ch 7 lecture slides

8/20/2019 Ch 7 Lecture Slides

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Engineering ElectromagneticsW.H. Hayt Jr. and J. A. Buck

Chapter 7:

The Steady agnetic !ield


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oti"ating the agnetic !ield Concept:

!orces Bet#een Currents

Ho# can #e descri$e a %orce %ield around #ire & that can $e used to determine the %orce on #ire '(

agnetic %orces arise #hene"er #e ha"e charges in motion. !orces $et#een current)carrying #ires

present %amiliar e*amples that #e can use to determine #hat a magnetic %orce %ield should look like:

Here are the easily)o$ser"ed %acts:


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agnetic !ieldThe geometry o% the magnetic %ield is set up to correctly model %orces $et#een currents that

allo# %or any relati"e orientation. The magnetic %ield intensity+ H+ circulates around its source+ I &+

in a direction most easily determined $y the right-hand rule: ,ight thum$ in the direction o% the

current+ %ingers curl in the direction o% H

-ote that in the third case perpendicular currents/+ I ' is in the same direction as H+ so that their

cross product and the resulting %orce/ is 0ero. The actual %orce computation in"ol"es a di%%erent

%ield 1uantity+ B+ #hich is related to H through B 2 µ 0H in %ree space. This #ill $e taken up in

a later lecture. 3ur immediate concern is ho# to %ind H %rom any gi"en current distri$ution.


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Biot)Sa"art 4a#

The Biot)Sa"art 4a# speci%ies the

magnetic %ield intensity+ H+ arising

%rom a 5point source6 current element

o% di%%erential length d L.

-ote in particular the in"erse)s1uare

distance dependence+ and the %act that

the cross product #ill yield a %ield "ector

that points into the page. This is a %ormal

statement o% the right)hand rule

Note the similarity to Coulomb’s Law+ in #hich

a point charge o% magnitude dQ& at oint & #ould

generate electric %ield at oint ' gi"en $y:

The units o% H are 8A9m


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agnetic !ield Arising !rom a Circulating Current

At point P, the magnetic %ield associated #ith

the di%%erential current element Id L is

The contri$ution to the %ield at P %rom any portion o% the current #ill $e ;ust the a$o"e integral e"alated

o"er ;ust that portion.

To determine the total %ield arising %rom the closed circuit path+

#e sum the contri$utions %rom the current elements that make up

the entire loop+ or


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T#o) and Three)


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E*ample o% the Biot)Sa"art 4a#

=n this e*ample+ #e e"aluate the magnetic %ield intensity on the y a*is e1ui"alently in the xy plane/

arising %rom a %ilament current o% in%inite length in on the z a*is.

>sing the dra#ing+ #e identi%y:

and so..

so that:


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E*ample: continued

We no# ha"e:

=ntegrate this o"er the entire #ire:

..a%ter carrying out the cross product


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E*ample: concluded

#e ha"e:

%inally:

Current is into the page.

agnetic %ield streamlines

are concentric circles+ #hose magnitudes

decrease as the in"erse distance %rom the z a*is

E"aluating the integral:


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!ield Arising %rom a !inite Current Segment=n this case+ the %ield is to $e %ound in the xy plane at oint '.

The Biot)Sa"art integral is taken o"er the #ire length:

..a%ter a %e# additional steps see ro$lem 7.?/+ #e %ind:


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Another E*ample: agnetic !ield %rom a

Current 4oopConsider a circular current loop o% radius a in the x-y plane+ #hich

carries steady current I. We #ish to %ind the magnetic %ield strength

any#here on the z a*is.

We #ill use the Biot)Sa"art 4a#:

#here:


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E*ample: Continued

Su$stituting the pre"ious e*pressions+ the Biot)Sa"art 4a# $ecomes:

carry out the cross products to %ind:

$ut #e must include the angle dependence in the radial

unit "ector:

#ith this su$stitution+ the radial component #ill integrate to 0ero+ meaning that all radial components #ill

cancel on the z a*is.


