terry (ect328) – homework 26 – Turner – (59130)
1
This printout should have 11 questions.
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beFore answering.
001
(part 1 oF 2) 10.0 points
A solenoid has 111 turns oF wire uniFormly
wrapped around an airflled core, which has
a diameter oF 15 mm and a length oF 7 cm.
The
permeability
oF
Free
space
is
1
.
25664
×
10
−
6
N
/
A
2
.
Calculate
the
selFinductance
oF
the
solenoid.
Correct answer: 3
.
90868
×
10
−
5
H.
Explanation:
Let :
N
= 111
,
D
= 15 mm = 0
.
015 m
,
ℓ
= 7 cm = 0
.
07 m
,
and
μ
0
= 1
.
25664
×
10
−
6
N
/
A
2
.
The selFinductance oF a solenoid is given
by
L
1
=
N
Φ
I
,
where Φ is the total ±ux inside the solenoid
and
I
is the current in the wire wrapped
around the solenoid.
By Ampere’s Law, the magnetic feld in the
solenoid is
B
=
μ
0
N I
ℓ
,
where
μ
=
μ
0
is the magnetic permeability oF
the air core which is the same as Free space.
The magnetic ±ux in the solenoid is
Φ =
B A
=
B π D
2
4
.
Using the above expressions For Φ and
B
,
we obtain For the inductance oF the solenoid
L
1
=
N B A
I
=
μ
0
N
2
π D
2
4
ℓ
= (1
.
25664
×
10
−
6
N
/
A
2
) (111)
2
×
π
(0
.
015 m)
2
4 (0
.
07 m)
=
3
.
90868
×
10
−
5
H
.
002
(part 2 oF 2) 10.0 points
The core is replaced with a soFt iron rod that
has the same dimensions, but a magnetic per
meability oF 800
μ
0
.
What is the new inductance?
Correct answer: 0
.
0312694 H.
Explanation:
Let :
μ
= 800
μ
0
.
L
2
=
μ N
2
π D
2
4
ℓ
=
μ
μ
0
L
1
= 800(3
.
90868
×
10
−
5
H)
=
0
.
0312694 H
.
003
(part 1 oF 2) 10.0 points
The current in a 102 mH inductor changes
with time as
I
=
b t
2

a t
.
With
a
= 16 A
/
s and
b
= 2 A
/
s
2
, fnd
the magnitude oF the induced
emf
,
E
, at
t
= 1
.
2 s.
Correct answer: 1
.
1424 V.
Explanation:
Let :
L
= 102 mH = 0
.
102 H
,
b
= 2 A
/
s
2
,
a
= 16 A
/
s
,
and
t
= 1
.
2 s
.
²rom ²araday’s Law, the induced
emf
E
is proportional to the rate oF change oF the
magnetic ±ux, which in turn is proportional
to the rate oF change oF the current. This is
expressed as
E
=
L
d I
dt
=
L
d
dt
(
b t
2

a t
)
=
L
(2
b t

a
)
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View Full Documentterry (ect328) – homework 26 – Turner – (59130)
2
At 1
.
2 s ,the magnitude of the induced
emf
is
E
= (0
.
102 H)
v
v
v
2(2 A
/
s
2
)(1
.
2 s)

16 A
/
s
v
v
v
=
1
.
1424 V
.
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 Fall '09
 Turner
 Magnetic Field, Correct Answer, Terry, N1 N2, N1 N2 I2

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