Phys 6561 HW 9 Solns.
November
=
IA c
(-^) y (3)
=m
(c) Since M (the total magnetization measured in the rest frame) involves a rest-frame volume element d3 r, but the polarization involves the lab frame volume element d3 r , we expect a factor d3 r/d3 r
HW 11 Sols. Phys 6561
November 29, 2011
1
Jackson 9.11
The charge density is = q(x)(y) [2(z) - (z - ) - (z + )] where = a cos 0 t and we are told to assume ka 1 so we can use equations 9.169 and 9.170 from Jackson. Sinc the magnetization is zero, Qm = Mm
HW 10 Solutions Phys 6561
November 28, 2011
1
Rotating fields into coincidence
E
We are given = E0 x ^ 2E0 (cos ^ + sin ^) x y (1)
B =
and want to make E |B . Note that any solution to this problem is not unique. Once we have found a frame with parallel
HW 6 solns. Phys 6561
October 16, 2011
1
Jackson 7.10
1 B c t 1 D c t
(a) No free currents, non permeable media E = -
B = Thus,
(1)
1 2D (2) c2 t2 Take E = E0 exp (i (k r - t) and D = E0 exp (i (k r - t) = D0 exp (i (k r - t). Substituting these definitio
HW 12 Solutions Phys 6561
December 4, 2011
1
Jackson 9.2
The charge density is given by (r - R) (cos ) ( - t) - ( - t - /2) + ( - t - ) - ( - t - 3/2) r2 (1) with R = a/ 2 distance center charges. Fourier transforming (1) with respect to t we get ^ 1 T in
Phys 6561 HW 7
November 1, 2011
1
Jackson 11.19
(a) The lab-frame is the center-of-mass frame, thus: P = (M, 0), p = (E1 , p1 ) 1 and p = (E2 , p2 ). 4-momentum conservation, P = p + p , gives M = 2 1 2 E1 + E2 E2 = M - E1 and p1 + p2 = 0 p1 = -p2 . We no
HW 5 Solutions Phys 6561
October 13, 2011
1
Jackson 7.19
u(x) = N e-
2
(b) The function is
x2 /4 ik0 x
e
(1)
The Fourier transform is defined by ^ ^ 2 2 1 N -ikx A(k) = e u(x)dx = e- x /4-i(k-k0 )x dx 2 - 2 - ^ N -(k-k0 )2 /2 -2 /4(x+2i(k-k0 )/2 )2 = e e
HW 4 Solutions Phys 6561
October 5, 2011
1
Jackson 7.4
Er Br kr air conductor kt Bt
Applying boundary conditions at the air conductor interface yields Ei + Er = Et E| continuous a (Ei - Er ) = c Et H| continuous a = /c
Ei Bi
ki
(1) (2)
a and c are given b
HW 3 Solutions Phys 6561
September 26, 2011
1
Problem 1
(1)
= 0, therefore the potential inside the cylinder will be G 1 (x) = - (x ) (x, x )d2 S 4 bdry n
Three forms of the Greens function were discussed in class and are listed in Jackson, problem 3.23.
Phys 6561 HW 2 Solutions
September 19, 2011
1
Jackson 3.1
0
V
a
0 b V
Given the Green's function for this geometry we know that ^ G 1 (x) = (x)G(x, x )d3 x - (x ) d2 x 4 n
(1)
In our case there is no interior charge distribution (x ) = 0, thus the volume
Physics 6561, Fall 2011. HW 1
September 6, 2011
1
Current carrying elements in empty space
W = 1 8 d 3 x B2 (1)
As suggested, we start with
and use B B = B ( A). We then employ the following vector identity B ( A) = (A B) + A ( B) to get, W = = 1 d3 x cfw
HW 8 solns. Phys 6561
October 20, 2011
1
Jackson 11.13
(a) Let's work in cylindrical coordinates (r, , z) with the wire along the z-axis. In K , B = 0. Get the E-field from Gauss's Law, using a Gaussian pillbox that is a cylinder of length L and radius r