar& for the purpose is +.10- × 10−- g.

Question 13.10:

'he half?life of is ) &ears. *hat is the disintegration rate of 15 mg of this isotope2

alf life of # $ ) &ears

$ ) × 3-5 × )4 × -0 × -0

$ .3 × 10 s

>ass of the isotope# m $ 15 mg

90 g of atom contains -.0)3 × 10)3 A(ogadroBs num%er atoms.

'herefore# 15 mg of containsI

,ate of disintegration#

*here#

X $ eca& constant



ence# the disintegration rate of 15 mg of the gi(en isotope is+.+ × 1010 atoms;s.

Question 13.11:

E%tain appro@imatel& the ratio of the nuclear radii of the gold isotope and the sil(erisotope .

uclear radius of the gold isotope $ 9Au

uclear radius of the sil(er isotope $ 9Ag

>ass num%er of gold# AAu $ 19+

>ass num%er of sil(er# AAg $ 10+

'he ratio of the radii of the two nuclei is related with their mass num%ers asI

ence# the ratio of the nuclear radii of the gold and sil(er isotopes is a%out 1.)3.

Question 13.12:

ind the S?(alue and the 7inetic energ& of the emitted P?particle in the q?deca& of a

and % .

=i(en $ ))-.0)540 u# $ ))).01+50 u#

$ ))0.0113+ u# $ )1-.0019 u.

(a) Alpha particle deca& of emits a helium nucleus. As a result# its mass num%erreduces to ))- − 4 ))) and its atomic num%er reduces to − ) -. 'his is shown in thefollowing nuclear reaction.



Q?(alue of

emitted P?particle $ Fum of initial mass − Fum of final mass + )

*here#

+ $ Fpeed of light

t is gi(en thatI

Q?(alue $ N))-.0)540 − ))).01+50 H 4.00)-03O u + ) $ 0.005)9+ u + )

8ut 1 u $ 931.5 >eM;+ )

∴Q $ 0.005)9+ × 931.5 4.94 >eM

Zinetic energ& of the P?particle

(b) Alpha particle deca& of is shown %& the following nuclear reaction.

t is gi(en thatI

>ass of $ ))0.0113+ u

>ass of $ )1-.0019 u

∴Q?(alue $

-41 >eM



Zinetic energ& of the P?particle

$ -.)9 >eM

Question 13.13:

'he radionuclide 11C deca&s according to

'he ma@imum energ& of the emitted positron is 0.9-0 >eM.

=i(en the mass (aluesI

calculate Q and compare it with the ma@imum energ& of the positron emitted

'he gi(en nuclear reaction isI

Atomic mass of $ 11.011434 u

Atomic mass of

>a@imum energ& possessed %& the emitted positron $ 0.9-0 >eM

'he change in the Q?(alue ]Q of the nuclear masses of the nucleus is gi(en asI

*here#

me $ >ass of an electron or positron $ 0.00054 u



+ $ Fpeed of light

m $ ,especti(e nuclear masses

f atomic masses are used instead of nuclear masses# then we ha(e to add - me in the case of

and 5 me in the case of .

ence# e!uation 1 reduces toI

∴]Q $ N11.011434 − 11.009305 − ) × 0.00054O + )

$ 0.001033 + ) u

8ut 1 u $ 931.5 >e(;+ )

∴]Q $ 0.001033 × 931.5 0.9-) >eM

'he (alue of Q is almost compara%le to the ma@imum energ& of the emitted positron.

Question 13.14:

'he nucleus deca&s %& emission. *rite down the deca& e!uation and determine the

ma@imum 7inetic energ& of the electrons emitted. =i(en thatI

$ )).9944-- u

$ )).99++0 u.

n emission# the num%er of protons increases %& 1# and one electron and an antineutrinoare emitted from the parent nucleus.

emission of the nucleus is gi(en asI

t is gi(en thatI

Atomic mass of $ )).9944-- u



Atomic mass of $ )).99++0 u

>ass of an electron# me $ 0.00054 u

Q?(alue of the gi(en reaction is gi(en asI

'here are 10 electrons in and 11 electrons in . ence# the mass of the electron iscancelled in the Q?(alue e!uation.

'he daughter nucleus is too hea(& as compared to and . ence# it carries negligi%leenerg&. 'he 7inetic energ& of the antineutrino is nearl& ero. ence# the ma@imum 7ineticenerg& of the emitted electrons is almost e!ual to the Q?(alue# i.e.# 4.3+4 >eM.

Question 13.15:

'he Q (alue of a nuclear reaction A H 5 → $ H d is defined %&

Q $ N m AH m5− m$ − md O+ ) where the masses refer to the respecti(e nuclei. etermine from

the gi(en data the Q?(alue of the following reactions and state whether the reactions are

e@othermic or endothermic.

i

ii

Atomic masses are gi(en to %e



(i) 'he gi(en nuclear reaction isI

t is gi(en thatI

Atomic mass

Atomic mass

Atomic mass

According to the !uestion# the Q?(alue of the reaction can %e written asI

'he negati(eS?(alue of the reaction shows that the reaction is endothermic.

