Kapoor (mk9499) – homework25 – Turner – (60230)
1
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10
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001
(part 1 of 2) 10.0 points
An
electric
motor
has
114
turns
of
wire
wrapped on a rectangular coil, of dimensions
3
.
13 cm by 4
.
35 cm.
Assume that the mo
tor uses 10
.
1 A of current and that a uniform
0
.
35 T magnetic field exists within the motor.
What is the maximum torque delivered by
the motor?
Correct answer: 0
.
548691 N m.
Explanation:
Let :
w
= 3
.
13 cm
,
= 4
.
35 cm
,
I
= 10
.
1 A
,
and
B
= 0
.
35 T
.
Force :
F
=
I
×
B
Torque :
τ
=
r
×
F
Power :
P
=
τ ω
.
Magnetic force exerted on the 4
.
35 cm side
is,
F
=
N I
B
= (114 turns) (10
.
1 A) (0
.
0435 m) (0
.
35 T)
= 17
.
5301 N
.
So, the maximum torque on the loop due to
the magnetic field is
τ
max
=
F
×
w
= (17
.
5301 N)
×
(0
.
0313 m)
=
0
.
548691 N m
.
002
(part 2 of 2) 10.0 points
If the motor rotates at 2740 rev
/
min, what is
the peak power produced by the motor?
Correct answer: 157
.
437 W.
Explanation:
P
=
τ ω
= (0
.
548691 N m) (286
.
932 s

1
)
=
157
.
437 W
.
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
2
μ
0
n N
d I
dt
2.
E
=

π
r μ
0
N
d I
dt
3.
E
=

π
r
2
μ
0
n
d I
dt
4.
E
=

π
R
2
μ
0
n N
d I
dt
correct
5.
E
=

π
r μ
0
n
d I
dt
6.
E
=

π
R μ
0
N
d I
dt
7.
E
=

π
R
2
μ
0
n
d I
dt
8.
E
=

π
R μ
0
n N
d I
dt
9.
E
=

π
r μ
0
n 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
)
,
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Kapoor (mk9499) – homework25 – Turner – (60230)
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
(part 1 of 2) 10.0 points
Two infinitely long solenoids (seen in cross
section in the figure below) thread a circuit.
Given:
a
= 0
.
9 m
,
r
1
= 0
.
2 m
,
r
2
= 0
.
1 m
,
Δ
B
Δ
t
= 160 T
/
s
,
R
l
= 9
Ω
,
R
m
= 4
Ω
,
and
R
r
= 2
.
6
Ω
,
as in the figure below.
The magnitude of
B
inside each solenoid is the same and is 300 T
at time
t
= 0.
B
in
B
out
I
l
I
m
I
r
r
2
R
m
R
l
R
r
r
1
a
a
a
What is the magnitude of the current,
I
m
,
in middle resistor,
R
m
?
Correct answer: 1
.
39706 A.
Explanation:
Basic Concept:
Faraday’s Law:
E
=

d
Φ
B
dt
Ohm’s Law:
I
=
V
R
Junction Rule:
n
i
=1
I
i
= 0
Solution:
Note:
The sidelength,
a
, of the
circuit loop is not necessary for this problem.
Neither is the magnitude of
B
at time
t
= 0.
From Faraday’s law, the induced emf in the
left loop (Loop 1) is
E
1

=
d
Φ
B
dt
=
A
1
d B
dt
=
π
r
2
1
d B
dt
= 20
.
1062 V
and the direction of this emf is counter
clockwise.
Similarly, the induced emf in the
right loop (Loop 2) is
E
2

=
d
Φ
B
dt
=
A
2
d B
dt
=
π
r
2
2
d B
dt
= 5
.
02655 V
and the direction of this emf is clockwise.
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 Spring '10
 Turner
 Magnetic Field, Kapoor, PX PX PX

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