This preview shows pages 1–2. Sign up to view the full content.
1
This printout should have 13 questions.
Multiplechoice questions may continue on
the next column or page – Fnd all choices
before answering.
001
(part 1 of 3) 10.0 points
A 27
.
7 mA current is carried by a uniformly
wound aircore solenoid with 285 turns as
shown in the Fgure below.
The
permeability
of
free
space
is
1
.
25664
×
10

6
N
/
A
2
.
27
.
7 mA
14
.
4 mm
16
.
1 cm
Compute the magnetic Feld inside the
solenoid.
Correct answer: 6
.
16181
×
10

5
T.
Explanation:
Let :
N
= 285
,
ℓ
= 16
.
1 cm
,
I
= 27
.
7 mA
,
and
μ
0
= 1
.
25664
×
10

6
N
/
A
2
.
I
d
ℓ
The magnetic Feld inside the solenoid is
B
=
μ
0
n I
=
μ
0
p
N
ℓ
P
I
= (1
.
25664
×
10

6
N
/
A
2
)
p
285
0
.
161 m
P
×
(0
.
0277 A)
=
6
.
16181
×
10

5
T
.
002
(part 2 of 3) 10.0 points
Compute the magnetic ±ux through each
turn.
Correct answer: 1
.
00351
×
10

8
Wb.
Explanation:
Let :
B
= 6
.
16181
×
10

5
T
,
and
d
= 14
.
4 mm = 0
.
0144 m
.
The magnetic ±ux through each turn is
Φ =
B A
=
B
±
π
4
d
2
²
= (6
.
16181
×
10

5
T)
π
4
(0
.
0144 m)
2
=
1
.
00351
×
10

8
Wb
.
003
(part 3 of 3) 10.0 points
Compute the inductance of the solenoid.
Correct answer: 0
.
10325 mH.
Explanation:
The inductance of the solenoid is
L
=
N
Φ
I
=
(285) (1
.
00351
×
10

8
Wb)
0
.
0277 A
=
0
.
10325 mH
.
004
(part 1 of 2) 10.0 points
A solenoid has 129 turns of wire uniformly
wrapped around an airFlled core, which has
a diameter of 8 mm and a length of 7
.
4 cm.
The
permeability
of
free
space
is
1
.
25664
×
10

6
N
/
A
2
.
Calculate
the
selfinductance
of
the
solenoid.
Correct answer: 1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue University.
 Spring '10
 Gavrine
 Current

Click to edit the document details