(ii) 'he gi(en nuclear reaction isI



t is gi(en thatI

Atomic mass of

Atomic mass of

Atomic mass of

'he Q?(alue of this reaction is gi(en asI

'he positi(e Q?(alue of the reaction shows that the reaction is e@othermic.

Question 13.16:

Fuppose# we thin7 of fission of a nucleus into two e!ual fragments# . s the fission

energeticall& possi%le2 Argue %& wor7ing out Q of the process. =i(en

and .

'he fission of can %e gi(en asI

t is gi(en thatI

Atomic mass of $ 55.93494 u

Atomic mass of

'he Q?(alue of this nuclear reaction is gi(en asI



'he Q?(alue of the fission is negati(e. 'herefore# the fission is not possi%le energeticall&. oran energeticall&?possi%le fission reaction# the Q?(alue must %e positi(e.

Question 13.17:

'he fission properties of are (er& similar to those of .

'he a(erage energ& released per fission is 10 >eM. ow much energ&# in >eM# is released ifall the atoms in 1 7g of pure undergo fission2

A(erage energ& released per fission of #

Amount of pure # m $ 1 7g $ 1000 g

A$ A(ogadro num%er $ -.0)3 × 10)3

>ass num%er of $ )39 g

1 mole of contains A atoms.

∴m g of contains

∴'otal energ& released during the fission of 1 7g of is calculated asI

ence# is released if all the atoms in 1 7g of pure undergo fission.



Question 13.18:

A 1000 >* fission reactor consumes half of its fuel in 5.00 &. ow much did it contain

initiall&2 Assume that the reactor operates 0_ of the time# that all the energ& generated

arises from the fission of and that this nuclide is consumed onl& %& the fission process

alf life of the fuel of the fission reactor# &ears

$ 5 × 3-5 × )4 × -0 × -0 s

*e 7now that in the fission of 1 g of nucleus# the energ& released is e!ual to )00 >eM.

1 mole# i.e.# )35 g of contains -.0)3 × 10)3 atoms.

∴1 g contains

'he total energ& generated per gram of is calculated asI

'he reactor operates onl& 0_ of the time.

ence# the amount of consumed in 5 &ears %& the 1000 >* fission reactor is calculatedasI

∴nitial amount of $ ) × 153 $ 30+- 7g

Question 13.19:



ow long can an electric lamp of 100* %e 7ept glowing %& fusion of ).0 7g of deuterium2 'a7e

the fusion reaction as

'he gi(en fusion reaction isI

Amount of deuterium# m $ ) 7g

1 mole# i.e.# ) g of deuterium contains -.0)3 × 10)3 atoms.

∴).0 7g of deuterium contains

t can %e inferred from the gi(en reaction that when two atoms of deuterium fuse# 3.)+ >eMenerg& is released.

∴'otal energ& per nucleus released in the fusion reactionI

/ower of the electric lamp# 6 $ 100 * $ 100 V;s

ence# the energ& consumed %& the lamp per second $ 100 V

'he total time for which the electric lamp will glow is calculated asI



Question 13.20:

Calculate the height of the potential %arrier for a head on collision of two deuterons. intI 'he

height of the potential %arrier is gi(en %& the Coulom% repulsion %etween the two deuterons

when the& Gust touch each other. Assume that the& can %e ta7en as hard spheres of radius ).0

fm.

*hen two deuterons collide head?on# the distance %etween their centres# d is gi(en asI

,adius of 1st deuteron H ,adius of )nd deuteron

,adius of a deuteron nucleus $ ) fm $ ) × 10 −15 m

∴d $ ) × 10−15 H ) × 10−15 $ 4 × 10−15 m

Charge on a deuteron nucleus $ Charge on an electron $ e $ 1.- × 10−19 C

/otential energ& of the two?deuteron s&stemI

*here#


ence# the height of the potential %arrier of the two?deuteron s&stem is

3-0 7eM.



Question 13.21:

rom the relation 9 $ 90 A1 ;3# where 90 is a constant and A is the mass num%er of a nucleus#

show that the nuclear matter densit& is nearl& constant i.e. independent of A.

*e ha(e the e@pression for nuclear radius asI

9 $ 90 A1 ;3

*here#

90 $ Constant.

A $ >ass num%er of the nucleus

uclear matter densit&#

<et m %e the a(erage mass of the nucleus.

ence# mass of the nucleus $ mA

ence# the nuclear matter densit& is independent of A. t is nearl& constant.

Question 13.22:

or the positron emission from a nucleus# there is another competing process 7nown as

electron capture electron from an inner or%it# sa&# the Z−shell# is captured %& the nucleus and

a neutrino is emitted.



Fhow that if emission is energeticall& allowed# electron capture is necessaril& allowed %ut

not (ice−(ersa.

<et the amount of energ& released during the electron capture process %e Q1. 'he nuclearreaction can %e written asI

<et the amount of energ& released during the positron capture process %e Q). 'he nuclearreaction can %e written asI

$ uclear mass of

$ uclear mass of

$ Atomic mass of

$ Atomic mass of

me $ >ass of an electron

+ $ Fpeed of light

Q-(alue of the electron capture reaction is gi(en asI

Q?(alue of the positron capture reaction is gi(en asI



t can %e inferred that if Q) [ 0# then Q1 [ 0L Also# if Q1[ 0# it does not necessaril& mean thatQ) [ 0.

n other words# this means that if emission is energeticall& allowed# then the electroncapture process is necessaril& allowed# %ut not (ice?(ersa. 'his is %ecause the Q?(alue must%e positi(e for an energeticall&?allowed nuclear reaction.

