Physics 21
Fall, 2011
Solution to HW18
307
At the instant when the current in an inductor is in
creasing at a rate of 6
.
45
×
10
−
2
A
/
s, the magnitude of the
selfinduced emf is 1
.
65
×
10
−
2
V. (a) What is the induc
tance of the inductor? (b) If the inductor is a solenoid with
405 turns, what is the average magnetic flux through each
turn when the current is 0
.
715A?
(a) In the class lecture we applied Faraday’s Law to a
solenoid with
N
turns. The selfinduced emf is
E
=

N
d
Φ
B
dt
,
and we defined the selfinductance
L
by
E
=

L
di
dt
Here we aren’t concerned with the direction of the emf, only
the magnitude. Therefore we can substitute the given values
to obtain
L
=
E
di/dt
=
1
.
65
×
10
−
2
V
6
.
45
×
10
−
2
A
/
s
= 0
.
256H
(b) For this part we must notice that since the flux through
one loop of a solenoid of length
l
is given by
Φ
B
=
BA
=
μ
0
N
l
iA,
we can rewrite the expression for the inductance
L
given on
the equation sheet in terms of the flux,
L
=
μ
0
N
2
l
A
=
Nμ
0
N
l
A
=
N
Φ
B
i
.
Note that the relation above,
L
=
N
Φ
B
i
,
is in fact the definition used in the book for
L
[Eq. (30.6)].
We can use it to answer this problem by solving for Φ
B
.
Φ
B
=
Li
N
=
(0
.
256H)(0
.
715A)
405
= 4
.
52
×
10
−
4
Wb
3019
An inductor with an inductance of 2.50 H and a
resistance of 8
.
00Ω is connected to the terminals of a battery
with an emf of 6.00 V and negligible internal resistance. Find
(a) the initial rate of increase of current in the circuit; (b) the
rate of increase of current at the instant when the current
is 0.500 A; (c) the current 0.250 s after the circuit is closed;
(d) the final steadystate current.
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 Spring '08
 Hickman
 Current, Inductance, Energy, Work, Inductor, 10−2 A/s

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