Infinite Cylinder (Mit)

MASSACHUSETTS INSTITUTE OF TECHNOLOGYPhysics Department

Physics 8.07: Electromagnetism II November 21, 2012Prof. Alan Guth

QUIZ 2 SOLUTIONSQUIZ DATE: NOVEMBER 15, 2012

PROBLEM 1: THE MAGNETIC FIELD OF A SPINNING, UNIFORMLYCHARGED SPHERE (25 points)

This problem is based on Problem 1 of Problem Set 8.

A uniformly charged solid sphere of radius R carries a total charge Q, and is setspinning with angular velocity about the z axis.

(a) (10 points) What is the magnetic dipole moment of the sphere?

(b) (5 points) Using the dipole approximation, what is the vector potential A(r ) at largedistances? (Remember that A is a vector, so it is not enough to merely specify itsmagnitude.)

(c) (10 points) Find the exact vector potential INSIDE the sphere. You may, if you wish,make use of the result of Example 5.11 from Griffiths book. There he considered aspherical shell, of radius R, carrying a uniform surface charge , spinning at angularvelocity directed along the z axis. He found the vector potential

0R r sin , (if r 3

R)A(r, , ) =

0R4 (1.1) sin , (if r

3 2 R) .

r

PROBLEM 1 SOLUTION:

(a) A uniformly charged solid sphere of radius R carries a total charge Q, hence it hascharge density = Q/( 4R3

3). To find the magnetic moment of sphere we can divide

the sphere into infinitesimal charges. Using spherical polar coordinates, we can takedq = d = r2 dr sin d d, with the contribution to the dipole moment given bydm = 1r2 J d . One method would be to write down the volume integral directly,using J = v = r. We can, however, integrate over before we start, so we arebreaking the sphere into rings, where a given ring is indicated by its coordinates rand , and its size dr and d. The volume of each ring is d = 2r2 dr sin d. Thecurrent dI in the ring is given by dq/T , where T = 2/ is the period, so

dq ddI = = = r2 dr sin d . (1.2)

T 2

8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 2

The magnetic dipole moment of each ring is then given by

1dm ring =

1r J d = dI

r d = dI(r2 sin2 ) z . (1.3)

2 ring 2 ring

The total magnetic dipole moment is then

m =

r2 sin (r2 sin2 ) dr d z

= R

r4 dr0

(1

0

cos2 ) sin d z

Q R5 4 1= 4 = QR

2 z . (1.4)R3 5 3 53

(b) The vector potential in dipole approximation is,

0 m r 0 |m siA = =4 r3

| n

= 0

QR2 sin . (1.5)

4 r2 4 5 r2

(c) To calculate the exact vector potential inside the sphere, we split the sphere intoshells. Let r be the integration variable and the radius of a shell, moreover letdr denote the thickness of the shell. Then we can use the results of Example 5.11(pp. 236-37) in Griffiths, if we replace by its value for this case. The value of isfound equating charges

(4r2Q

) = 4 (4r2)dr (1.6)

R33

and therefore we must replace

Q dr .4 R33

Making this replacement in Griffiths Eq. (5.67), quoted above as Eq. (1.1), we nowhave

rr if r < rQ

dA(r, , ) =0

dr 44 sin

R3 33

r (1.7)

if r > rr2

.

Note that the R of Griffiths has been replaced

by r, which is the radius of the

integration shell. Now we can calculate the vector potential inside the sphere at


some radius r < R. The integration will require two pieces, a piece where 0 < r < rand the other where r < r < R, thus using the two options in Eq. (1.7):

Q r 4A(

0 rR

r, , ) = sin dr + drrr . (1.8)4 R3

[r20

r

]

Doing the integrals one finds

0 Q[ 3r3 rR2

A(r, , ) = sin 4 R3 +

10 2

]. (1.9)

PROBLEM 2: SPHERE WITH VARIABLE DIELECTRIC CONSTANT (35points)

A dielectric sphere of radius R has variable permittivity, so the permittivity throughoutspace is described by

r)20(R/ if r < R(r) =

{(2.1)

0 , if r > R .