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E*ample: Continued

-o#+ only the z component remains+ and the integral

e"aluates easily:

-ote the %orm o% the numerator: the product o%

the current and the loop area. We de%ine this as

the magnetic moment :


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Ampere@s Circuital 4a#

Ampere@s Circuital 4a# states that the line integral o% H a$out any closed path

is e*actly e1ual to the direct current enclosed $y that path.

=n the %igure at right+ the integral o% H a$out closed paths a and b gi"es

the total current I, #hile the integral o"er path c gi"es only that portion

o% the current that lies #ithin c


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Ampere@s 4a# Applied to a 4ong Wire

ρ

Choosing path a, and integrating H around the circle

o% radius ρ gi"es the enclosed current+ I :

so that: as $e%ore.

Symmetry suggests that H #ill $e circular+ constant)"alued

at constant radius+ and centered on the current z / a*is.


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Coa*ial Transmission 4ine

=n the coa* line+ #e ha"e t#o concentric

solid conductors that carry e1ual and opposite

currents+ I.

The line is assumed to $e in%initely long+ and the

circular symmetry suggests that H #ill $e entirely

φ ) directed+ and #ill "ary only #ith radius ρ .

3ur o$;ecti"e is to %ind the magnetic %ield

%or all "alues o% ρ


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!ield Bet#een Conductors

The inner conductor can $e thought o% as made up o% a

$undle o% %ilament currents+ each o% #hich produces the

%ield o% a long #ire.

Consider t#o such %ilaments+ located at the same

radius %rom the z a*is+ ρ 1+ $ut #hich lie at symmetric

φ coordinates+ φ 1and )φ

1.Their %ield contri$utions

superpose to gi"e a net H φ component as sho#n. The

same happens %or e"ery pair o% symmetrically)located

%ilaments+ #hich taken as a #hole+ make up the entire

center conductor.

The %ield $et#een conductors is thus %ound to $e the same

as that o% %ilament conductor on the z a*is that carries current+ I. Speci%ically:

a


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!ield Within the =nner Conductor

With current uni%ormly distri$uted inside the conductors+ the H can $e assumed circular e"ery#here.

=nside the inner conductor+ and at radius ρ, #e again ha"e:

But no#+ the current enclosed is

so that or %inally:


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!ield 3utside Both Conducors

3utside the transmission line+ #here ρ c+

no current is enclosed $y the integration path+

and so

As the current is uni%ormly distri$uted+ and since #e

ha"e circular symmetry+ the %ield #ould ha"e to

$e constant o"er the circular integration path+ and so it

must $e true that:


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!ield =nside the 3uter Conductor

=nside the outer conductor+ the enclosed current consists

o% that #ithin the inner conductor plus that portion o% the

outer conductor current e*isting at radii less than ρ

Ampere@s Circuital 4a# $ecomes

..and so %inally:


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agnetic !ield Strength as a !unction o% ,adius in

the Coa* 4ine

Com$ining the pre"ious results+ and assigning dimensions as sho#n in the inset $elo#+ #e %ind:


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agnetic !ield Arising %rom a Current Sheet

!or a uni%orm plane current in the y direction+ #e e*pect an x)directed H %ield %rom symmetry.

Applying Ampere@s circuital la# to the path #e %ind:

or

=n other #ords+ the magnetic %ield is discontinuous across the current sheet $y the magnitude o% the sur%ace

current density.


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=% instead+ the upper path is ele"ated to the line $et#een and + the same current is enclosed and #e #ould ha"e

%rom #hich #e conclude that

By symmetry+ the %ield a$o"e the sheet must $e

the same in magnitude as the %ield $elo# the sheet.

There%ore+ #e may state that

and

so the %ield is constant in each region a$o"e and $elo# the current plane/


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The actual %ield con%iguration is sho#n $elo#+ in #hich magnetic %ield a$o"e the current sheet is

e1ual in magnitude+ $ut in the direction opposite to the %ield $elo# the sheet.

The %ield in either region is %ound $y the cross product:

#here a N is the unit "ector that is normal to thecurrent sheet+ and that points into the region in

#hich the magnetic %ield is to $e e"aluated.