Question 13.23:

n a periodic ta%le the a(erage atomic mass of magnesium is gi(en as )4.31) u. 'he a(erage

(alue is %ased on their relati(e natural a%undance on earth. 'he three isotopes and their

masses are )3.9504u# )4.954u and )5.9)59u. 'he natural

a%undance of is +.99_ %& mass. Calculate the a%undances of other two isotopes.

A(erage atomic mass of magnesium# m $ )4.31) u

>ass of magnesium isotope # m1 $ )3.9504 u

>ass of magnesium isotope # m) $ )4.954 u

>ass of magnesium isotope # m3 $ )5.9)59 u

A%undance of # L1$ +.99_

A%undance of # L) $ x _

ence# a%undance of # L3 $ 100 − x − +.99_ $ )1.01 − x _

*e ha(e the relation for the a(erage atomic mass asI



ence# the a%undance of is 9.3_ and that of is 11.+1_.

Question 13.24:

'he neutron separation energ& is defined as the energ& re!uired to remo(e a neutron from the

nucleus. E%tain the neutron separation energies of the nuclei and from the

following dataI

$ 39.9-)591 u

$ 40.9-))+ u

$ )5.9-95 u

$ )-.91541 u

or

or

A neutron is remo(ed from a nucleus. 'he corresponding nuclear reaction can %ewritten asI

t is gi(en thatI



>ass $ 39.9-)591 u

>ass $ 40.9-))+ u

>ass $ 1.00--5 u

'he mass defect of this reaction is gi(en asI

]m $

∴]m $ 0.009+ × 931.5 >eM;+ )

ence# the energ& re!uired for neutron remo(al is calculated asI

or # the neutron remo(al reaction can %e written asI

t is gi(en thatI

>ass $ )-.91541 u

>ass $ )5.9-95 u

'he mass defect of this reaction is gi(en asI



ence# the energ& re!uired for neutron remo(al is calculated asI

Question 13.25:

A source contains two phosphorous radio nuclides 4 1;) $ 14.3d and 4 1;) $ )5.3d.

nitiall&# 10_ of the deca&s come from . ow long one must wait until 90_ do so2

alf life of # 4 1;) $ 14.3 da&s

alf life of # 4 1;) $ )5.3 da&s

nucleus deca& is 10_ of the total amount of deca&.

'he source has initiall& 10_ of nucleus and 90_ of nucleus.

Fuppose after t da&s# the source has 10_ of nucleus and 90_ of nucleus.

nitiall&I

um%er of nucleus $ B

um%er of nucleus $ 9 B

inall&I

um%er of

um%er of

or nucleus# we can write the num%er ratio asI



or # we can write the num%er ratio asI

En di(iding e!uation 1 %& e!uation )# we getI

ence# it will ta7e a%out )0.5 da&s for 90_ deca& of .

Question 13.26:

Pnder certain circumstances# a nucleus can deca& %& emitting a particle more massi(e than an

α?particle. Consider the following deca& processesI



Calculate the Q?(alues for these deca&s and determine that %oth are energeticall& allowed.

'a7e a emission nuclear reactionI

*e 7now thatI

>ass of m1 $ ))3.0150 u

>ass of m) $ )0.910+ u

>ass of # m3 $ 14.003)4 u

ence# the Q?(alue of the reaction is gi(en asI

Q $ m1 − m) − m3 + )

$ ))3.0150 − )0.910+ − 14.003)4 + )

$ 0.03419 + ) u

8ut 1 u $ 931.5 >eM;+ )

∴Q $ 0.03419 × 931.5

$ 31.4 >eM

ence# the Q?(alue of the nuclear reaction is 31.4 >eM. Fince the (alue is positi(e# thereaction is energeticall& allowed.

ow ta7e a emission nuclear reactionI

*e 7now thatI

>ass of m1 $ ))3.0150

>ass of m) $ )19.0094

>ass of # m3 $ 4.00)-0



Q?(alue of this nuclear reaction is gi(en asI

Q $ m1 − m) − m3 + )

$ ))3.0150 − )19.0094 − 4.00)-0 C)

$ 0.00-4) + ) u

$ 0.00-4) × 931.5 $ 5.9 >eM

ence# the Q (alue of the second nuclear reaction is 5.9 >eM. Fince the (alue is positi(e# thereaction is energeticall& allowed.

Question 13.27:

Consider the fission of %& fast neutrons. n one fission e(ent# no neutrons are emitted

and the final end products# after the %eta deca& of the primar& fragments# are and

. Calculate S for this fission process. 'he rele(ant atomic and particle masses are

m $)3.050+9 u

m $139.90543 u

m $ 9.90594 un the fission of # 10 − particles deca& from the parent nucleus.'he nuclear reaction can %e written asI

t is gi(en thatI

>ass of a nucleus m1 $ )3.050+9 u

>ass of a nucleus m) $ 139.90543 u

>ass of a nucleus # m3 $ 9.90594 u

>ass of a neutron m4 $ 1.00--5 u

Q?(alue of the a%o(e e!uation#



*here#

mB $ ,epresents the corresponding atomic masses of the nuclei

$ m1 − 9)me

$ m) − 5me

$ m3 − 44me

$ m4

ence# the Q?(alue of the fission process is )31.00+ >eM.