There are no free charges anywhere in this problem. The sphere is embedded in a constantexternal electric field E = E0z, which means that V (r ) E0r cos for r R.(a) (9 points) Show that V (r ) obeys the differential equation

d ln V2V + = 0 . (2.2)dr r

(b) (4 points) Explain why the solution can be written as

V (r, ) =

V(r) zi1 . . . zi ri1 . . . ri , (2.3a)

=0

{ }

or equivalently (your choice)

V (r, ) =

V(r)P(cos ) , (2.3b)

=0

where { . . .} denotes the traceless symmetric part of . . . , and P(cos ) is the Leg-endre polynomial. (Your answer here should depend only on general mathematicalprinciples, and should not rely on the explicit solution that you will find in parts (c)and (d).)


(c) (9 points) Derive the ordinary differential equation obeyed by V(r) (separately forr < R and r > R) and give its two independent solutions in each region. Hint: theyare powers of r. You may want to know that

dd

(dP

sin (cos )

d

)= (+ 1) sin P(cos ) . (2.4)

The relevant formulas for the traceless symmetric tensor formalism are in the formulasheets.

(d) (9 points) Using appropriate boundary conditions on V (r, ) at r = 0, r = R, andr , determine V (r, ) for r < R and r > R.

(e) (4 points) What is the net dipole moment of the polarized sphere?

PROBLEM 2 SOLUTION:

(a) Since we dont have free charges anywhere,

D = ( E),= E ( ) + E = 0 . (2.5)

dThe permittivity only depends on r, so we can write = er. Then putting thisdrresult into Eq. (2.5) with E = V , we find

d0 = ( V ) er + dr

2V

V d 1= +2V

r dr

V d ln = 0 = +

r dr2V . (2.6)

(b) With an external field along the z-axis, the problem has azimuthal symmetry, imply-ing V/ = 0, so V = V (r, ). The Legendre polynomials P(cos ) are a completeset of functions of the polar angle for 0 , implying that at each value ofr, V (r, ) can be expanded in a Legendre series. In general, the coefficients may befunctions of r, so we can write

V (r, ) =

V(r)P(cos ) . (2.7)

=0


The same argument holds for an expansion in { zi1 . . . zi } ri1 . . . ri , since these arein fact the same functions, up to a multiplicative constant. Note that if dependedon as well as r, then the completeness argument would still be valid, and it wouldstill be possible to write V (r, ) as in Eqs. (2.3). In that case, however, the equationsfor the functions V(r) would become coupled to each other, making them much moredifficult to solve.

d ln 2(c) For r < R we have = . Using the hint, Eq. (2.4) in the problem statement,

dr rwe write

2 V d ln [ 1 2 V dV 2 (+ 1) V + = P r +

r d (cos ) V = 0 .

r r2 r

(r

)dr

r

)

r2=0

( ](2.8)

For this equation to hold for all r < R and for all , the term inside the squarebrackets should be zero. (To show this, one would multiply by P(cos ) sin andthen integrate from = 0 to = 2. By the orthonormality of the Legendrepolynomials, only the = term would survive, so it would have to vanish for every.) Thus,

1 (

V 2 r2

V)

d (+ 1) d2V (+ 1)+ V (2.9)

r r d =

r

( V = 0 .

r2 r

)

r2 dr2

r2

The general solution to Eq. (2.9) is

V (r) = A r+1B

+. (2.10)

r

(This can be verified by inspection, but it can also be found by assuming a trialfunction in the form of a power, V rp. Inserting the trial function into thedifferential equation, one finds p(p1) = (+1) . One might see by inspection thatthis is solved by p = + 1 or p = , or one can solve it as a quadratic equation,finding

1p =

(2+ 1)= + 1 or

2 .)