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agnetic !ield Arising %rom T#o Current Sheets

Here are t#o parallel currents+ e1ual and opposite+ as you #ould %ind in a parallel)plate

transmission line. =% the sheets are much #ider than their spacing+ then the magnetic %ield

#ill $e contained in the region $et#een plates+ and #ill $e nearly 0ero outside.

K & 2 ) K y a y

K ' 2 ) K

ya

y

H x& z )d 9' /

H x& -d 9' z d 9' /

H x' -d 9' z d 9' /

H x' z )d 9' /

H x& z d 9' /

H x' z d 9' / These %ields cancel %or current sheets o%

in%inite #idth.

These %ields cancel %or current sheets o%

in%inite #idth.

These %ields are e1ual and add to gi"e

H 2 K x a N )d 9' z d 9' /

#here K is either K & or K '


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Current 4oop !ield

>sing the Biot)Sa"art 4a#+ #e pre"iously %ound the magnetic

%ield on the z a*is %rom a circular current loop:

We #ill no# use this result as a $uilding $lock

to construct the magnetic %ield on the a*is o%

a solenoid )) %ormed $y a stack o% identical current

loops+ centered on the z a*is.


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3n)A*is !ield Within a Solenoid

We consider the single current loop %ield as a di%%erential

contri$ution to the total %ield %rom a stack o% N closely)spaced

loops+ each o% #hich carries current I . The length o% the stack

solenoid/ is d, so there%ore the density o turns #ill $e N!d.

-o# the current in the turns #ithin a di%%erential length+ dz + #ill $e

z

-d! '

d! '

so that the pre"ious result %or H %rom a single loop:

no# $ecomes:

in #hich z is measured %rom the center o% the coil+

#here #e #ish to e"aluate the %ield.

We consider this as our di%%erential 5loop current6


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Solenoid !ield+ Continued

z

-d! '

d! '

The total %ield on the z a*is at z " #ill $e the sum o% the

%ield contri$utions %rom all turns in the coil )) or the integral

o% d H o"er the length o% the solenoid.


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Appro*imation %or 4ong Solenoids

z

-d! '

d! '

We no# ha"e the on)a*is %ield at the solenoid midpoint z 2 /:

-ote that %or long solenoids+ %or #hich + the

result simpli%ies to:

/

This result is "alid at all on)a*is positions deep #ithin long coils )) at distances %rom each end o% se"eral radii.


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Another =nterpretation: Continuous Sur%ace Current

The solenoid o% our pre"ious e*ample #as assumed to ha"e many tightly)#ound turns+ #ith se"eral

e*isting #ithin a di%%erential length+ dz . We could model such a current con%iguration as a continuous

sur%ace current o% density K 2 K a aφ A9m.

There%ore:

=n other #ords+ the on)a*is %ield magnitude near the center o% a cylindrical

current sheet+ #here current circulates around the z a*is+ and #hose length

is much greater than its radius+ is ;ust the sur%ace current density.

d! '

-d! '


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Solenoid !ield )) 3%%)A*is

To %ind the %ield #ithin a solenoid+ $ut o%% the z a*is+ #e apply Ampere@s Circuital 4a# in the %ollo#ing #ay:

The illustration $elo# sho#s the solenoid cross)section+ %rom a length#ise cut through the z a*is. Current in

the #indings %lo#s in and out o% the screen in the circular current path. Each turn carries current I. The magnetic

%ield along the z a*is is NI!d as #e %ound earlier.


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Application o% Ampere@s 4a#

Applying Ampere@s 4a# to the rectangular path sho#n $elo# leads to the %ollo#ing:

Where allo#ance is made %or the e*istence o% a radial H component+


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,adial ath Segments

The radial integrals #ill no# cancel+ $ecause they are oppositely)directed+ and $ecause in the long coil+

is not e*pected to di%%er $et#een the t#o radial path segments.


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Completing the E"aluation

What is le%t no# are the t#o z integrations+ the %irst o% #hich #e can e"aluate as sho#n. Since

this %irst integral result is e1ual to the enclosed current+ it must %ollo# that the second integral )) and

the outside magnetic %ield )) are 0ero.