Question 13.28:

Consider the −' reaction deuterium−tritium fusion

(a) Calculate the energ& released in >eM in this reaction from the dataI

$ ).01410) u

$ 3.01-049 u



(b)Consider the radius of %oth deuterium and tritium to %e appro@imatel& ).0 fm. *hat is the

7inetic energ& needed to o(ercome the coulom% repulsion %etween the two nuclei2 'o what

temperature must the gas %e heated to initiate the reaction2 intI Zinetic energ& re!uired for

one fusion e(ent $a(erage thermal 7inetic energ& a(aila%le with the interacting particles $

)3k4 ;)L k $ 8oltmanBs constant# 4 $ a%solute temperature.

(a) 'a7e the ?' nuclear reactionI

t is gi(en thatI

>ass of # m1$ ).01410) u

>ass of # m) $ 3.01-049 u

>ass of m3 $ 4.00)-03 u

>ass of # m4 $ 1.00--5 u

Q?(alue of the gi(en ?' reaction isI

Q $ Nm1 H m)− m3 ; m4O + )

$ N).01410) H 3.01-049 − 4.00)-03 − 1.00--5O + )

$ N0.013 + )O u

8ut 1 u $ 931.5 >eM;+ )

∴Q $ 0.013 × 931.5 $ 1+.59 >eM

(b) ,adius of deuterium and tritium# r ).0 fm $ ) × 10−15 m

istance %etween the two nuclei at the moment when the& touch each other# d = r < r $ 4 ×10−15 m

Charge on the deuterium nucleus $ e

Charge on the tritium nucleus $ e

ence# the repulsi(e potential energ& %etween the two nuclei is gi(en asI

*here#




ence# 5.+- × 10−14 V or of 7inetic energ& Z" is needed to o(ercome the Coulom%repulsion %etween the two nuclei.

owe(er# it is gi(en thatI

Z"

*here#

k $ 8oltmann constant $ 1.3 × 10−)3 m) 7g s−) Z−1

' $ 'emperature re!uired for triggering the reaction

ence# the gas must %e heated to a temperature of 1.39 × 109 Z to initiate the reaction.

Question 13.29:

E%tain the ma@imum 7inetic energ& of ?particles# and the radiation fre!uencies of γ deca&s in

the deca& scheme shown in ig. 13.-. 6ou are gi(en that

m 19Au $ 19+.9-)33 u

m 19g $19+.9--+-0 u



t can %e o%ser(ed from the gi(en γ ?deca& diagram that γ 1 deca&s from the 1.0 >eM energ&le(el to the 0 >eM energ& le(el.

ence# the energ& corresponding to γ 1?deca& is gi(en asI

E 1 $ 1.0 − 0 $ 1.0 >eM

'ν 1$ 1.0 × 1.- × 10−19 × 10- V

*here#


ν1 $ re!uenc& of radiation radiated %& γ 1?deca&

t can %e o%ser(ed from the gi(en γ ?deca& diagram that γ ) deca&s from the 0.41) >eM energ&le(el to the 0 >eM energ& le(el.

ence# the energ& corresponding to γ )?deca& is gi(en asI

E ) $ 0.41) − 0 $ 0.41) >eM

'ν )$ 0.41) × 1.- × 10

−19

× 10

-

V

*here#

ν) $ re!uenc& of radiation radiated %& γ )?deca&



t can %e o%ser(ed from the gi(en γ ?deca& diagram that γ 3 deca&s from the 1.0 >eM energ&le(el to the 0.41) >eM energ& le(el.

ence# the energ& corresponding to γ 3?deca& is gi(en asI

E 3 $ 1.0 − 0.41) $ 0.-+- >eM

'ν 3$ 0.-+- × 10−19 × 10- V

*here#

ν3 $ re!uenc& of radiation radiated %& γ 3?deca&

>ass of $ 19+.9-)33 u

>ass of $ 19+.9--+-0 u

1 u $ 931.5 >eM;+ )

"nerg& of the highest le(el is gi(en asI

1 deca&s from the 1.3+)0995 >eM le(el to the 1.0 >eM le(el

∴>a@imum 7inetic energ& of the 1 particle $ 1.3+)0995 − 1.0

$ 0.)40995 >eM

) deca&s from the 1.3+)0995 >eM le(el to the 0.41) >eM le(el



∴>a@imum 7inetic energ& of the ) particle $ 1.3+)0995 − 0.41)

$ 0.9-00995 >eM

Question 13.30:

Calculate and compare the energ& released %& a fusion of 1.0 7g of h&drogen deep within Fun

and % the fission of 1.0 7g of )35P in a fission reactor.

(a) Amount of h&drogen# m $ 1 7g $ 1000 g

1 mole# i.e.# 1 g of h&drogen contains -.0)3 × 10)3 atoms.

