ch5. 靜磁學 (magnetostatics)

Y.M. Hu, Assistant Professor, Department of Applied Physics, National University

of Kaohsiung

Ch5. 靜磁學 (Magnetostatics)

5.1 羅倫茲力定律 (The Lorentz Force Law)

5.2 必歐沙伐定律 (The Biot-Savart Law)

5.3 磁場的散度與旋度 (The Divergence and Curl of B)

5.4 磁向量位勢 (Magnetic Vector Potential)


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5.1.1 The Lorentz Force Law-磁場 (Magnetic Fields)

Bar magnet Conducting wire

Parallel wires carrying currents in the same direction attract each other

Parallel wires carrying currents in opposite direction repel each other


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5.1.2.1 磁力 (Magnetic Forces)

BvqEqFFFBE

The Lorentz force

Additional speed parallel to B


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5.1.2.2 磁力 (Magnetic Forces) z


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0)0(z)0(y But the particle started from rest ( ), at the origin ( );0)0(z)0(y

EtB

EtsinAtcosB)t(y

0

0

CtsinBtcosA)t(z

EtB

EtsinAtcosB)t(y

0

0 0

0

B

EtcosAtsinB)t(y

0EB)0(y 0B

EA)0(y

0

0

CtsinBtcosA)t(z tcosBtsinA)t(z

0CA)0(z 0B)0(z

0BE CB

EA

0

0


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)tsint(B

EEt

B

EtsinAtcosB)t(y

0

0

0

0

)tcos1(B

ECtsinBtcosA)t(z

0

0

0

0

B

ER

tsinRtR)t(y

tcosRR)t(z

tsinR]tR)t(y[ 222

tcosR]R)t(z[ 222

222 R]R)t(z[]tR)t(y[

This is the formula for a circle, of radius R, whose center (0, Rt, R) travels in the y-direction at a constant speed

0

0

B

ERv


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5.1.3.1 電流 (Currents)

Example 5.3 : A rectangular loop of wire, supporting a mass m, hangs vertically with one end in a uniform magnetic field B, which points into the page in the shaded region. For what current I, in the loop, would the magnetic force upward exactly balance the gravitational force downward?

mgIBaFmag

Ba

mgI


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What happens if we now increase the current?

mgIBaFmag The loop rises a height of h

?0IBahhFW magmag

u

w

vmagF qwB

quB

IaBawBqwBFvert Rise the loop

auBquBFhoriz

IBahuwdtaBauBwdtwdtFW horizagent


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Example 5.4

a) A current I is uniformly distributed over a wire of circular cross section, with radius a. Find the volume current density J.

2a

IJ

b) Suppose the current density in the wire is proportional to the distance from the axis, J = ks. Find the total current in the wire.

The current in the shaded path dsdks)ds)(sd(JJda 2

3

ka2dssk2dsdksJdaI

3a

0

22


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SS

adJJdaI

The current crossing a surface S can be written as

The total charge per unit time leaving a volume V is

VS

d)J(adJ

Because charge is conserved

VVV

d)t

(ddt

dd)J(

tJ

Continuity equation


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5.2.1 穩定電流 (Steady Currents)

0t

J

Continuity equation

Actually, it is not necessary that the charges be stationary, but only that the charge density at each point be constant ( be independent with time).


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5.2.2 穩定電流的磁場 (The Magnetic Field of a Steady Current)

Example 5.5

Find the magnetic field a distance s from a long straight wire carrying a steady current I

I

sr

ℓ dℓ


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20

r

rd

4

IB

I

sr

ℓ dℓ

dcos

scosdsindrd

tans d

cos

sd 2

2

2

2 s

cos

r

1

2

1

2

1

dcoss4

I

s

cosd

cos

s

4

I

r

rd

4

IB 0

2

20

20

)sin(sins4

I12

0

s2

I)]1(1[

s4

I 00

2

,2 21

1

2


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The field at 2 due to 1 is

d2

IB 10

into the page

BdIF

The force on 2 due to 1 is d)d2

I(IF 10

2

The force per unit lengthd2

IIF 210


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Example 5.6 Magnetic Field on the Axis of a Circular Current Loop

)Rx(

ds

4

I

r

rsd

4

IdB

22

0

2

0

)Rx(

cosds

4

IcosdBB

22

0

x

2/122 )Rx(

Rcos

2/322

2

0

2/322

0

x )Rx(2

IRds

)Rx(4

IRB


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For surface and volume currents, the Biot-Savart Law becomes

'daˆ)'r(K

4)r(B 2

0

'dˆ)'r(J

4)r(B 2

0

Problem 5.8

a) Find the magnetic field at the center of a square loop, which carries a steady current I. Let R be the distance from center to side.

