PHY 5347
Midterm Exam, Problem 1, Solutions - D. Kimel
2 (50 points) The system is described by
a) We know from our work with periodic systems that the possible radiated frequencies are
integral multiples of the fundamental frequency g, which is the lowes
PHY 5347
Midterm Exam, Problem 1, Solutions - D. Kimel
1 (50 points) The system is described by
Let us summarize what we know about waveguides from the class notes and the textbook (we will
be assuming propation in the positive z direction everywhere): Th
PHY 5347
Final Exam
April 24, 2001
4. (a)
3 = ax cos g 0 t +
!
r
sin g 0 t
3
!
K = ag 0 ?x sin g 0 t +
c
cos g 0 t
ag 2
%
!
K = ? c 0 x cos g 0 t +
sin g 0 t
The problem gives us
2
dP = q
4^c
dI
%
!
!3
n n ? K K
!3
1 ? n 6 K 5
2
We can write the numerator
PHY 5346
Final Exam Solutions - D. Kimel
April 24, 2001
3.
a) We are given
3x
A3, t =
3
qKt v
3
!
1 ? Kt v 6 nRt v
t v = t?Rt v /c
As usual
3
43
4!
B rad = 3 A, where in the radiation zone, 3 = n /
/R
v
v
v
v
3 = ?n / . Therefore
!v
But from t = t ? Rt
PHY 5346
Final Exam Solutions - D. Kimel
April 24, 2001
2.
a)
x0
x
vW
=
x1
x2
x3
v
L
=
?KL 0 0
x0
Lx 0 ? KLx 1
?KLx 0 + Lx 1
?KL
L
00
x1
0
0
10
x2
0
0
01
x3
=
x2
x3
where K = v/c, and L = 1/ 1 ? K 2 .
b) Generaly for a boost along x with speed cK,
E v1 =
PHY 5346
Final Exam Solutions - D. Kimel
April 24, 2001
1. The system is described by
3!
a) We are given E = xE 0 sin ^y sin ^z e Tgt , and Maxwells equations read
a
a
3
3
3 6 E = 0, 3 6 H = 0, 3 H = P /E , 3 E = ?W /H . Thus we can use the last equation
PHY 5347
Homework Set 11 Solutions Kimel
2. 14.12
a) From Jackson, Eq (14.38)
dP = e 2
4^c
dI
%
!
!3
n n ? K K
!3
1 ? n 6 K 5
2
!
Using azimuthal symmetry, we can choose n in the x-z plane.
From the figure
!
!
!
n = cos Sz + sin Sx
3
!
!
Kt v = ? a g 0 si
PHY 5347
Homework Set 11 Solutions Kimel
1. (i)
3 = ax cos g 0 t +
!
r
sin g 0 t
3
!
K = ag 0 ?x sin g 0 t +
c
cos g 0 t
ag 2
%
!
K = ? c 0 x cos g 0 t +
sin g 0 t
From Jackson, Eq (14.38)
2
dP = q
dI
4^c
%
!
!3
n n ? K K
!3
1 ? n 6 K 5
2
We can write the
PHY 5347
Homework Set 9 Solutions Kimel
3. 12.14
a) We are given
L = ? 1 /JAK/JAK ? 1 JJAJ
c
8^
which can be rewritten
L = ? 1 /KAJ/KAJ ? 1 JJAJ
c
8^
Using the Euler-Lagrange equations of motion,
/K
/L
? /L = 0
/A J
/ K A J
Noting
/L
= ? 1 /KAJ
4^
/ K A
PHY 5347
Homework Set 9 Solutions Kimel
2. 12.5
a) The system is described by
3
3
3
3
3
Background: particle having m, e. Choose 3 to B and E. We want E v = 0 = LE + u B.
u
c
3
3
3
3
33
Thus u B = ?E; now B 3 B = cE B. Using BAC - CAB on the lhs of the eq
PHY 5347
Homework Set 6 Solutions Kimel
2. a) The system is described by
!
where the incident wave is polarized in the x direction. We showed in the class notes that in the
radiation zone (kr > 1,
ikr
3 xv
3x
! 3x
E3 = ie 3 X n E3v e ?ik63 da v
k
2^r
S
bu
PHY 5347
Homework Set 6 Solutions Kimel
1. 10.1
!
!
a) Let us first simplify the expression we want to get for the cross section. Using n 0 = z,
da P , n , n = k 4 a 6 5 ? |P 0 6 n |2 ? 1 |n 6 z P 0 |2 ? z 6 n
!
!
!
!
4
dI 0 0
4
Orienting the system as
an
PHY 5347
Homework Set 3 Solutions Kimel
3. 9.3
Since the problem has azimuthal symmetry, we can expand V3, t (in the radiation zone) in terms
r
of Legendre polynomials:
V3, t =
r
> b l tr ?l?1 Pl cos S
l
Using the orthogonality of the Legendre polynomials
PHY 5347
Homework Set 3 Solutions Kimel
3. 9.2 First consider a rotating charge which is at an angle J at time t = 0.
Compared to the lecture notes for this problem, where we assumed J = 0, we should let
gt gt + J. Thus using the result developed in class
PHY 5347
Homework Set 3 Solutions Kimel
3. 9.1 a)
_3, t = qNzNy ? sin g 0 tNx ? d cos gt
x
To illustrate the equivalence of the two methods, Ill consider the lowest two moments.
n = 0 : Qt =
n = 1 : 3t =
p
x
X _3, td 3 x = q = Reqe ?i06gt
xx
X _3, t3d 3
PHY 5347
Homework Set 2 Solutions Kimel
3. As we saw in class, the time-averaged power through the cylindrical waveguide in the TE mode
is
P=
W
2 PW
g
gV
2
1?
g2
V
g2
1/2
XA f D fda
where in this case g V = g v11 , given in the previous problem, and
f = H
PHY 5347
Homework Set 2 Solutions Kimel
2. As we saw in class and in problem 1 of Set #1, the lowest cutoff frequency for a cylindrical
waveguide (circular cross section of radius R) is the TE 11 mode. The cutoff frequency found in that
problem was
g v11