∴1000 g of contains -.0)3 × 10)3 × 1000 atoms.

*ithin the sun# four nuclei com%ine and form one nucleus. n this process )- >eM ofenerg& is released.

ence# the energ& released from the fusion of 1 7g isI

(b) Amount of $ 1 7g $ 1000 g

1 mole# i.e.# )35 g of contains -.0)3 × 10)3 atoms.

∴1000 g of contains

t is 7nown that the amount of energ& released in the fission of one atom of is )00 >eM.

ence# energ& released from the fission of 1 7g of isI



∴

'herefore# the energ& released in the fusion of 1 7g of h&drogen is nearl& times the energ&released in the fission of 1 7g of uranium.

Question 13.31:

Fuppose ndia had a target of producing %& )0)0 A# )00#000 >* of electric power# ten

percent of which was to %e o%tained from nuclear power plants. Fuppose we are gi(en that# on

an a(erage# the efficienc& of utiliation i.e. con(ersion to electric energ& of thermal energ&

produced in a reactor was )5_. ow much amount of fissiona%le uranium would our countr&

need per &ear %& )0)02 'a7e the heat energ& per fission of )35P to %e a%out )00>eM.

Amount of electric power to %e generated# 6 $ ) × 105 >*

10_ of this amount has to %e o%tained from nuclear power plants.

∴Amount of nuclear power#

$ ) × 104 >*

$ ) × 104 × 10- V;s

$ ) × 1010 × -0 × -0 × )4 × 3-5 V;&

eat energ& released per fission of a )35P nucleus# E $ )00 >eM

"fficienc& of a reactor $ )5_

ence# the amount of energ& con(erted into the electrical energ& per fission is calculated asI

um%er of atoms re!uired for fission per &earI

1 mole# i.e.# )35 g of P)35 contains -.0)3 × 10)3 atoms.

∴>ass of -.0)3 × 10)3 atoms of P)35 $ )35 g $ )35 × 10−3 7g



∴>ass of +40 × 10)4 atoms of P)35

ence# the mass of uranium needed per &ear is 3.0+- × 104 7g.

Chapter 14 - Seic!&d:c"!$ Eec"$!&ic7# a"e$ia79 Device7 A&d Si%e Ci$c:i"7

Question 14.1:

n an n?t&pe silicon# which of the following statement is trueI

(a) "lectrons are maGorit& carriers and tri(alent atoms are the dopants.

(b) "lectrons are minorit& carriers and penta(alent atoms are the dopants.

(c) oles are minorit& carriers and penta(alent atoms are the dopants.

(d) oles are maGorit& carriers and tri(alent atoms are the dopants

'he correct statement is (c).

n an n?t&pe silicon# the electrons are the maGorit& carriers# while the holes are the minorit&carriers. An n?t&pe semiconductor is o%tained when penta(alent atoms# such as phosphorus#are doped in silicon atoms.

Question 14.2:

*hich of the statements gi(en in "@ercise 14.1 is true for p?t&pe semiconductors

'he correct statement is (d).



n a p?t&pe semiconductor# the holes are the maGorit& carriers# while the electrons are theminorit& carriers. A p?t&pe semiconductor is o%tained when tri(alent atoms# such asaluminium# are doped in silicon atoms.

Question 14.3:

Car%on# silicon and germanium ha(e four (alence electrons each. 'hese are characterised %&

(alence and conduction %ands separated %& energ& %and gap respecti(el& e!ual to E gC# E gFi

and E g=e. *hich of the following statements is true2

(a) E gFi E g=e E gC

(b) E gC E g=e [ E gFi

(c) E gC [ E gFi [ E g=e

(d) E gC $ E gFi $ E g=e


Ef the three gi(en elements# the energ& %and gap of car%on is the ma@imum and that ofgermanium is the least.

'he energ& %and gap of these elements are related asI E gC [ E gFi [ E g=e

Question 14.4:

n an un%iased p?n Gunction# holes diffuse from the p?region to n?region %ecause

(a) free electrons in the n?region attract them.

(b) the& mo(e across the Gunction %& the potential difference.

(c) hole concentration in p?region is more as compared to n?region.

(d) All the a%o(e.


'he diffusion of charge carriers across a Gunction ta7es place from the region of higherconcentration to the region of lower concentration. n this case# the p?region has greaterconcentration of holes than the n?region. ence# in an un%iased p?n Gunction# holes diffusefrom the p?region to the n?region.



Question 14.5:

*hen a forward %ias is applied to a p?n Gunction# it

(a) raises the potential %arrier.

(b) reduces the maGorit& carrier current to ero.

(c) lowers the potential %arrier.

(d) one of the a%o(e.


*hen a forward %ias is applied to a p?n Gunction# it lowers the (alue of potential %arrier. n thecase of a forward %ias# the potential %arrier opposes the applied (oltage. ence# the potential

%arrier across the Gunction gets reduced.

Question 14.6:

or transistor action# which of the following statements are correctI

(a) 8ase# emitter and collector regions should ha(e similar sie and doping concentrations.

(b) 'he %ase region must %e (er& thin and lightl& doped.

(c) 'he emitter Gunction is forward %iased and collector Gunction is re(erse %iased.

