hussain (wh4892) – hw 12 – Opyrchal – (121104)
1
This printout should have 11 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
Calculate the magnetic ±ux through a 400
turn, 8
.
06 mH coil when the current in the
coil is 11 mA.
Correct answer: 2
.
2165
×
10
−
7
Wb.
Explanation:
Let :
N
= 400
,
L
= 8
.
06 mH = 0
.
00806 H
,
and
I
= 11 mA = 0
.
011 A
.
The inductance oF a coil is related to the
magnetic ±ux as
L
=
N
Φ
I
Φ =
L I
N
=
(0
.
00806 H) (0
.
011 A)
400
=
2
.
2165
×
10
−
7
Wb
.
002
10.0 points
An inductor in the Form oF an aircore solenoid
contains 276 turns, is oF length 14 cm, and has
a crosssectional area oF 1
.
2 cm
2
.
The
permeability
oF
Free
space
is
1
.
25664
×
10
−
6
N
/
A
2
.
What is the magnitude oF the uniForm rate
oF change in current through the inductor that
induces an
emf
oF 130
μ
V?
Correct answer: 1
.
58439 A
/
s.
Explanation:
Let :
N
= 276 turns
,
μ
0
= 1
.
25664
×
10
−
6
N
/
A
2
,
ℓ
= 14 cm
,
and
A
= 1
.
2 cm
2
= 0
.
00012 m
2
.
The selFinductance oF a solenoid is
L
=
μ
0
N
2
A
ℓ
= 1
.
25664
×
10
−
6
N
/
A
2
×
(276 turns)
2
(0
.
00012 m
2
)
0
.
14 m
= 8
.
20505
×
10
−
5
H
.
The induced emF
E
is given by
E
=
v
v
v
v

L
d I
dt
v
v
v
v
d I
dt
=
v
v
v
v
E
L
v
v
v
v
=
v
v
v
v

130
μ
V
8
.
20505
×
10
−
5
H
v
v
v
v
=
1
.
58439 A
/
s
.
003
10.0 points
A solenoid inductor is 30 cm long and has
a crosssectional area oF 6 cm
2
. When the
current through the solenoid decreases at a
rate oF 0
.
625 A
/
s, the induced
emf
is 150
μ
V.
The
permeability
oF
Free
space
is
1
.
25664
×
10
−
6
N
/
A
2
.
²ind the number oF turns per meter oF the
solenoid.
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 Spring '08
 moro
 Inductance, Correct Answer, Inductor, RL circuit

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