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Ch. 30
PHYSICS 2B
Spring 2011
Homework Solutions – Chapter 30
Author: Evan Grohs 
[email protected]
Ch. 30 Problems: 1, 5, 8, 19, 24, 27, 33, 36, 50, 66, 68
Page 1/6
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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.
30.5.
I
DENTIFY
and
S
ET
U
P
:
Apply Eq.(30.5).
E
XECUTE
:
(a)
22
1
400 0.0320 Wb
1.96 H
6.52 A
B
N
M
i
(b)
3
11
2
1
21
(1.96 H)(2.54 A)
so
7.11 10 Wb
700
B
B
NM
i
M
iN
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.
30.8.
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
E
XECUTE
:
72
4
2
4
(4
10
T m/A)(500) (6.25 10 m )
7.81 10 H
2 (0.0400 m)
L
(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
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.
E
VALUATE
:
This is a reasonable value for selfinductance, in the range of a mH.
30.19. I
DENTIFY
:
A currentcarrying inductor has a magnetic field inside of itself and
hence stores magnetic energy.
(a) S
ET
U
P
:
The magnetic field inside a solenoid is
0
.
B
nI
E
XECUTE
:
7
(4
10 T m/A)(400)(80.0 A)
= 0.161 T
0.250 m
B
(b) S
ET
U
P
:
The energy density in a magnetic field is
2
0
.
2
B
u
30
Ch. 30
page 3/6
E
XECUTE
:
2
43
7
(0.161
T)
1.03 10 J/m
2(4
10 T m/A)
u
(c) S
ET
U
P
:
The total stored energy is
U
=
uV
.
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This note was uploaded on 12/18/2011 for the course PHYS 2B 2b taught by Professor Hirsch during the Spring '10 term at UCSD.
 Spring '10
 Hirsch

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