(d) 8oth the emitter Gunction as well as the collector Gunction are forward %iased.

'he correct statement is (b)# (c).

or a transistor action# the Gunction must %e lightl& doped so that the %ase region is (er& thin.Also# the emitter Gunction must %e forward?%iased and collector Gunction should %e re(erse?%iased.

Question 14.7:

or a transistor amplifier# the (oltage gain

(a) remains constant for all fre!uencies.

(b) is high at high and low fre!uencies and constant in the middle fre!uenc& range.



(c) is low at high and low fre!uencies and constant at mid fre!uencies.

(d) one of the a%o(e.


'he (oltage gain of a transistor amplifier is constant at mid fre!uenc& range onl&. t is low athigh and low fre!uencies.

Question 14.8:

n half?wa(e rectification# what is the output fre!uenc& if the input fre!uenc& is 50 . *hat is

the output fre!uenc& of a full?wa(e rectifier for the same input fre!uenc&.

nput fre!uenc& $ 50

or a half?wa(e rectifier# the output fre!uenc& is e!ual to the input fre!uenc&.

∴Eutput fre!uenc& $ 50

or a full?wa(e rectifier# the output fre!uenc& is twice the input fre!uenc&.

∴Eutput fre!uenc& $ ) × 50 $ 100

Question 14.9:

or a C"?transistor amplifier# the audio signal (oltage across the collected resistance of ) 7 is

) M. Fuppose the current amplification factor of the transistor is 100# find the input signal

(oltage and %ase current# if the %ase resistance is 1 7.

Collector resistance# 9C $ ) 7 $ )000

Audio signal (oltage across the collector resistance# 7 $ ) M

Current amplification factor of the transistor# $ 100

8ase resistance# 98 $ 1 7 $ 1000

nput signal (oltage $ 7 i

8ase current $ 3 8

*e ha(e the amplification relation asI

Moltage amplification



'herefore# the input signal (oltage of the amplifier is 0.01 M.

8ase resistance is gi(en %& the relationI

'herefore# the %ase current of the amplifier is 10 :A.

Question 14.10:

'wo amplifiers are connected one after the other in series cascaded. 'he first amplifier has a

(oltage gain of 10 and the second has a (oltage gain of )0. f the input signal is 0.01 (olt#

calculate the output ac signal.

Moltage gain of the first amplifier# 7 1 $ 10

Moltage gain of the second amplifier# 7 ) $ )0

nput signal (oltage# 7 i $ 0.01 M

Eutput AC signal (oltage $ 7 o

'he total (oltage gain of a two?stage cascaded amplifier is gi(en %& the product of (oltagegains of %oth the stages# i.e.#

7 $ 7 1 × 7 )

$ 10 × )0 $ )00

*e ha(e the relationI

7 0 $ 7 × 7 i



$ )00 × 0.01 $ ) M

'herefore# the output AC signal of the gi(en amplifier is ) M.

Question 14.11:

A p?n photodiode is fa%ricated from a semiconductor with %and gap of ). eM. Can it detect a

wa(elength of -000 nm2

"nerg& %and gap of the gi(en photodiode# E g $ ). eM

*a(elength# X $ -000 nm $ -000 × 10−9 m

'he energ& of a signal is gi(en %& the relationI

E $

*here#


$ -.-)- × 10−34 Vs

c $ Fpeed of light

$ 3 × 10 m;s

E

$ 3.313 × 10−)0 V

8ut 1.- × 10−19 V $ 1 eM

∴E $ 3.313 × 10−)0 V

'he energ& of a signal of wa(elength -000 nm is 0.)0+ eM# which is less than ). eM − theenerg& %and gap of a photodiode. ence# the photodiode cannot detect the signal.

Question 14.12:



'he num%er of silicon atoms per m3 is 5 × 10). 'his is doped simultaneousl& with 5 × 10)) atoms per m3 of Arsenic and 5 × 10)0 per m3 atoms of ndium. Calculate the num%er ofelectrons and holes. =i(en that ni $ 1.5 × 101- m−3. s the material n?t&pe or p?t&pe2

um%er of silicon atoms# B $ 5 × 10) atoms;m3

um%er of arsenic atoms# nAs $ 5 × 10)) atoms;m3

um%er of indium atoms# nn $ 5 × 10)0 atoms;m3

um%er of thermall&?generated electrons# ni $ 1.5 × 101- electrons;m3

um%er of electrons# ne $ 5 × 10)) − 1.5 × 101- 4.99 × 10))

um%er of holes $ n'

n thermal e!uili%rium# the concentrations of electrons and holes in a semiconductor are

related asI

nen' $ ni )

'herefore# the num%er of electrons is appro@imatel& 4.99 × 10)) and the num%er of holes isa%out 4.51 × 109. Fince the num%er of electrons is more than the num%er of holes# thematerial is an n?t&pe semiconductor.

Question 14.13:

n an intrinsic semiconductor the energ& gap E gis 1.) eM. ts hole mo%ilit& is much smaller than

electron mo%ilit& and independent of temperature. *hat is the ratio %etween conducti(it& at

-00Z and that at 300Z2 Assume that the temperature dependence of intrinsic carrier

concentration ni is gi(en %&

where n0 is a constant.