R R4

I2)

2

2

2

2(

R4

I)sin(sin

s4

IB 00

120

for 10

2 45,Rs

Four sidesR

I2

R4

I24B 00


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b) Find the field at the center of a rectangular n-sided polygon, carring a steady current I. Let R be the distance from center to any side.

nsin

R2

I)

nsin

n(sin

R4

I)sin(sin

s4

IB 00

120

for 12 n,Rs

n sidesn

sinR2

InB 0

c) Check that your formula reduces to the field at the center of a circular loop, in the limit n ∞.

n ∞nn

sinsinsin 12

R2

I

nR2

In

nsin

R2

InB 000


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5.2.8 穩定電流的磁場 (The Magnetic Field of a Steady Current)Problem 5.9

Find the magnetic field at point P for each of the steady current configurations.

P

I b

a

The vertical and horizontal lines produce no field at P.

The two quarter-circles

)b

1

a

1(

8

I)

b2

I

a2

I(

4

1B 000

R I

IP

The two half-lines are the same as one infinite line

R2

IB 0

The half-circle: R4

IB 0

Total:

)2

1(R4

IB 0


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5.3.1.1 磁場的散度與旋度– 直線電流 (Straight-Line Currents)

The magnetic field of an infinite straight wire

Ids2

Id

s2

IdB 0

00

Notice that the answer is independent of s; that’s because B decreases at the same rate as the circumference increases.

In fact, it doesn’t have to be a circle; any old loop that encloses the wire would give the same answer. For if we use cylindrical coordinates (s,,z), with this current flowing along the z axis,

ˆ

s2

IB 0

zdzˆsdsdsd

Id2

Isd

s

1

2

IdB 0

2

0

00

This assumes the loop encircles the wire exactly once.


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5.3.1.2 磁場的散度與旋度– 直線電流 (Straight-Line Currents)

Suppose we have a bundle of straight wires. Each wire that posses through our loop contributes 0I, and those outside contribute nothing.

I1

I2

I3

I4

enc0IdB

I enc : total current enclosed by the integration path

If the flow of charge is represented by a volume current density J, the enclosed current is

adJIenc

Applying Stokes’ theorem

adJIad)B(dB 0enc0

JB 0


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5.3.2.1 磁場的散度與旋度 (The Divergence and Curl of B)

(x,y,z)r

(x’,y’,z’)

d’

The Biot-Savart law for the general case of a volume current reads

'd]ˆ)'r(J

[4

)r(B 20

)B(A)A(B)BA(

)ˆ

()'r(J)]'r(J[ˆ

]ˆ

)'r(J[ 222

0)'r(J

Because J doesn’t depend on the unprimed variables (x,y,z)

0ˆ2

0]ˆ

)'r(J[ 2

0)r(B

The divergence of the magnetic field is zero!

'dˆ)'r(J

4)r(B 2

0


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'd]ˆ)'r(J

[4

)r(B 20

)A(B)B(AB)A(A)B()BA(

)]'r(J[ˆ

)ˆ

)('r(Jˆ

])'r(J[)'r(J)ˆ

(]ˆ

)'r(J[ 22222

)'rr(4)(4)ˆ

( 332

= 0 = 0

)ˆ

)('r(Jˆ

])'r(J[]ˆ

)'r(J[ 222

'dˆ)'r(J

4)r(B 2

0


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22

ˆ]')'r(J[

ˆ])'r(J[

)'xx(f'x

)'xx(fx

The x component )'xx

](')'r(J[ 3

)f(A)A(f)Af(

)A(f)Af()f(A

)]'r(J')['xx

()]'r(J'xx

[')'xx

](')'r(J[ 333

For steady currents the divergence of J is zero 0)'r(J'

)]'r(J'xx

[')'xx

](')'r(J[ 33

This contribution to the integral can be written

0'ad)'r(J'xx

'd)]'r(J'xx

['S

3V

3

On the boundary J = 0


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'd]ˆ)'r(J

[4

)r(B 20

)'rr()'r(J4)()'r(J4)ˆ

)('r(J]ˆ

)'r(J[ 3322

'd)'rr()'r(J44

'd]ˆ)'r(J

[4

)r(B 302

0

)r(J0

)r(J)r(B 0

Ampère’s law in differential form


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5.3.3.1 安培定律的應用 (Applications of Ampère’s Law)

)r(J)r(B 0

Ampère’s law in differential form

enc00 IadJdBad)B(

Ampère’s law in integral form

Example 5.8

Find the magnetic field of an infinite uniform surface current , flowing over the xy plane

xKK


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What is the direction of B?