For r > R,1 ( + 1)r2 r

The general solution to Eq. (2.11)

( Vr2

r

) V = 0. (2.11)r2

is,

DV ( ) = r Cr + . (2.12)

r+1


(d) The coefficients B are zero, B = 0, to avoid a singularity at r = 0. The potentialgoes as V (r) = E0r cos for r R; this gives C = 0 except for C1 = E0. Thepotential V (r, ) is continuous at r = R, implying that

+1AR D= for = 1R+1 (2.13)

D1A1R

2 = E0R+ for = 1 .R2

In addition, the norma

l component of the displacement vector is continuous on the

boundary of the sphere. Since is continuous at r = R, this means that Er =V/r is continuous, which one could also have deduced from Eq. (2.2), since anydiscontinuity in V/r would produce a -function in 2V/r2. Setting V/r atr = R equal to its value at r = R+, we find (+ 1)A R D= (+ 1) for = 1

R+2 (2.14) D 2A1R = 2 13 E0 for = 1 .RSolving Eq. (2.13) and Eq. (2.14) as two equations (for each ) for the two unknownsA and D, we see that A = D = 0 for = 1, and that

3E 3A1 = 0 E, C1 = E , and 0R0 D =4 1 . (2.15)R 4

Then we find the potential as

3E 0r2 cos for r < R

V (r, ) = 4RE0 cos

(R3

(2.16) r fo

4r

)r r < R .

2

(e) Eq. (2.16) tells us that for r > R, the potential is equal to that of the applied externalfield, Vext = E0r cos , plus a term that we attribute to the sphere:

EVsphere(

0R3

r, ) = cos . (2.17)4r2

This has exactly the form of an electric dipole,

1 p rVdip =

, (2.18)

4 r20if we identify

p = 0R3E0 z . (2.19)


PROBLEM 3: PAIR OF MAGNETIC DIPOLES (20 points)

Suppose there are two magnetic dipoles. One has dipole moment m 1 = m0z andis located at r = +11 a z2 ; the other has dipole moment m 2 = m0z, and is located atr 2 = 1a z2 .(a) (10 points) For a point on the z axis at large z, find the leading (in powers of 1/z)

behavior for the vector potential A(0, 0, z) and the magnetic field B(0, 0, z).

(b) (3 points) In the language of monopole ( = 0), dipole ( = 1), quadrupole ( = 2),octupole ( = 3), etc., what type of field is produced at large distances by thiscurrent configuration? In future parts, the answer to this question will be called awhatapole.

(c) (3 points) We can construct an ideal whatapole a whatapole of zero size bytaking the limit as a 0, keeping m0an fixed, for some power n. What is the correctvalue of n?

(d) (4 points) Given the formula for the current density of a dipole,

J 3dip(r ) = m r (r rd) , (3.1)where r d is the position of the dipole, find an expression for the current densityof the whatapole constructed in part (c). Like the above equation, it should beexpressed in terms of -functions and/or derivatives of -functions, and maybe evenhigher derivatives of -functions.

PROBLEM 3 SOLUTION:

(a) For the vector potential, we have from the formula sheet that

A(r ) = 0m r

, (3.2)4 r2

which vanishes on axis, since m = m0z, and r = z on axis. Thus,

A(0, 0, z) = 0 . (3.3)

This does not mean that B = 0, however, since B depends on derivatives of A withrespect to x and y. From the formula sheet we have

0 3(m r)r mBdip(r ) = , (3.4)4 r3where we have dropped the -function because we are interested only in r = 0.Evaluating this expression on the positive z axis, where r = z, we find

0 2m0z Bdip(0, 0, z) = =0 m0z

. (3.5)4 r3 2 r3


For 2 dipoles, we have

mB2 dip(0, 0, z) =0 0

[1 1

z3 3]

2 z 1a z + 1a2 2

= 0m0

( ) ( )

2z3

[1 1(

1 1 a2 z)3

]z

0m0 1

(1 + 31 a2 z

1

) z

2z3

[(1

3 a2 z

) (1 + 3 a

2 z

]

) 0m0

[(3 a

(3 a

1 +) 1

)]z

2z3 2 z 2 z 0m0

[ a3

]z

2z3 z

3= 0

m0az . (3.6)

4z4

(b) Since it falls off as 1/z4, it is undoubtedly a quadrupole ( = 2) . For either the E

or B fields, the monopole falls off as 1/r2, the dipole as 1/r3, and the quadrupole as1/r4.

(c) We wish to take the limit as a 0 in such a way that the field at large z approachesa constant, without blowing up or going to zero. From Eq. (3.6), we see that thisgoal will be accomplished by keeping m0a fixed, which means n = 1 .

