toupal (rgt374) – Homework 10 – Chiu – (58295)
1
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001
10.0 points
A powerful electromagnet has a field of 1
.
56 T
and a cross sectional area of 0
.
429 m
2
. Now we
place a coil of 179 turns with a total resistance
of 2
.
06 Ω around the electromagnet and turn
off the power to the electromagnet in 0
.
0476 s.
What will be the induced current in the
coil?
Correct answer: 1221
.
69 a.
Explanation:
Basic Concepts:
Faraday’s Law of Induc
tion:
E
=
−
d
Φ
B
dt
Ohm’s law:
I
=
V
R
For coil of many turns, the emf is the sum
of induced emf in each turn.
Besides, when
calculating the induced emf in the coil, we
actually are try to get a average emf. For one
turn of coil, the induced emf is:
E
1
=
−
d
Φ
B
dt
=
ΔΦ
B
Δ
t
=
Δ
B
·
A
Δ
t
= 14
.
0597 V
So, the emf in the whole coil is:
E
=
n
· E
1
= 179 turns
×
14
.
0597 V
= 2516
.
68 V
Apply Ohm’s law, we get the current as fol
lowing:
I
=
V
R
=
E
R
=
2516
.
68 V
2
.
06 Ω
= 1221
.
69 a
002
10.0 points
A coil with
N
1
= 21
.
5 turns and radius
r
1
=
6
.
82 cm surrounds a long solenoid of radius
r
2
= 1
.
73 cm and
N
2
ℓ
2
=
n
2
= 1300 m
−
1
(see
the figure below). The current in the solenoid
changes as
I
=
I
0
sin(
ω t
), where
I
0
= 5 A
and
ω
= 120 rad
/
s.
R
a
b
E
ℓ
2
ℓ
1
Outside solenoid has
N
1
turns
Inside solenoid has
N
2
turns
A
1
A
2
What is the magnitude of the induced
emf
,
E
AB
, across the 21
.
5 turn coil at
t
= 1000 s?
Correct answer: 0
.
0165152 V.
Explanation:
Faraday’s Law for solenoid
E
=
−
N
d
Φ
dt
=
−
N
d
(
BA
)
dt
.
Magnetic field induced by solenoid
B
=
μ
0
n I .
Faraday’s Law for solenoid
E
=
−
N
d
Φ
dt
=
−
N
d
(
BA
)
dt
.
Magnetic field induced by inner solenoid
B
=
μ
0
n I .
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toupal (rgt374) – Homework 10 – Chiu – (58295)
2
So the induced
emf
is
E
=
−
N
1
d
Φ
dt
=
−
N
1
d
(
B A
2
)
dt
=
−
N
1
A
2
d B
dt
=
−
N
1
A
2
d
(
μ
0
n
2
I
)
dt
=
−
μ
0
N
1
n
2
A
2
d I
dt
=
0
.
0165152 V
.
003
10.0 points
A long solenoid has a coil made of fine wire
inside it and coaxial with it.
I
Outside solenoid has
n
turns per meter
Inside coil has
N
turns
r
R
Given a varying current
I
in the outer
solenoid, what is the
emf
induced in the inner
loop?
1.
E
=
−
π R μ
0
N
d I
dt
2.
E
=
−
π r μ
0
N
d I
dt
3.
E
=
−
π r μ
0
n N
d I
dt
4.
E
=
−
π R
2
μ
0
n N
d I
dt
correct
5.
E
=
−
π R
2
μ
0
n
d I
dt
6.
E
=
−
π r
2
μ
0
n
d I
dt
7.
E
=
−
π r
2
μ
0
n N
d I
dt
8.
E
=
−
π R μ
0
n N
d I
dt
9.
E
=
−
π R μ
0
n
d I
dt
10.
E
=
−
π r μ
0
n
d I
dt
Explanation:
The magnetic field of a solenoid is
B
=
μ
0
n I .
The magnetic flux is
Φ
B
=
B
·
A
= (
μ
0
n I
) (
π R
2
)
,
so the
emf
is
E
=
−
d
Φ
B
dt
=
−
π R
2
μ
0
n N
d I
dt
.
We are interested in the
emf
in the inner coil,
so we use the smaller area of the inner coil
rather than the larger solenoid area.
keywords:
004
10.0 points
A long solenoid has a tilted coil inside it made
of fine wire, as shown in the figure below.
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 Spring '09
 Chiu
 Physics, Inductance, Power, Work, Correct Answer, Inductor, Vrms Irms

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