"nerg& gap of the gi(en intrinsic semiconductor# E g $ 1.) eM



'he temperature dependence of the intrinsic carrier?concentration is written asI

*here#

k 8 $ 8oltmann constant $ .-) × 10−5 eM;Z

' $ 'emperature

n0 $ Constant

nitial temperature# 4 1 $ 300 Z

'he intrinsic carrier?concentration at this temperature can %e written asI

D 1

inal temperature# 4 ) $ -00 Z

'he intrinsic carrier?concentration at this temperature can %e written asI

D )

'he ratio %etween the conducti(ities at -00 Z and at 300 Z is e!ual to the ratio %etween therespecti(e intrinsic carrier?concentrations at these temperatures.

'herefore# the ratio %etween the conducti(ities is 1.09 × 105.

Question 14.14:



n a p?n Gunction diode# the current can %e e@pressed as

where 3 0 is called the re(erse saturation current# 7 is the (oltage across the diode and is

positi(e for forward %ias and negati(e for re(erse %ias# and 3 is the current through the diode#

k 8is the 8oltmann constant .-×10−5 eM;Z and ' is the a%solute temperature. f for a gi(en

diode 3 0 $ 5 × 10−1) A and ' $ 300 Z# then

(a) *hat will %e the forward current at a forward (oltage of 0.- M2

(b) *hat will %e the increase in the current if the (oltage across the diode is increased to 0.+

M2

(c) *hat is the d&namic resistance2

(d) *hat will %e the current if re(erse %ias (oltage changes from 1 M to ) M2

n a p?n Gunction diode# the e@pression for current is gi(en asI

*here#

3 0 $ ,e(erse saturation current $ 5 × 10−1) A

4 $ A%solute temperature $ 300 Z

k 8 $ 8oltmann constant $ .- × 10−5 eM;Z $ 1.3+- × 10−)3 V Z−1

M $ Moltage across the diode

(a) orward (oltage# 7 $ 0.- M

∴Current# 3

'herefore# the forward current is a%out 0.0)5- A.



(b) or forward (oltage# 7 B $ 0.+ M# we can writeI

ence# the increase in current# ]3 $ 3 k − 3

$ 1.)5+ − 0.0)5- $ 1.)3 A

(c) &namic resistance

(d) f the re(erse %ias (oltage changes from 1 M to ) M# then the current 3 will almost remaine!ual to 3 0 in %oth cases. 'herefore# the d&namic resistance in the re(erse %ias will %e infinite.

Question 14.15:

6ou are gi(en the two circuits as shown in ig. 14.44. Fhow that circuit a acts as E, gate

while the circuit % acts as A gate.

(a) A and are the inputs and ? is the output of the gi(en circuit. 'he left half of the gi(enfigure acts as the E, =ate# while the right half acts as the E' =ate. 'his is shown in the

following figure.

ence# the output of the E, =ate $



'his will %e the input for the E' =ate. ts output will %e $ A H

∴? $ A H

ence# this circuit functions as an E, =ate.

(b) A and are the inputs and ? is the output of the gi(en circuit. t can %e o%ser(ed from thefollowing figure that the inputs of the right half E, =ate are the outputs of the two E'=ates.

ence# the output of the gi(en circuit can %e written asI

ence# this circuit functions as an A =ate.

Question 14.16:

*rite the truth ta%le for a A gate connected as gi(en in ig. 14.45.

ence identif& the e@act logic operation carried out %& this circuit.

A acts as the two inputs of the A gate and ? is the output# as shown in the followingfigure.

ence# the output can %e written asI

'he truth ta%le for e!uation i can %e drawn asI



AY

0 1

1 0

'his circuit functions as a E' gate. 'he s&m%ol for this logic circuit is shown asI

n %oth the gi(en circuits# A and are the inputs and ? is the output.

(a) 'he output of the left A gate will %e # as shown in the following figure.

ence# the output of the com%ination of the two A gates is gi(en asI

ence# this circuit functions as an A gate.

(b) is the output of the upper left of the A gate and is the output of the lower half of the A gate# as shown in the following figure.

ence# the output of the com%ination of the A gates will %e gi(en asI



ence# this circuit functions as an E, gate.

Question 14.18:

*rite the truth ta%le for circuit gi(en in ig. 14.4+ %elow consisting of E, gates and identif&

the logic operation E,# A# E' which this circuit is performing.

intI A $ 0# 8 $ 1 then A and 8 inputs of second E, gate will %e 0 and hence 6$1. Fimilarl&

wor7 out the (alues of 6 for other com%inations of A and 8. Compare with the truth ta%le of

E,# A# E' gates and find the correct one.

A and are the inputs of the gi(en circuit. 'he output of the first E, gate is . t can%e o%ser(ed from the following figure that the inputs of the second E, gate %ecome the output of the first one.

ence# the output of the com%ination is gi(en asI

'he truth ta%le for this operation is gi(en asI

A B Y ( A @ B)

0 0 0

0 1 1

1 0 1



1 1 1

'his is the truth ta%le of an E, gate. ence# this circuit functions as an E, gate.

Question 14.19:

*rite the truth ta%le for the circuits gi(en in ig. 14.4 consisting of E, gates onl&. dentif&

the logic operations E,# A# E' performed %& the two circuits.