KB

From the Biot-Savart law

Could it have a z-component ? no (symmetry)

The magnetic field points to the left above the plane and to the right below it

KIB2dB 0enc0 2

KB 0

0zfory2

K

0zfory2

K

B0

0


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Example 5.9

Find the magnetic field of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R and carrying a steady current I.

BdsBsdBsdB

1side1side

NIBsdB0

Where N is the number of turns in the length nII

NB

00


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Example 5.10

Find the magnetic field of a toroidal coil, consisting of a circular ring around which a long wire is wrapped.

NI)r2(BdsBsdB0

r2

NIB 0


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Problem 5.13

A steady current I flows down a long cylindrical wire of radius a. Find the magnetic field, both inside and outside the wire, if(a). The current is uniformly distributed over the outside surface of the wire.(b). The current is distributed in such a way that J is proportional to s, the distance from the axis.

Ia

(a)

enc0IsB2dB

asforˆs2

Iasfor0

B0


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(b)

enc0IsB2dB

ksJ 3

ka2ds)s2(ksJdaI

3a

0

3a2

I3k

3a2

Is3ksJ

For s < a

3

303

30

s

030enc0 a

Iss

3

1

a

I3sds2

a2

Is3IsB2dB

3

20

a2

IsB

For s > a

Ia3

1

a

I3sds2

a2

Is3IsB2dB 0

330

a

030enc0

s2

IB 0


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5.3.4. 靜磁學與靜電學之比較 (Comparison of Magnetostatics and Electrostatics)

The divergence and curl of the electrostatic field are

0E

1E

0

Gauss’s law

The divergence and curl of the magnetostatic field are

JB

0B

0

Ampère’s law

“The electric force is stronger than the magnetic force. Only when both the source charge and the test charge are moving at velocities comparable to the speed of light, the magnetic force approaches the electric force.”


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5.4.1.1. 磁向量位勢 (Magnetic Vector Potential)

0E

VE

V: electric scalar potential

You can add to V any function whose gradient is zero

VfV)fV(

0B

AB

: magnetic vector potentialA

You can add to any function whose curl is zeroA

AA)A(

AB

JA)A()A(B 02

0A

We will prove that :


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Suppose that our original vector potential is not divergenceless0A

0A0

Because we can add to any function whose curl is zero0A

0AA

200 AAA

If a function can be found that satisfies

02 A

0A

Mathematically identical to Poisson’s equation0

2V

In particular, if goes to zero at infinity, then the solution is

'd4

1V

0


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By the same token, if goes to zero at infinity, then 0A

'dA

4

1 0

It is always possible to make the vector potential divergenceless

0A

so JAA)A(B 022

This again is a Poisson’s equation

Assuming goes to zero at infinity, then J

'd)'r(J

4)r(A 0

For line and surface currents

'd)'r(I

4)r(A 0

'da)'r(K

4)r(A 0


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Example 5.11

A spherical shell, of radius R, carrying a uniform surface charge , is set spinning at angular velocity . Find the vector potential it produces at point . r

The integration is easier if we let lie on the z axis, so that is titled at an angle .

r

r

'r


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'da)'r(K

4)r(A 0

where vK

'cosRr2rR 22

'cosR'sin'sinR'cos'sinR

cos0sin

zyx

'rv

]z)'sin'sin(siny)'cossin'cos'sin(cosx)'sin'sin(cos[R

Because 0'd'cos'd'sin2

0

2

0

We just consider y'cossinRv

y)'d'cosRr2rR

'cos'sin(

2

sinR

y'd'd'sinR'cosRr2rR

)'cossinR(

4)r(A

022

30

2

22

0


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'cosu

)]rR)(RrrR(rR)RrrR[(rR3

1

]Rru2rRrR3

)RrurR([

])Rru2rR(rR3

1Rru2rR

Rr

u[

duRru2rR

u'd

'cosRr2rR

'cos'sinI

222222

1

1

2222

22

1

1

2/32222

22

1

122

022

Letting , the integral becomes

If the point lies inside the sphere, Then R > r. r

2R3

r2I

If the point lies outside the sphere, Then R < r. r

2r3

R2I


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yI2

sinRy)'d

'cosRr2rR

'cos'sin(

2

sinR)r(A

30

022

30

If the point lies inside the sphere, Then R > r. r

2R3

r2I

If the point lies outside the sphere, Then R < r. r

2r3

R2I

Noting that ysinr)r(

For the point inside the sphere

)r(3

Ry

R3

r2

2

sinRyI

2

sinR)r(A 0

2

30

30

For the point outside the sphere

)r(r3

Ry

r3

R2

2

sinRyI

2

sinR)r(A 3

40

2

30

30


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We revert to the original coordinates, in which coincides with the z axis and the point is at (r,,)

