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8.02 Sample Final C Solutions (Modified from Spring 1993)
p. 1/5
1.
Displacement Current
(a) The energy stored is in the electric field.
Since E is
nearly constant we can just multiply the energy density
by the volume inside the capacitor:
R
2
2
0
2
2
0
2
2
E
d
R
d
R
E
V
u
U
E
E
π
ε
=
=
⋅
=
(b) With the electric field increasing, we have an upwards displacement current:
( )
P
at
page
of
out
2
1
2
0
0
2
0
0
0
2
0
2
0
0
dt
dE
r
B
dt
dE
r
I
r
B
dl
B
dt
dE
r
dt
r
E
d
dt
d
I
nt
displaceme
E
nt
displaceme
µ
=
⇒
=
=
⋅
=
⋅
=
=
Φ
=
∫
(c)
dt
dE
rE
dt
dE
r
E
B
E
S
0
0
0
0
0
2
1
2
1
1
=
=
×
=
G
G
G
(to the right/inwards!)
(d) To find the total energy flowing in consider that the band at
r
=
R
has an area
A
=
2
π
Rd
, so
()
dt
dE
dE
R
Rd
dt
dE
RE
A
R
r
S
dt
dU
2
0
0
2
2
1
=
⎟
⎠
⎞
⎜
⎝
⎛
=
⋅
=
=
G
G
.
Notice that this is
indeed the time derivative of U
E
that we calculated in part (a).
2.
Waves
E
= E(x,t)
;
B
= B(x,t)
, Faraday:
y
ˆ
z
ˆ
t
B
x
E
∂
∂
−
=
∂
∂
;
Ampere:
t
E
x
B
∂
∂
−
=
∂
∂
0
0
(a) Take the partial of Faraday’s law w.r.t. x and the time derivative of Ampere’s Law
and substitute:
2
2
0
0
2
2
2
2
0
0
2
2
2
t
E
x
E
t
E
t
x
B
x
E
∂
∂
=
∂
∂
⇒
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
−
=
∂
∂
∂
−
=
∂
∂
(b) To see if
E(x,t) = E
0
(axbt)
2
holds, plug it into the differential eqn. of (a):
0
0
2
0
2
0
0
0
2
0
2
0
0
0
0
0
2
2
0
0
?
0
2
0
2
2
2
2
:
Condition
2
2
2
2
=
⇒
=
=
−
−
∂
∂
=
∂
∂
=
=
−
∂
∂
=
∂
∂
b
a
E
b
E
a
E
b
bt
ax
bE
x
t
E
E
a
bt
ax
aE
x
x
E
(c) Faraday:
2
0
0
2
bt
ax
b
aE
B
t
B
bt
ax
aE
x
E
−
=
⇒
∂
∂
−
=
−
=
∂
∂
.
Check in Ampere:
bt
ax
bE
b
a
t
E
bt
ax
b
E
a
x
B
−
−
⎟
⎠
⎞
⎜
⎝
⎛
−
=
∂
∂
−
=
−
=
∂
∂
0
2
0
0
?
0
2
2
2
Yes!
Using the condition we derived in (b) we find this works.
(d) Putting this wave in a dielectric medium with dielectric constant
κ
e
changes Ampere’s
law (since
ε
0
becomes
κ
e
ε
0
).
The speed of propagation (
b
/
a
) decreases by
e
/
1
r
d
P
E
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View Full Document8.02 Sample Final C Solutions (Modified from Spring 1993)
p. 2/5
3.
Wave
B
= 10
7
sin [ x  3
·10
8
t]
Tesla
y
ˆ
(a) What is the wavelength
λ
of the wave?
m
k
π
λ
2
1
2
2
=
=
=
(b) What is the amplitude
E
0
of the associated magnetic field? E
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 Spring '07
 Hudson

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