(d) For the two-dipole system we add together the two contributions to the currentdensity, using the appropriate values of r d and m :

J2 dip(r ) = m0z r 3 r az 02 +m z r 3 r az .2 (3.7)

Rewriting,

( ) ( )

3(r + a z) 3(r a z) J2 dip(r ) = m0az 2 2r[

a

]. (3.8)

Now we can define Q m0a, and if we take the limit a 0 with Q fixed, the aboveexpression becomes

J2 dip(r ) = Qz r 3(r ) . (3.9)z


Since partial derivatives commute, this could alternatively be written as

J2 dip(r ) = Qz r 3(r ) . (3.10)z

PROBLEM 4: UNIFORMLY MAGNETIZED INFINITE CYLINDER (10points)

Consider a uniformly magnetized infinite circular cylinder, of radius R, with its axiscoinciding with the z axis. The magnetization inside the cylinder is M = M0z.

(a) (5 points) Find H(r ) everywhere in space.

(b) (5 points) Find B(r ) everywhere in space.

PROBLEM 4 SOLUTION:

(a) The magnetization inside the cylinder is M = M0z. The curl of the H(r) field is

H(r) = Jfree = 0 , (4.1)and the divergence is

) ((

B(r 1) = ( H r M r)

)= B M = 0 . (4.2)

0 0

Note that for a finite length cylinder, the divergence would be nonzero because of theabrupt change in M at the boundaries. Since H(r ) is divergenceless and curl-free,we can say

H(r ) = 0 everywhere in space. (4.3)

(b) Having H(r ) = 0 everywhere in space, we can find magnetic field as

B(r ) M z for r < R , ( ) = ( r B( 0 0H M r ) = 0 = r ) = (4.4)0

{0 for r > R .

In this question we could alternatively find the bound currents as Jb = M = 0 and

Kb = M n = M 0. Then, using Amperes law as we did for a solenoid, we could findthe magnetic field and then also H, obtaining the same answers as above.


PROBLEM 5: ELECTRIC AND MAGNETIC UNIFORMLY POLARIZEDSPHERES (10 points)

Compare the electric field of a uniformly polarized sphere with the magnetic field ofa uniformly magnetized sphere; in each case the dipole moment per unit volume pointsalong z. Multiple choice: which of the following is true?

(a) The E and B field lines point in the same direction both inside and outside thespheres.

(b) The E and B field lines point in the same direction inside the spheres but in oppositedirections outside.

(c) The E and B field lines point in opposite directions inside the spheres but in thesame direction outside.

(d) The E and B field lines point in opposite directions both inside and outside thespheres.

PROBLEM 5 SOLUTION:

E field of a uniformly B field of a uniformlypolarized sphere magnetized sphere

The answer is (c), E and B field lines point in opposite directions inside the spheres butin the same direction outside, as shown in the diagrams, which were scanned from thefirst edition of Jackson. Note that the diagram on the left shows clearly that E = 0at the boundary of the sphere, so it could not possibly be a picture of B. It is at leastvisually consistent with E = 0, or equivalently E d = 0 for any closed loop, as itmust be to describe an electrostatic field. The diagram on the right, on the other hand,shows clearly that B = 0, or equivalently B

d = 0, so it could not possibly be a

picture of an electrostatic field. It is at least qu

alitatively consistent with B = 0, as

it must be.

8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 11

MASSACHUSETTS INSTITUTE OF TECHNOLOGYPhysics Department

Physics 8.07: Electromagnetism II November 13, 2012Prof. Alan Guth

FORMULA SHEET FOR QUIZ 2, V. 2Exam Date: November 15, 2012

Some sections below are marked with asterisks, as this section is. The asterisksindicate that you wont need this material for the quiz, and need not understand it. It isincluded, however, for completeness, and because some people might want to make useof it to solve problems by methods other than the intended ones.

Index Notation:

A B = AiBi , ABi = ijkAjBk , ijkpqk = ipjq iqjpdetA = i1i2inA1,i1A2,i2 An,in

Rotation of a Vector:Ai = RijAj , Orthogonality: RijRik =

Tjk (R T = I)

j=1 j=2 j=3

i=1 cos sin 0Rotation about z-axis by : Rz()ij = i=2 sin cos 0

i=3

0 0 1

Rotation about axis n by :

R(n, )ij = ij cos+ ninj(1

cos) ijknk sin .