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!inding the 3%%)A*is !ield

The situation does not change i% the lo#er z-directed path is raised a$o"e the z a*is. The "ertical

paths still cancel+ and the outside %ield is still 0ero. The %ield along the path # to $ is there%ore NI!d

as $e%ore.

Conclusion: The magnetic %ield #ithin a long solenoid is appro*imately constant throughout the coil

cross)section+ and is H z 2 NI!d .


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Toroid agnetic !ieldA toroid is a doughnut)shaped set o% #indings around a core material. The cross)section could $e

circular as sho#n here+ #ith radius a/ or any other shape.

Belo#+ a slice o% the toroid is sho#n+ #ith current

emerging %rom the screen around the inner periphery

in the positi"e z direction/. The #indings are modeled

as N indi"idual current loops+ each o% #hich carries current I.


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Ampere@s 4a# as Applied to a Toroid

Ampere@s Circuital 4a# can $e applied to a toroid $y taking a closed loop integral

around the circular contour % at radius ρ . agnetic %ield H is presumed to $e circular+

and a %unction o% radius only at locations #ithin the toroid that are not too close to the

indi"idual #indings. >nder this condition+ #e #ould assume:

Ampere@s 4a# no# takes the %orm:

so thatD.

er%orming the same integrals o"er contours dra#n

in the regions or #ill

lead to zero magnetic %ield there+ $ecause no current

is enclosed in either case.

This appro*imation impro"es as the density o% turns gets higher

using more turns #ith %iner #ire/.


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Sur%ace Current odel o% a Toroid

Consider a sheet current molded into a doughnut shape+ as sho#n.

The current density at radius crosses the xy plane in the z

direction and is gi"en in magnitude $y K a

Ampere@s 4a# applied to a circular contour % inside the

toroid as in the pre"ious e*ample/ #ill take the %orm:

leading toD

inside the toroidD. and the %ield is 0ero outside as $e%ore.


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Ampere@s 4a# as Applied to a Small Closed 4oop.

Consider magnetic %ield H e"aluated at the point

sho#n in the %igure. We can appro*imate the %ield

o"er the closed path &'F $y making appropriate

ad;ustments in the "alue o% H along each segment.

The o$;ecti"e is to take the closed path integral

and ultimately o$tain the point %orm o% Ampere@s 4a#.


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Appro*imation o% H Along 3ne Segment

Along path &)'+ #e may #rite:

#here:

And there%ore:


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Contri$utions o% y)


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Contri$utions o% x)


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-et Closed ath =ntegral

The total integral #ill no# $e the sum:

and using our pre"ious results+ the $ecomes:


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,elation to the Current


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3ther 4oop 3rientations

The same e*ercise can $e carried #ith the rectangular loop in the other t#o orthogonal orientations.

The results are:

4oop in yz plane:

4oop in xz plane:

4oop in xy plane:

This gi"es all three components o% the current density %ield.


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Curl o% a Gector !ield

The pre"ious e*ercise resulted in the rectangular coordinate representation o% the %url o% H.

=n general+ the curl o% a "ector %ield is another %ield that is normal to the original %ield.

The curl component in the direction N + normal to the plane o% the integration loop is:

The direction o% N is taken using the right)hand con"ention: With %ingers o% the right hand oriented

in the direction o% the path integral+ the thum$ points in the direction o% the normal or curl/.


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Curl in ,ectangular Coordinates

Assem$ling the results o% the rectangular loop integration e*ercise+ #e %ind the "ector %ield

that comprises curl H:

An easy #ay to calculate this is to e"aluate the %ollo#ing determinant:

#hich #e see is e1ui"alent to the cross product o% the del operator #ith the %ield:


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Curl in 3ther Coordinate Systems

Da little more complicated

4ook these up as neededD.


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Gisuali0ation o% Curl

Consider placing a small 5paddle #heel6 in a %lo#ing stream o% #ater+ as sho#n $elo#. The #heel

a*is points into the screen+ and the #ater "elocity decreases #ith increasing depth.

