301
I
NDUCTANCE
30.1.
I
DENTIFY
and
S
ET
U
P
:
Apply Eq.(30.4).
E
XECUTE
:
(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
E
VALUATE
:
The induced emf is the same in either case. A constant
/
di dt
produces a constant emf.
30.2.
I
DENTIFY
:
2
1
i
M
t
Δ
=
Δ
E
and
1
2
.
i
M
t
Δ
=
Δ
E
2
2
1
B
N
M
i
Φ
=
, where
2
B
Φ
is the flux through one turn of the second
coil.
S
ET
U
P
:
M
is the same whether we consider an emf induced in coil 1 or in coil 2.
E
XECUTE
:
(a)
3
3
2
1
1.65
10
V
6.82
10
H
6.82 mH
/
0.242 A/s
M
i
t
−
−
×
=
=
=
×
=
Δ
Δ
E
(b)
3
4
1
2
2
(6.82
10
H)(1.20 A)
3.27
10
Wb
25
B
Mi
N
−
−
×
Φ
=
=
=
×
(c)
3
3
2
1
(6.82
10
H)(0.360 A/s)
2.46
10
V
2.46 mV
i
M
t
−
−
Δ
=
=
×
=
×
=
Δ
E
E
VALUATE
:
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.
I
DENTIFY
:
Replace units of Wb, A and
Ω
by their equivalents.
S
ET
U
P
:
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
E
XECUTE
:
2
2
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.
=
=
⋅
=
⋅
=
=
⋅
=
=
⋅
E
VALUATE
:
We may use whichever equivalent unit is the most convenient in a particular problem.
30.4.
I
DENTIFY
:
Changing flux from one object induces an emf in another object.
(a) S
ET
U
P
:
The magnetic field due to a solenoid is
0
.
B
nI
μ
=
E
XECUTE
:
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
(
)
4
2
8
1
1.81
10
T
(0.0100 m)
= 5.68
10
Wb
B
B A
π
−
−
Φ
=
=
×
×
(b) S
ET
U
P
:
The flux is the same through each turn of both solenoids due to the geometry, so
2
,2
2
,1
1
1
B
B
N
N
M
i
i
Φ
Φ
=
=
E
XECUTE
:
(
)
8
5
(25)
5.68
10
Wb
1.18
10
H
0.120 A
M
−
−
×
=
=
×
(c) S
ET
U
P
:
The induced emf is
1
2
.
di
M
dt
= −
E
E
XECUTE
:
(
)
5
2
1.18
10
H
(1750 A/s)
0.0207 V
−
= −
×
= −
E
E
VALUATE
:
A mutual inductance around
5
10
−
H is not unreasonable.
30
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302
Chapter 30
30.5.
I
DENTIFY
and
S
ET
U
P
:
Apply Eq.(30.5).
E
XECUTE
:
(a)
(
)
2
2
1
400
0.0320 Wb
1.96 H
6.52 A
B
N
M
i
Φ
=
=
=
(b)
3
1
1
2
1
2
1
(1.96 H)(2.54 A)
so
7.11
10
Wb
700
B
B
N
Mi
M
i
N
−
Φ
=
Φ
=
=
=
×
E
VALUATE
:
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.
I
DENTIFY
:
A changing current in an inductor induces an emf in it.
(a) S
ET
U
P
:
The selfinductance of a toroidal solenoid is
2
0
.
2
N A
L
r
μ
π
=
E
XECUTE
:
7
2
4
2
4
(4
10
T
m/A)(500)
(6.25
10
m )
7.81
10
H
2
(0.0400 m)
L
π
π
−
−
−
×
⋅
×
=
=
×
(b) S
ET
U
P
:
The magnitude of the induced emf is
.
di
L
dt
=
E
E
XECUTE
:
(
)
4
3
5.00 A
2.00 A
7.81
10
H
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.
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 Spring '10
 Satpathy
 Energy, Inductor, dt, electrical energy, Energy density, IMAX

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