Vector Calculus:

Gradient: ( )i = i , i xi

Divergence: A iAiCurl: ( A)i = ijkjAk

Laplacian: 2

2

= ( ) =xixi

Fundamental Theorems of Vector Calculus:

Gradient: b

d = (b)a

(a)

Divergence:

Ad3x =V

A

S

da

where S is the boundary of V

Curl:

(S

A) da =where P is th

A

P

de boundary of S


Delta Functions:(x)(x x) dx = (x) ,

(r )3(r r ) d3x = (r )

d d(x) (x x) dx =

dx

dx

(

x=x

x x( i

)g(x)) =

, g(x 0|g i) =(xi)

i

|

(

r r = 2

1= 43(r r )|r r 3

) | |r r |(

rj) (xj ) 1 ij 3r r 4

3i = i ji ij = + ij (r)r2 r3

(r

)r3 3

3(d r)r d 8= (

r3 d

3 )3(r )

3(d r) r d 4 = dr3 3

3(r )

Electrostatics:

F = qE , where1 (r r ) qi 1 (r r )E(r ) = =

r r |3 (r) d3x

40 | 40 | 3r ri |0 =permittivity of free space = 8.854 1012 C2/(Nm2)1

= 8.988 109 Nm2/C240 r 1 (r )V (r ) = V (r 0) E(r ) d = d3x

r 0 40 |r r |

E = 0

E = 0 , , E = V

2 V = (Poissons Eq.) , = 0 = 2V = 0 (Laplaces Eq.)0

Laplacian Mean Value Theorem (no generally accepted name): If 2V = 0, then

the average value of V on a spherical surface equals its value at the center.

Energy:

1 1 qiqj 1 1 3 3 (r )(r )W = = d x d x2 40 rij 2 40

iji=j

|r r |

1 1W =

d3 2x(r)V (r ) =

2 2 0

E d3x


Conductors:

Just outside, E = n0

Pressure on surface: 12 | E|outsideTwo-conductor system with charges Q and Q: Q = CV , W = 1CV 2

2

N isolated conductors:

Vi =

PijQj , Pij = elastance matrix, or reciprocal capacitance matrixj

Qi =

CijVj , Cij = capacitance matrixj

a a2Image charge in sphere of radius a: Image of Q at R is q = Q, r =

R R

Separation of Variables for Laplaces Equation in Cartesian Coordinates:

V ={cosx

}{cosy

}{cosh z

}where 2 = 2 + 2sinx siny sinh z

Separation of Variables for Laplaces Equation in Spherical Coordinates:

Traceless Symmetric Tensor expansion:

2 1 1(r, , ) =r2 r

(r2

r

)+

r22 = 0 ,

where the angular part is given by

2 1 1 2

sin +sin (

)sin2 2

2 () = ( + 1) () Ci i ...i ni1ni2 . . . ni Ci i ...i ni1 ni i1 2 . . . n2 1 2 ,where ()Ci1i2...i is a symmetric traceless tensor and

n = sin cos e1 + sin sin e2 + cos e3 .

General solution to Laplaces equation:

(( ) =

()

() C + i1i2...iV r Ci i ...i r

)r

1 2 ir+1 1ri2 . . . ri , where r = rr

=0


Azimuthal Symmetry:

V (r ) =(

A r B+

){ zi1 . . . zi rir } ri1 . . .+1

=0

where { . . .} denotes the traceless symmetric part of . . . .Special cases:

{ 1 } = 1{ zi } = zi{ zizj } = zizj 1ij3{ zizj zk } = zizj zk 1 zijk + zjik + zkij5{ zizj zkzm } = ziz 1j zk zm

( (zizjkm + zizk

)mj + zizmjk + zjzk7 im

+ zj zmik + zk zm 1ij + ijkm jk35 + ikjm + im

Legendre Polynomial / Spherical H

)armo

(nic expansion:

)

General solution to Laplaces equation:

V (r ) = ( Bm

Am r + Ym(, )

r+1=0 m=

)

Orthonormality: 2

d

sin d Ym (, Y ) m(, ) = mm

0 0

Azimuthal Symmetry:

V (r ) =(

A rB

+

r+1=0

)P(cos )

Electric Multipole Expansion:

First several terms:1 [ Q p r 1 r

V (r ) = + + irj

Q4 r r2 2 ij

+ , wherer30

]

Q =

d3x (r) , p =

d3 3i x (r ) xi Qij =

d x (r)(3x 2ixjij |r | ) ,

1 p r 1 3(p r)r p 1 ( Edip r ) =

( )= r

2

p 3(4 r 4 i

) r3

0 0

30

1 1 Edip( r ) = 0 , Edip(r ) = dip(r ) =0

p 0

3(r )


Traceless Symmetric Tensor version:

1 V (r ) =

1 ()Ci ...i ri1 . . . ri ,4 +1 10 r

=0

where

() (2Ci ...i =

1)!! (r ) { x d3i1 . . . xi } x (r1 ! rr xiei)

1 (2 1)!! r=

{ ri1 . . . ri } r . . . r , for r < r|r r | ! r+1 i1 i=0

(2)!(2 1)!! (2 1)(2 3)(2 5) . . .1 = , with (

2!1)!! 1 .

Reminder: { . . .} denotes the traceless symmetric part of . . . .

Griffiths version:

1 V (r ) =

1 r (r )P (c 34

os ) d x0 r+1

=0

where = angle between r and r .

1 =

r< 1P(cos ) , = P x|r r | r+1 )> 1=0 2x+ (2 =01 d

P(x) = (x2 1) , (Rodrigues formula)2!

(dx

)

1

P(1) = 1 P(x) = (1) 2P(x)

dxP(x)P(x) = 1 2+ 1

Spherical Harmonic version:

1 4 qmV (r ) = Ym(, )40 2+ 1 r+1

=0 m=

where q =

Y r 3m m (r) d x

1 4 r= Ym

(, )Ym(, ) , for r < r|r r | 2+ 1 r+1

=0 m=


Electric Fields in Matter:

Electric Dipoles:

p =

d3x (r )r

dip(r ) = p r 3(r rd) , where rd = position of dipoleF = (p ) E = (p E) (force on a dipole) = p E (torque on a dipole)U = p E

Electrically Polarizable Materials:P (r ) = polarization = electric dipole moment per unit volume

bo = und P , bound = P nD 0E + P , D = free , E = 0 (for statics)Boundary conditions:

Eab

ove Ebe low = Dab ove0 Dbe

low = free

E E = 0 D D above below above below = Pabove P below

Linear Dielectrics:P = 0eE, e = electric susceptibility 0(1 + e) = permittivity, D = E

r = = 1 + e = relative permittivity, or dielectric constant

0

N/Clausius-Mossotti equation: 0e = ,N where N = number density of atoms1

30

or (nonpolar) molecules, = atomic/molecular polarizability (P = E)

1Energy: W =

D E

2 d3x (linear materials only)

Force on a dielectric: F = W (Even if one or more potential differences areheld fixed, the force can be found by computing the gradient with the totalcharge on each conductor fixed.)

Magnetostatics:

Magnetic Force:dp 1 = ( + F q E v B) = , where p = m0v , =dt 1 v2c2


F =

d B =

I J B d3x

Current Density:

Current through a surface S: IS =

JS

da

Charge conservation: = J

t

Moving density of charge: J = v

Biot-Savart Law:

0

d (r r ) K(r B(r ) = 0 ) (r r)

I = da4 |r r |3

4

|r r |3

= 0

J(r ) (r r )d3x

4 |r r |3

where 0 = permeability of free space 4 107 N/A2

Examples: I

Infinitely long straight wire: B = 0 2r

Infintely long tightly wound solenoid: B = 0nI0 z , where n = turns perunit length

IR2Loop of current on axis: B(0, 0, z) = 0 z

2(z2 +R2)3/2

1Infinite current sheet: B(r ) = 0K n , n = unit normal toward r2

Vector Potential:

( ) = 0J

co

(r )

A r ul d3 , B = x A ,4 |r r A| coul = 0

B = 0 (Subject to modification if magnetic monopoles are discovered)Gauge Transformations: A( r ) = A(r ) + (r ) for any (r ). B = A is

unchanged.