The #heel #ill rotate clock#ise+ and gi"e a curl component that points into the screen right)hand rule/.

ositioning the #heel at all three orthogonal orientations #ill yield measurements o%

all three components o% the curl. -ote that the curl is directed normal to $oth the %ield

and the direction o% its "ariation.


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Another a*#ell E1uation

=t has ;ust $een demonstrated that:

D..#hich is in %act one o% a*#ell@s e1uations %or static %ields:

This is Ampere@s Circuital 4a# in point %orm.


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D.and Another a*#ell E1uation

We already kno# that %or a static electric %ield:

This means that:

,ecall the condition %or a conser"ati"e %ield: that is+ its closed path integral is 0ero e"ery#here.

There%ore+ a %ield is conser"ati"e i% it has zero curl at all points o"er #hich the %ield is de%ined.

applies to a static electric %ield/


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Curl Applied to artitions o% a 4arge Sur%ace

Sur%ace & is paritioned into su$)regions+ each o% small area

The curl component that is normal to a sur%ace element can

$e #ritten using the de%inition o% curl:

or:

We no# apply this to e"ery partition on the sur%ace+ and add the resultsD.


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Adding the Contri$utions

.

Cancellation here:

We no# e"aluate and add the curl contri$utions

%rom all sur%ace elements+ and note that

ad;acent path integrals #ill all cancel

This means that the only contri$ution to the

o"erall path integral #ill $e around the outer

periphery o% sur%ace &.

No cancellation here:

>sing our pre"ious result+ #e no# #rite:


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Stokes@ Theorem

.

We no# take our pre"ious result+ and take the limit as

=n the limit+ this side

$ecomes the path integral

o% H o"er the outer perimeter

$ecause all interior paths

cancel

=n the limit+ this side

$ecomes the integral

o% the curl o% H o"er

sur%ace &

The result is Stokes@ Theorem

This is a "alua$le tool to ha"e at our disposal+ $ecause it gi"es us t#o #ays to e"aluate the same thing


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3$taining Ampere@s Circuital 4a# in =ntegral !orm+

using Stokes@ Theorem

Begin #ith the point %orm o% Ampere@s 4a# %or static %ields:

=ntegrate $oth sides o"er sur%ace & :

..in #hich the %ar right hand side is %ound %rom the le%t hand side

using Stokes@ Theorem. The closed path integral is taken around the

perimeter o% & . Again+ note that #e use the right)hand con"ention inchoosing the direction o% the path integral.

The center e*pression is ;ust the net current through sur%ace & +

so #e are le%t #ith the integral %orm o% Ampere@s 4a#:


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agnetic !lu* and !lu*


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A Iey roperty o% B

=% the %lu* is e"aluated through a closed sur%ace+ #e ha"e in the case o% electric %lu*+ auss@ 4a#:

=% the same #ere to $e done #ith magnetic %lu* density+ #e #ould %ind:

The implication is that %or our purposes/ there are no magnetic charges

)) speci%ically+ no point sources o magnetic ield exist . A hint o% this has already

$een o$ser"ed+ in that magnetic %ield lines al#ays close on themsel"es.


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Another a*#ell E1uation

We may re#rite the closed sur%ace integral o% B using the di"ergence theorem+ in #hich the

right hand integral is taken o"er the "olume surrounded $y the closed sur%ace:

Because the result is 0ero+ it %ollo#s that

This result is kno#n as auss@ 4a# %or the magnetic %ield in point %orm.


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a*#ell@s E1uations %or Static !ields

We ha"e no# completed the deri"ation o% a*#ell@s e1uations %or no time "ariation. =n point %orm+ these are:

auss@ 4a# %or the electric %ield

Conser"ati"e property o% the static electric %ield


auss@ 4a# %or the agnetic !ield

#here+ in %ree space:

Signi%icant changes in the a$o"e %our

e1uations #ill occur #hen the %ields are

allo#ed to "ary #ith time+ as #e@ll see later.


