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RECITATION 12
5.
(a) Table 262 gives the resistivity of copper. Thus,
(
29
(
29
(
29
8
3
2
3
0.10m
1.69 10
m
1.1 10
.
1.25 10 m
L
R
A
π
ρ



=
=
×
Ω ⋅
=
×
Ω
×
(b) The current induced in the loop has a magnitude of
mag
2
1
,
d
r
dB
i
R
R
dt
R
dt
Φ
=
=
=
E
where
r
is the radius of the loop.
Therefore, to produce a current of 10 A requires that the field change at
a rate of
(
29
(
29
(
29
3
2
2
10A 1.1 10
1.4T s.
0.05m
i R
dB
dt
r

×
Ω
=
=
=
9.
Since a square's area is related to the length of its sides by
2
,
A
=
l
we have
2
.
dA dt
d
dt
=
l l
Thus,
Faraday's law gives the magnitude of the emf as
(
29
(
29
(
29
mag
2
2 0.24T 0.12m 0.050m/s
0.0029V.
d
dA
d
B
B
dt
dt
dt
Φ
=
=
=
=
=
l
l
E
17.
(a) Each side of the square has a length of
2.00 m.
L
=
The magnetic flux through the circuit is
mag
2
2,
BL
Φ
=
and the induced emf has a magnitude of
(
29
(
29
mag
2
2
2.00 m
0.870T s
1.74V .
2
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 Spring '08
 Goksu
 Physics, Current

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