PHYS1200
PHYSICS II
SPRING 2005
HOMEWORK SOLUTIONS
CLASS 8.
30.02, 30.03, 30.06, 30.08, 30.22
30.02.
a)
mV
0
.
7
0
.
12
mWb]
)
0
.
7
0
.
6
[(
2
+
=
+
=
Φ
=
t
t
t
dt
d
dt
d
B
E
At
t
= 2.0 s,
0
.
7
)
0
.
2
(
0
.
12
+
=
E
mV
E
= 31 mV
b)
The flux is increasing, so Lenz’s law tells us that the field produced by the induced
current must point into the page to oppose the increase. The current must flow
clockwise around the circuit, or:
to the left in resistor
R
.
30.03.
dt
dB
r
dt
dB
r
dt
d
r
B
BA
dt
d
B
B
B
2
2
2
and
,
so
,
where
,
π
π
π

=
=
Φ
=
=
Φ
Φ

=
E
E
mV
11
Then,
s.
/
T
25
.
0
s
0
.
4
s
0
.
6
T
5
.
0
T
0
.
0
s,
6
<
<
s
4
For
c)
mV
0
.
0
Then,
s.
/
T
0
.
0
s
0
.
2
s
0
.
4
T
5
.
0
T
5
.
0
s,
4
<
<
s
2
For
b)
mV
11
Then,
s.
/
T
25
.
0
s
0
.
0
s
0
.
2
T
0
.
0
T
5
.
0
s,
2
<
<
0
For
a)
T)
/
s
mV
45
(
T)
/
s
V
045
.
0
(
)
m
12
.
0
(
Then,
2
=

=


=
=
=


=

=
=


=
⋅

=
⋅

=

=
E
E
E
E
dt
dB
t
dt
dB
t
dt
dB
t
dt
dB
dt
dB
dt
dB
π
30.06.
dt
dr
rB
dt
dr
B
dt
r
d
B
dt
dA
B
dt
BA
d
dt
d
B
π
π
π
2
)
(
)
(
2
2

=

=

=

=

=
Φ

=
E
m/s)
0.750
T)(
m)(0.800
120
.
0
(
2
2


=

=
π
π
dt
dr
rB
E
E
= 0.452 V
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30.08.
The field (due to the current in the straight wire) is outofthepage in the upper half of
the circle and is into the page in the lower half of the circle. Due to the symmetry, the
magnitude of the flux in the upper half equals that in the lower half producing zero net
flux, at any time. There is no induced current in the circle.
i
= 0
30.22.
a)
we
so
position,
of
function
a
is
loop.
gh the
flux throu
net
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 Spring '06
 Stoler
 Work, DT DT DT

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