Physics 505 Electricity and Magnetism Fall 2003Prof. G. Raithel

Problem Set 7

Maximal score: 25 Points

1. Jackson, Problem 5.1 6 Points

Consider the i-th cartesian component of the B-Field,

40I

B(x) xi =

S

xi [dl (x x

)|x x|

]

=

S

xi [dl x 1|x x|

]

=

S

dl [x 1|x x| xi

]| Stokes law

=

S

da {x

[x 1|x x| xi

]}

=

S

da {[x 1|x x|

](x xi) xi

[2x

1|x x|

]

+(xi x)[x 1|x x|

] (

[x 1|x x|

] x)xi

}use x 6= x always

=

S

da {

0 0 + (xi x)[x 1|x x|

] 0

}

=

S

da {

0 0 + (xi x)[x 1|x x|

] 0

}

=

S

da {

xi

[x 1|x x|

]}=

S

da {

xi

[x 1|x x|

]}

= xi

S

da {[x 1|x x|

]}| see Eqn. after 1.25 on page 33 of textbook

=

xi

S

d =

xi(x)

Thus, Bi = 0I4

xi(x), and

B(x) =0I

4x(x) q.e.d.

1

2. Jackson, Problem 5.3 6 Points

Consider a loop current with radius R around the z-axis. The loop is centered at location zz. Then, themagnetic field at an observation point zz is

B(0, 0, z) =0I

4

dl x x

|x x|3

Insert x =

R cos

R sin

z

and dl =

R sin R cos

0

d to find

B(0, 0, z) =0I

41

R2 + z2

3

20

Rz cos Rz sin

R2

d = z0I

2R2

R2 + z2

3

Now, consider a solenoid the axis of which coincides with the z-axis. The solenoid has N windings per length,current I, and end points z1 and z2. Then, the amount of current flowing across a length dz is dI = INdz,and

B(0, 0, z) =

z0dI

2R2

R2 + z2

3

= z0R

2IN

2

z2

z1

dz

R2 + z23

= z0R

2IN

2

[z

R2

R2 + z23

]z2

z1

= z0IN

2[cos 1 + cos 2] ,

with angles 1 and 2 as shown in the problem statement.

2

3. Jackson, Problem 5.8 7 Points

a): In the Coulomb gauge, 2A = 0J(r, ). Using a variable separation method, we construct asolution of the form A = A(r, ). (By finding the solution of that form, it is shown that it exists.) Thefollowing derivatives of spherical unit vectors are useful:

r = 0

= 0

r = sin

= cos

= r sin cos 2 = .

Thus, writing out 2(A(r, )

)= 0J(r, ) in spherical coordinates yields:

(1r2

rr2r +

1r2 sin

sin +1

r2 sin2 2

) (A(r, )

)= 0J(r, )

([1r2

rr2r +

1r2 sin

sin

]A(r, )

)

(1

r2 sin2 A(r, )

) = 0J(r, )

[1r2

rr2r +

1r2 sin

sin 1r2 sin2

]A(r, ) = 0J(r, )

This is a 2-nd order, linear, inhomogeneous PDE for A(r, ), similar to the Poisson equation, which issolvable. To identify the behavior inside and outside the current distribution, we solve the homogeneousequation by separation of variables. Writing A(r, ) =

U(r)r (), it is

r2 d2

r2 U(r)U(r)

+1

sin dd sin

dd () ()sin2

() = 0

= l(l + 1) = l(l + 1)

The angular equation is the generalized Legendre differential equation,

[1

sin d

dsin

d

d 1

sin2 + l(l + 1)

]() = 0 ,

which has the regular solution P 1l (cos ); note that P1l (cos ) is linearly dependent. This finding justifies a

posteriori that l(l + 1) with l = 1, 2... is a good choice for the separation variable. The radial equation,

d2

r2U(r) l(l + 1)

r2U(r) = 0

has the solution

3

U(r) = Alrl+1 + Blrl .

Summarizing, the interior and exterior solutions are found to be

A,interior =

l=1

AlrlP 1l (cos ) =

04

l=1

mlrlP 1l (cos )

A,exterior =

l=1

Blrl1P 1l (cos ) =

04

l=1

lrl1P 1l (cos ) q.e.d. (1)

There, we also define the multipole moments ml and l. Note that the{P 1l (x), l = 1, 2, 3..

}form a complete

orthogonal set on the interval [1, 1].

b): In analogy with electrostatics, spherical multipole moments are obtained by expanding 1|xx| in sphericalharmonics. For azimuthal current distributions it is, for an observation point with = 0,

A(r, ) =04

J(r, , ) |x x| d

3x =04

J(r, )[

]

|x x| d3x =

04

J(r, ) cos

|x x| d3x

=04

l,m

42l + 1

Ylm(, = 0)

rl

Y lm(, )

12

[exp(i) + exp(i)] J(r, )d3x

Upon integration over , only m = 1 give non-zero contributions. For each l, the m = 1 and m = 1terms are equal; to show this, use the fact that = 0, and Eqs. 3.51 and 3.53 of the textbook. Thus,

A(r, ) =04

l,m=1

42l + 1

2l + 1

4

(l 1)!(l + 1)!

P 1l (cos )

rl

2l + 1

4

(l 1)!(l + 1)!

P 1l (cos )

12J(r, )d3x 2

=04

l,m=1

1l(l + 1)

P 1l (cos )

rl

P 1l (cos )J(r, )d3x (2)

For the interior region, r< = r and r> = r, and

A,interior(r, ) = 04

l

{ 1

l(l + 1)

1

rl+1P 1l (cos

)J(r, )d3x}

rlP 1l (cos )

Comparison with Eq. 1 shows that

ml = 1l(l + 1)

1

rl+1P 1l (cos

)J(r, )d3x q.e.d.

4

Similarly, for the exterior region, r> = r and r< = r, and

A,interior(r, ) = 04

l

{ 1

l(l + 1)

rlP 1l (cos

)J(r, )d3x}

1rl+1

P 1l (cos )

Comparison with Eq. 1 shows that

l = 1l(l + 1)

rlP 1l (cos

)J(r, )d3x q.e.d.

5

4. Jackson, Problem 5.13 7 Points

There is an azimuthal surface current K() = sin a. The corresponding three-dimensional currentdensity is

j(r, ) = K()(r a) = sin a(r a) = J(r, ) .

Using Eq. 2 of the previous problem and 11 P

ml (x)P

ml (x)dx =

22l+1

(l+m)!(lm)!l,l and P

1l = sin , it is

A(r, ) =04

l,m=1

1l(l + 1)

P 1l (cos )

rl

P 1l (cos )J(r, )d3x

=04

l,m=1

1l(l + 1)

P 1l (cos )

rl

P 1l (cos ) sin (r a) a r2d cos d

= 04

l,m=1

1l(l + 1)

P 1l (cos )2a3rl

P 1l (x)P

11 (x)dx

= 0a3

4P 1l (cos )

r

43

=0a

3

3sin

r

Thus, it is outside the sphere

Aexterior(r, ) = 0a

4

31r2

sin

and inside

Ainterior(r, ) = 0a

3r sin

Using that for azimuthal A it is B = A = r 1r sin [sin A] 1r r [rA] it is found:

Bexterior(r, ) =0a

4

3

[r2 cos

r3+

sin r3

],

which is the field of a magnetic dipole m = z 4a4

3 , and

Binterior(r, ) =20a

3

[r cos sin

],

which is a homogeneous magnetic field in z-direction.

6