301
I
NDUCTANCE
30.1.
IDENTIFY
and
SET UP:
Apply Eq.(30.4).
EXECUTE:
(a)
4
1
2
(3.25 10 H)(830 A/s)
0.270 V;
di
M
dt
−
==
×
=
E
yes, it is constant.
(b)
2
1
;
di
M
dt
=
E
M
is a property of the pair of coils so is the same as in part (a). Thus
1
0.270 V.
=
E
EVALUATE:
The induced emf is the same in either case. A constant
/
di dt
produces a constant emf.
30.2.
IDENTIFY:
2
1
i
M
t
Δ
=
Δ
E
and
1
2
.
i
M
t
Δ
=
Δ
E
22
1
B
N
M
i
Φ
=
, where
2
B
Φ
is the flux through one turn of the second
coil.
SET UP:
M
is the same whether we consider an emf induced in coil 1 or in coil 2.
EXECUTE:
(a)
3
3
2
1
1.65 10 V
6.82 10 H
6.82 mH
/0
.
2
4
2
A
/
s
M
it
−
−
×
=
×
=
ΔΔ
E
(b)
3
4
1
2
2
(6.82 10 H)(1.20 A)
3.27 10 Wb
25
B
Mi
N
−
−
×
Φ=
=
=
×
(c)
33
2
1
(6.82 10 H)(0.360 A/s)
2.46 10 V
2.46 mV
i
M
t
−−
Δ
×
=
×
=
Δ
E
EVALUATE:
We can express
M
either in terms of the total flux through one coil produced by a current in the
other coil, or in terms of the emf induced in one coil by a changing current in the other coil.
30.3.
IDENTIFY:
Replace units of Wb, A and
Ω
by their equivalents.
SET UP:
2
1 Wb
1 T m .
=⋅
1 T
1 N/(A m).
1 N m
1 J.
⋅=
1 A
1 C/s.
=
1 V
1 J/C.
=
1 V/A
1
.
=Ω
EXECUTE:
2
1H
1 Wb/A
1T m /A
1 N m/A
1J/A
1(J/[A C])s
1(V/A)s
1
Ω
s.
⋅
=
⋅
=
=
⋅
=
=
⋅
EVALUATE:
We may use whichever equivalent unit is the most convenient in a particular problem.
30.4.
IDENTIFY:
Changing flux from one object induces an emf in another object.
(a) SET UP:
The magnetic field due to a solenoid is
0
.
Bn
I
μ
=
EXECUTE:
The above formula gives
()
7
4
1
4
10 T m/A (300)(0.120 A)
=1.81 10 T
0.250 m
B
π
−
−
×⋅
=×
The average flux through each turn of the inner solenoid is therefore
42
8
1
1.81 10 T
(0.0100 m) = 5.68 10 Wb
B
BA
=
×
×
(b) SET UP:
The flux is the same through each turn of both solenoids due to the geometry, so
2,
2
1
11
BB
NN
M
ii
ΦΦ
EXECUTE:
8
5
(25) 5.68 10 Wb
1.18 10 H
0.120 A
M
−
−
×
×
(c) SET UP:
The induced emf is
1
2
.
di
M
dt
=−
E
EXECUTE:
5
2
1.18 10 H (1750 A/s)
0.0207 V
−
×
E
EVALUATE:
A mutual inductance around
5
10
−
H is not unreasonable.
30
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Chapter 30
30.5.
IDENTIFY
and
SET UP:
Apply Eq.(30.5).
EXECUTE:
(a)
()
22
1
400 0.0320 Wb
1.96 H
6.52 A
B
N
M
i
Φ
==
=
(b)
3
11
2
1
21
(1.96 H)(2.54 A)
so
7.11 10 Wb
700
B
B
NM
i
M
iN
−
Φ
=Φ
=
=
=
×
EVALUATE:
M
relates the current in one coil to the flux through the other coil. Eq.(30.5) shows that
M
is the
same for a pair of coils, no matter which one has the current and which one has the flux.
30.6.
IDENTIFY:
A changing current in an inductor induces an emf in it.
(a) SET UP:
The selfinductance of a toroidal solenoid is
2
0
.
2
NA
L
r
μ
π
=
EXECUTE:
72
4
2
4
(4
10
T m/A)(500) (6.25 10 m )
7.81 10 H
2 (0.0400 m)
L
−−
−
×⋅
×
×
(b) SET UP:
The magnitude of the induced emf is
.
di
L
dt
=
E
EXECUTE:
4
3
5.00 A
2.00 A
0.781 V
3.00 10 s
−
−
−
⎛⎞
=×
=
⎜⎟
×
⎝⎠
E
(c)
The current is decreasing, so the induced emf will be in the same direction as the current, which is from
a
to
b,
making
b
at a higher potential than
a.
EVALUATE:
This is a reasonable value for selfinductance, in the range of a mH.
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 Spring '11
 Shaefer
 Energy, Inductor, dt, electrical energy, Energy density, IMAX

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