r

For the point inside the sphere

ˆsinr

3

R)r(

3

R)r(A 00

For the point outside the sphere

ˆsin

r3

R)r(

r3

R)r(A 2

40

3

40

z3

R2)ˆsinr(cos

3

R2)r(AB 00

The magnetic field inside this spherical shell is uniform!


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Example 5.12

Find the vector potential of an infinite solenoid with n turns per unit length, radius R, and current I

We cannot use because the current itself extends to infinity.

'd)'r(I

4)r(A 0

Notice that adBad)A(dA

Since the magnetic field is uniform inside the solenoid : znIB 0

)s(nIadB)s2(AdA 20

ˆ2

nIsA 0

For s < R

For an amperian outside the solenoid

)R(nIadB)s2(AdA 20

ˆ

s2

nIRA

20

For s > R


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5.4.2.1. 總結 ;靜磁邊界條件 (Summary; Magnetostatic Boundary Conditions)

B

J

A

'd)'r(J

4)r(A 0

JA 02

'dˆ)'r(J

4)r(B 2

0

JB 0

BA

?


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0B

0adBd)B(V

belowabove BB

KI)BB(dB 0enc0//below

//above KBB 0

//below

//above

For an amperian loop running perpendicular to the current

For an amperian loop running parallel to the current

0I)BB(dB enc0//below

//above

Perpendicular to the current

//below

//above BB

Parallel to the current

)nK(BB 0belowabove

Where is a unit vector perpendicular to the surface, pointing “upward”.n


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The vector potential is continuous across any boundary

danBdan)A(dA

For an amperian loop of vanishing thickness 0

belowabove AA

KBB 0//below

//above B

z

A

00A

z/y/x/

zyx

)A(

y

y

Kn

A

n

A0

belowabove


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5.4.3.1. 向量位勢的多極展開 (Multipole Expansion of the Vector Potential)

I

r

'r

'd'rd

'

A multipole expansion which is an approximate formula and valid at distant points) for the vector potential of a localized current distribution.

0n

nn

22)'(cosP)

r

'r(

r

1

'cos'rr2)'r(r

11


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0nn

n1n

00 'd)'(cosP)'r(r

1

4

I'd

1

4

I)r(A

....]'d)2

1'cos

2

3()'r(

r

1'd'cos'r

r

1'd

r

1[

4

I)r(A 22

320

monopole dipole quadrupole

The magnetic monopole tern is always zero0'd

'd)'rr(r4

I'd'cos'r

r4

I)r(A 2

02

0dip

Let , where is a constant vector.Tcv

c

area

'da'n)v'('dv

area

'da'n)Tc'('dTc

0B


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T'cT'cT)c'(cT'T)c'()Tc('

areaarea

'da)'nT'(c'da)T'c('n'dTc

area

'da)'nT'('Td

Let 'rrT

area

'da]'n)'rr('['d)'rr(

ar'da'nr'da)'nr('d)'rr(areaarea

'da'n'd'r2

1a

)'d'r

2

1(r'd)'rr(

20

dip r

rm

4)r(A

where aIadIm

)ra(r4

I)ar(

r4

I'd)'rr(

r4

I)r(A 2

02

02

0dip


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Example 5.13

Find the magnetic dipole moment of the “Bookend-shaped” loop shown in Figure below. All sides have length w, and it carries a current I.

w

w

w

The wire could be considered the superposition of two plane square loops shown in Fig. 5.53 of text book.

The combined (net) magnetic dipole moment is

zIwyIwm 22


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The magnetic dipole moment is independent of the choice of origin.

The magnetic field of a (pure) dipole is easier to calculate if we put the dipole moment at the origin and let it point in the z-direction.

20

dip r

rm

4)r(A

ˆ

r

sinm

4)r(A 2

0dip

)ˆsinrcos2(r4

m

)r4

sinm(sinr00

//r/

ˆsinrˆrr

sinr

1)r(A)r(B

30

20

2dipole

This is identical in structure to the field of an electric dipole!

ch5. 靜磁學 (magnetostatics)

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