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Problem
12. The current in an inductor is changing at the rate of 100 A/s, and the inductor emf is 40 V. What is its
self-inductance?
Solution
From Equation 32-5,
L
=
−
E
=
(
dI
=
dt
)
=
40 V
=
(
100 A
/
s
)
=
0
.
4 H
.
Problem
15. A cardboard tube measures 15 cm long by 2.2 cm in diameter. How many turns of wire must be wound
on the full length of the tube to make a 5.8-mH inductor?
Solution
From Equation 32-4,
N
=
l
=
μ
0
A
=
[(
5
.
8 mH
)(
15 cm
)
=
(
4
π
×
10
−
7
H
/
m
)
(
1
.
1 cm
)
2
]
1
=
2
=
1
.
35
×
10
3
turns.
Problem
16. The current in a 2.0-H inductor is given by
I
=
3
t
2
+
15
t
+
8,
where
t
is in seconds and
I
in amperes.
Find an expression for the magnitude of the inductor emf.
Solution
From Equation 32-5,
E
=
−
L
dI
=
dt
=
2
(
6
t
+
15
),
where
E
is in volts and
t
is in seconds.
Problem
18. The current in a 40-mH inductor is given by
I
=
I
0
e
−
bt
,
where
I
0
=
10 A
and
b
=
20 s
−
1
.
What is
the magnitude of the inductor emf at (a)
t
=
0,
(b)
t
=
25 ms,
and (c)
t
=
50 ms?
Solution
The magnitude of the inductor emf is
E
=
L dI
=
dt
=
LI
0
be
−
bt
,
from Equation 32-5. (a) At
t
=
E
(
0
)
=
(
40 mH
)
×
(10 A)(20 s
−
1
)
=
8 V,
so (b) at
t
=
25 ms
,
E
(
t
)
=
(
8 V
)
e
−
0
.
5
=
4
.
85 V,
and (c)
E
(
50 ms
)
=
2
.
94 V
.
Problem
21. The emf in a 50-mH inductor is given by
E
=
E
p
sin
ω
t
,
where
E
p
=
75 V
and
=
140 s
−
1
.
What
is the peak current in the inductor? (Assume the current swings symmetrically about zero.)
Solution
From Equation 32-5,
dI
=
dt
=
−
(
E
p
=
L
) sin
t
,
so integration yields
I
(
t
)
=
(
E
p
=
L
) cos
t
.
(Since
I
(
t
)
is
symmetric about
I
=
the constant of integration is zero.) The peak current is
I
p
=
E
p
=
L
=
75 V
=
(
140 s
−
1
×
50 mH
)
=
10
.
7 A
.
Problem
4.
Two coils have a mutual inductance of 580 mH. One coil is supplied with a current given by
I
=
3
t
2
−
2
t
+
4,
where
I
is in amperes and
t
in seconds. What is the induced emf in the other coil at
time
t
=
2.5 s?

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