Amperes Law:

B = 0J , or equivalently

BP

d = 0Ienc


Magnetic Multipole Expansion:

Traceless Symmetric Tensor version:

Aj(r ) =0

4

{

M() ri1 . . . rij;i1i2...i}

r+1=0

where M() (2j;i1i2...i = 1)!!

( . . .!

d3xJj r ){ xi1 xi }

Current conservation restriction:

d3x Sym(xi1 . . . xi Ji) = 01i1...i

where Sym means to symmetrize i.e. average over alli1...i

ordering

s in the indices i1 . . . iSpecial cases:

= 1:

d3x Ji = 0

= 2:

d3x (Jixj + Jjxi) = 0

Leading term (dipole): ( ) = 0

m A r

r,

4 r2where

1mi = 2 ijkM

(1)j;k

1

1m = I r d =

d3xr

2 P 2 J = Ia ,

where a =

da for any surface S spanning PS

0 m r 0 3(m r)r m 2B ( ) = = + 0dip r m 3(r )4 r2 4 r3 3 Bdip(r ) = 0 , Bdip(r ) = 0Jdip(r ) = 0m 3(r )

Griffiths version:

( ) = 0I

1 ( ) (cos )dA r r P

4

r+1=0

Magnetic Fields in Matter:

Magnetic Dipoles:

1m = I

2

1

rP

d =2

d3xr J = Ia


Jdip(r ) = m r 3(r r d), where r d = position of dipole F = (m B) (force on a dipole) = m B (torque on a dipole)U = m B

Magnetically Polarizable Materials:M(r ) = magnetization = magnetic dipole moment per unit volumeJbound = M , Kbound = M n

1H B M , H = Jfree , B = 00

Boundary conditions:Bab

ove Bbe low = 0 Hab ove Hbe low = (Mab ove Mbe low)

Babove Bbelow = 0(K n) Habove Hbelow = Kfree nLinear Magnetic Materials:

M = mH, m = magnetic susceptibility = 0(1 + m) = permeability, B = H

Magnetic Monopoles: ( ) = 0

qmB r = r ; Force on a static monopole: F q

4 mB

r2 q q

Angular momentum of monopole/charge system: = 0 e mL r , where r points4

from qe to qm q q 1

Dirac quantization condition: 0 e m = h4 2

integer

Connection Between Traceless Symmetric Tensors and Legendre Polynomialsor Spherical Harmonics:

(2)!P(cos ) = { zi1 . . . z2(!) i2 } ni1 . . . niFor m 0,

( ) = (,m)Ym , Ci ...i ni1 . . . ni ,1

where (,m)C = d { u+ . . . u+i1i2...i m i1 i zm im+1 . . . zi } ,(1)m(2)! + 1

with dm = 2

2m (2 )

,! 4 (+m)! (m)!

and u+1

= (ex + iey)2

Form m < 0, Y,m(, ) = (1)mYm (, )


More Information about Spherical Harmonics:2+ 1 (

Ym(, ) =m)!

Pm4 (+m)!

(cos )eim

where Pm (cos ) is the associated Legendre function, which can be defined bym

Pm( 1)m d+

(x) =

(1!

x2)m/2 (x2 1)2 dx+m

Legendre Polynomials:

Image by MIT OpenCourseWare.

SPHERICAL HARMONICS Ylm( , )

Y00 = 1

4

l = 1

l = 0

cos Y10 = 43

sin ei Y11 = - 83

l = 2 sin cosei Y21 = - 8

15

( cos2 Y20 = 45

sin2 e2i Y22 = 2151

4

32

1 2 )

l = 3

sin2 cos e2i Y32 = 2105

( cos3

Y31 = 421

sin3 e3i Y33 = - 4351

4

52

14

14

3 2 cos )

- sin (5cos2 -1)ei

Y30 = 47

MIT OpenCourseWarehttp://ocw.mit.edu

8.07 Electromagnetism IIFall 2012 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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