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a*#ell@s E1uations in 4arge Scale !orm

The di"ergence theorem and Stokes@ theorem can $e applied to the pre"ious %our point %orm e1uations

to yield the integral %orm o% a*#ell@s e1uations %or static %ields:

auss@ 4a# %or the electric %ield

Conser"ati"e property o% the static electric %ield


auss@ 4a# %or the magnetic %ield


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E*ample: agnetic !lu* Within a Coa*ial 4ine

d B

Consider a length d o% coa*+ as sho#n here. The magnetic %ield strength $et#een conductors is:

and so:

The magnetic %lu* is no# the integral o% B o"er the

%lat sur%ace $et#een radii a and b+ and o% length d along z :

The result is:

The coa* line thus 5stores6 this amount o% magnetic %lu* in the region $et#een conductors.

This #ill ha"e importance #hen #e discuss inductance in a later lecture.


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Scalar agnetic otential

We are already %amiliar #ith the relation $et#een the scalar electric potential and electric %ield:

So it is tempting to de%ine a scalar magnetic potential such that:

This rule must $e consistent #ith a*#ell@s e1uations+ so there%ore:

But the curl o% the gradient o% any %unction is identically 0ero There%ore+ the scalar magnetic potential

is "alid only in regions #here the current density is 0ero such as in %ree space/.

So #e de%ine scalar magnetic

potential #ith a condition:


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!urther ,e1uirements on the Scalar agnetic otential

The other a*#ell e1uation in"ol"ing magnetic %ield must also $e satis%ied. This is:

in %ree space

There%ore:

..and so the scalar magnetic potential satis%ies 4aplace@s e1uation again #ith the restriction

that current density must $e 0ero:


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E*ample: Coa*ial Transmission 4ine

With the center conductor current %lo#ing out o% the screen+ #e ha"e

Thus:

So #e sol"e:

.. and o$tain:

#here the integration constant has $een set to 0ero


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Am$iguities in the Scalar otential

The scalar potential is no#:

#here the potential is 0ero at

At point P / the potential is

But #ait As increases to

#e ha"e returned to the same physical location+ and

the potential has a ne# "alue o% ) I.

=n general+ the potential at P #ill $e multi"alued+ and #ill

ac1uire a ne# "alue a%ter each %ull rotation in the xy plane:


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3"ercoming the Am$iguity

Barrier at

To remo"e the am$iguity+ #e construct a mathematical $arrier at any "alue o% phi. The angle domain

cannot cross this $arrier in either direction+ and so the potential %unction is restricted to angles on either

side. =n the present case #e choose the $arrier to lie at so that

The potential at point P is no# single)"alued:


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Gector agnetic otential

We make use o% the a*#ell e1uation:

.. and the %act that the di"ergence o% the curl o% any "ector %ield is identically 0ero sho# this/

This leads to the de%inition o% the magnetic 'ector potential, A:

Thus:

and Ampere@s 4a# $ecomes


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E1uation %or the Gector otential

We start #ith:

Then+ introduce a "ector identity that de%ines the 'ector (aplacian:

>sing a lengthy/ procedure see Sec. 7.7/ it can $e pro"en that

We are there%ore le%t #ith


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The


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E*pressions %or otential

Consider a di%%erential elements+ sho#n here. 3n the le%t is a point charge represented $y a di%%erential length o% line charge. 3n the right is a di%%erential current element. The setups

%or o$taining potential are identical $et#een the t#o cases.

Line Charge Line Current

Scalar Electrostatic otential Gector agnetic otential


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eneral E*pressions %or Gector otential

!or large scale charge or current distri$utions+ #e #ould sum the di%%erential

contri$utions $y integrating o"er the charge or current+ thus:

and

The closed path integral is taken $ecause the current must

close on itsel% to %orm a complete circuit.

!or sur%ace or "olume current distri$utions+ #e #ould ha"e+ respecti"ely:

or

in the same manner that #e used %or scalar electric potential.


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E*ample

We continue #ith the di%%erential current element as sho#n here:

=n this case

$ecomes at point P :

-o#+ the curl is taken in cylindrical coordinates:

This is the same result as %ound using the Biot)Sa"art 4a# as it should $e/

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