(a) A acts as the two inputs of the E, gate and ? is the output# as shown in the following

figure. ence# the output of the circuit is .

'he truth ta%le for the same is gi(en asI

AY



0 1

1 0

'his is the truth ta%le of a E' gate. ence# this circuit functions as a E' gate.

(b) A and are the inputs and ? is the output of the gi(en circuit. 8& using the resulto%tained in solution (a)# we can infer that the outputs of the first two E, gates are

as shown in the following figure.

are the inputs for the last E, gate. ence# the output for the circuit can %e writtenasI

'he truth ta%le for the same can %e written asI

A B Y

( A

B)

0 0 0

0 1 0

1 0 0

1 1 1



'his is the truth ta%le of an A gate. ence# this circuit functions as an A gate.

Chapter 15 ? Communication F&stems

Question 15.1:

*hich of the following fre!uencies will %e suita%le for %e&ond?the?horion communication

using s7& wa(es2

(a) 10 7

(b) 10 >

(c) 1 =

(d) 1000 =

(b) A&7*e$#

10 >

or %e&ond?the?horion communication# it is necessar& for the signal wa(es to tra(el a largedistance. 10 Z signals cannot %e radiated efficientl& %ecause of the antenna sie. 'he highenerg& signal wa(es 1= − 1000 = penetrate the ionosphere. 10 > fre!uencies getreflected easil& from the ionosphere. ence# signal wa(es of such fre!uencies are suita%le for%e&ond?the?horion communication.



Question 15.2:

re!uencies in the P range normall& propagate %& means ofI

(a) =round wa(es.

(b) F7& wa(es.

(c) Furface wa(es.

(d) Fpace wa(es

(d) A&7*e$#

Fpace wa(es

Ewing to its high fre!uenc&# an ultra high fre!uenc& P wa(e can neither tra(el along thetraGector& of the ground nor can it get reflected %& the ionosphere. 'he signals ha(ing P arepropagated through line?of?sight communication# which is nothing %ut space wa(epropagation.

Question 15.3:

igital signals

i o not pro(ide a continuous set of (alues#

ii ,epresent (alues as discrete steps#

iii Can utilie %inar& s&stem# and

i( Can utilie decimal as well as %inar& s&stems.

*hich of the a%o(e statements are true2

a i and ii onl&

% ii and iii onl&

c i# ii and iii %ut not i(

d All of i# ii# iii and i(.

(c) A&7*e$#



A digital signal uses the %inar& 0 and 1 s&stem for transferring message signals. Fuch as&stem cannot utilise the decimal s&stem which corresponds to analogue signals. igitalsignals represent discontinuous (alues.

Question 15.4:

s it necessar& for a transmitting antenna to %e at the same height as that of the recei(ing

antenna for line?of?sight communication2 A 'M transmitting antenna is 1m tall. ow much

ser(ice area can it co(er if the recei(ing antenna is at the ground le(el2

<ine?of?sight communication means that there is no ph&sical o%struction %etween thetransmitter and the recei(er. n such communications it is not necessar& for the transmittingand recei(ing antennas to %e at the same height.

eight of the gi(en antenna# ' $ 1 m

,adius of earth# 9 $ -.4 × 10- m

or range# d $ )9'I\# the ser(ice area of the antenna is gi(en %& the relationI

A $ ^d )

$ ^ )9'

$ 3.14 × ) × -.4 × 10- × 1

$ 3)55.55 × 10- m)

$ 3)55.55

∼ 3)5- 7m)

Question 15.5:

A carrier wa(e of pea7 (oltage 1) M is used to transmit a message signal. *hat should %e the

pea7 (oltage of the modulating signal in order to ha(e a modulation inde@ of +5_2

Amplitude of the carrier wa(e# Ac $ 1) M

>odulation inde@# m $ +5_ $ 0.+5

Amplitude of the modulating wa(e $ Am

Psing the relation for modulation inde@I



Question 15.6:

A modulating signal is a s!uare wa(e# as shown in ig. 15.14.

'he carrier wa(e is gi(en %&

(i) F7etch the amplitude modulated wa(eform

(ii) *hat is the modulation inde@2

t can %e o%ser(ed from the gi(en modulating signal that the amplitude of the modulating

signal# Am $ 1 M

t is gi(en that the carrier wa(e + t $ ) sin ^t

∴Amplitude of the carrier wa(e# Ac $ ) M

'ime period of the modulating signal 4 m $ 1 s

'he angular fre!uenc& of the modulating signal is calculated asI

'he angular fre!uenc& of the carrier signal is calculated asI



rom e!uations i and ii # we getI

'he amplitude modulated wa(eform of the modulating signal is shown in the following figure.

(ii)>odulation inde@#

Question 15.7:

or an amplitude modulated wa(e# the ma@imum amplitude is found to %e 10 M while the

minimum amplitude is found to %e ) M. etermine the modulation inde@ K. *hat would %e the

(alue of K if the minimum amplitude is ero (olt2

>a@imum amplitude# Ama@ $ 10 M

>inimum amplitude# Amin $ ) M

>odulation inde@ K# is gi(en %& the relationI

electric charge

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