Patel, Kinal – Homework 15 – Due: Mar 28 2008, 7:00 pm – Inst: Weathers
1
This
printout
should
have
18
questions.
Multiplechoice questions may continue on
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before answering.
The due time is Central
time.
001
(part 1 of 2) 10 points
Assume that the length of the solenoid is much
larger than the solenoid’s radius and that the
core of the solenoid is air.
Calculate the inductance of a uniformly
wound solenoid having 260 turns if the length
of the solenoid is 31 cm and its crosssectional
area is 4 cm
2
.
Correct answer: 0
.
109611 mH.
Explanation:
Let :
N
= 260
,
μ
0
= 1
.
25664
×
10

6
N
/
A
2
,
‘
= 31 cm = 0
.
31 m
,
and
A
= 4 cm
2
= 0
.
0004 m
2
.
In this case, we can take the interior magnetic
field to be uniform and given by the equation
B
=
μ
0
n I
=
μ
0
N
‘
I ,
where
n
is the number of turns per unit length,
N
‘
.
The magnetic flux through each turn is
Φ
B
=
B A
=
μ
0
N A
‘
I ,
where
A
is the crosssectional area of the
solenoid. Using this expression and the equa
tion for the inductance,
L
=
N
Φ
B
I
,
we find that
L
=
μ
0
N
2
A
‘
= (1
.
25664
×
10

6
N
/
A
2
)
×
(260)
2
(0
.
0004 m
2
)
0
.
31 m
= 0
.
000109611 H
=
0
.
109611 mH
.
002
(part 2 of 2) 10 points
Calculate the selfinduced
emf
in the solenoid
described
in
the
first
part
if
the
current
through it is decreasing at the rate of 66 A
/
s.
Correct answer: 7
.
23434 mV.
Explanation:
Using the equation
E
=

N
d
Φ
B
d t
,
and given that
d I
d t
=

66 A
/
s
,
we get
E
=

L
d I
d t
=

(0
.
000109611 H ) (

66 A
/
s )
= 0
.
00723434 V =
7
.
23434 mV
.
keywords:
003
(part 1 of 1) 10 points
An automobile starter motor draws a current
of 4 A from a 9
.
7 V battery when operating at
normal speed. A broken pulley locks the ar
mature in position, and the current increases
to 24
.
8 A.
What was the back
emf
of the motor when
operating normally?
Correct answer: 8
.
13548 V.
Explanation:
Let :
I
= 24
.
8 A
,
I
0
= 4 A
,
and
E
= 9
.
7 V
.
When not rotating,
E
=
I R
, and from this,
R
=
E
I
=
9
.
7 V
24
.
8 A
= 0
.
391129 Ω
,
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Patel, Kinal – Homework 15 – Due: Mar 28 2008, 7:00 pm – Inst: Weathers
2
When rotating,
E  E
back
=
I
0
R
, or
E
back
=
E 
I
0
R
= 9
.
7 V

(4 A) (0
.
391129 Ω)
=
8
.
13548 V
.
keywords:
004
(part 1 of 3) 10 points
An inductor and a resistor are connected
with a double pole switch to a battery as
shown in the figure.
The switch has been in position
b
for a long
period of time.
137 mH
5
.
85 Ω
6
.
4 V
S
b
a
If the switch is thrown from position
b
to position
a
(connecting the battery), how
much time elapses before the current reaches
89 mA?
Correct answer: 1
.
98713 ms.
Explanation:
Let :
R
= 5
.
85 Ω
,
L
= 137 mH
,
and
E
= 6
.
4 V
.
L
R
E
S
b
a
The time constant of an
RL
circuit is
τ
=
L
R
=
0
.
137 H
5
.
85 Ω
= 0
.
0234188 s
.
The final current reached in the circuit is
I
0
=
E
R
=
6
.
4 V
5
.
85 Ω
= 1
.
09402 A
.
The switch is in position
a
in an
RL
circuit
connected to a battery at
t
= 0 when
I
= 0.
Then the current vs. time is
I
=
I
0
‡
1

e

t / τ
·
.
Solving the above expression for
t
, when
I
=
I
1
gives
t
1
=

τ
ln
1

I
1
I
0
¶
=

(0
.
0234188 s) ln
1

0
.
089 A
1
.
09402 A
¶
=
1
.
98713 ms
.
005
(part 2 of 3) 10 points
What is the maximum current in the inductor
a long time after the switch is in position
a
?
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 Spring '08
 Weathers
 Inductance, Work, Correct Answer, Inductor, patel

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