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 multi
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 assumin
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
%
!
!
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 /
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 .
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 =
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 f
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
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
/
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;
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
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 |
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
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
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 : 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
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