Nguyen, Thanh – Homework 15 – Due: Mar 28 2008, 7:00 pm – Inst: Weathers
1
This printout should have 18 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
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 240 turns iF the length
oF the solenoid is 21 cm and its crosssectional
area is 3 cm
2
.
Correct answer: 0
.
103403 mH.
Explanation:
Let :
N
= 240
,
μ
0
= 1
.
25664
×
10

6
N
/
A
2
,
‘
= 21 cm = 0
.
21 m
,
and
A
= 3 cm
2
= 0
.
0003 m
2
.
In this case, we can take the interior magnetic
feld 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 ±ux 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 fnd that
L
=
μ
0
N
2
A
‘
= (1
.
25664
×
10

6
N
/
A
2
)
×
(240)
2
(0
.
0003 m
2
)
0
.
21 m
= 0
.
000103403 H
=
0
.
103403 mH
.
002
(part 2 oF 2) 10 points
Calculate the selFinduced
emf
in the solenoid
described in the frst part iF the current
through it is decreasing at the rate oF 44 A
/
s.
Correct answer: 4
.
54974 mV.
Explanation:
Using the equation
E
=

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

44 A
/
s
,
we get
E
=

L
d I
d t
=

(0
.
000103403 H ) (

44 A
/
s )
= 0
.
00454974 V =
4
.
54974 mV
.
keywords:
003
(part 1 oF 1) 10 points
An automobile starter motor draws a current
oF 1
.
4 A From a 16
.
6 V battery when operating
at normal speed. A broken pulley locks the ar
mature in position, and the current increases
to 15
.
8 A.
What was the back
emf
oF the motor when
operating normally?
Correct answer: 15
.
1291 V.
Explanation:
Let :
I
= 15
.
8 A
,
I
0
= 1
.
4 A
,
and
E
= 16
.
6 V
.
When not rotating,
E
=
I R
, and From this,
R
=
E
I
=
16
.
6 V
15
.
8 A
= 1
.
05063 Ω
,
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View Full DocumentNguyen, Thanh – 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
= 16
.
6 V

(1
.
4 A) (1
.
05063 Ω)
=
15
.
1291 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 Fgure.
The switch has been in position
b
for a long
period of time.
80 mH
3
.
31 Ω
8
.
6 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
173 mA?
Correct answer: 1
.
66538 ms.
Explanation:
Let :
R
= 3
.
31 Ω
,
L
= 80 mH
,
and
E
= 8
.
6 V
.
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 Spring '00
 Littler
 Inductance, Work, Correct Answer, Inductor, dt, DI

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