Husain, Zeena – Homework 9 – Due: Mar 29 2004, 4:00 am – Inst: Sonia Paban
1
This
printout
should
have
21
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before making your selection. The due time is
Central time.
001
(part 1 of 1) 10 points
The recent discovery of “high temperature”
superconductors
with
negligible
resistance
has raised many possibilities for technological
applications (magnetically levitated trains,
zerodissipation power transmission lines, in
credibly strong refrigerator magnets,
etc.)
However, these materials lose their supercon
ductivity if they are subjected to excessive
magnetic fields.
This limits the amount of
current that a superconductor can carry, since
the current results in a magnetic field at the
surface of the wire.
Calculate this “critical current” for a
d
=
2
.
08 mm diameter superconducting wire if
the superconductivity is destroyed when the
surface magnetic field exceeds
B
= 0
.
0348 T.
Given
μ
0
= 1
.
25664
×
10

6
N
/
A
2
.
Correct answer: 180
.
96 A.
Explanation:
Solution:
This problem can be solved with Ampere’s
law,
I
~
B
·
d~s
=
μ
0
I
Choose the integration path to be a loop
around the surface. This gives us
B
(
π d
) =
μ
0
I
where
d
is the diameter. Or,
I
=
π B d
μ
0
=
π
(0
.
0348 T) (0
.
00208 m)
1
.
25664
×
10

6
N
/
A
2
= 180
.
96 A
002
(part 1 of 1) 10 points
An infinitely long straight wire carrying a
current
I
1
= 59
.
6 A is partially surrounded
by a loop as in figure. The loop has a length
L
= 33
.
2 cm, a radius
R
= 12
.
6 cm, and
carries a current
I
2
= 25
.
2 A. The axis of the
loop coincides with the wire.
R
L
I
1
I
2
Calculate the force exerted on the loop.
Correct answer: 1582
.
98
μ
N.
Explanation:
The central wire creates field
~
B
=
μ
0
I
1
2
π R
counterclockwise
.
The curved portions of the loop feels zero
force since
~
l
×
~
B
= 0 there.
The straight
portions both feel
I
2
~
l
×
~
B
forces to the right,
amounting to
~
F
=
I
2
2
L
μ
0
I
1
2
π R
=
μ
0
I
1
I
2
L
π R
to the right
k
~
F
k
=
μ
0
(59
.
6 A) (25
.
2 A) (0
.
332 m)
π
(12
.
6 cm)
= 1582
.
98
μ
N
.
003
(part 1 of 2) 10 points
A superconducting solenoid is to generate a
magnetic field of 4
.
51 T. The solenoid winding
has 2290 turns per meter.
What is the required current?
Correct answer: 1567
.
22 A.
Explanation:
Magnetic field at the center of solenoid is
B
=
μ
0
N I
l
=
μ
0
n I ,
then
I
=
B l
μ
0
N
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Husain, Zeena – Homework 9 – Due: Mar 29 2004, 4:00 am – Inst: Sonia Paban
2
=
(4
.
51 T)
μ
0
(2290)
= 1567
.
22 A
.
004
(part 2 of 2) 0 points
What force per unit length is exerted on the
windings by this magnetic field?
Correct answer: 7068
.
18 N
/
m.
Explanation:
Force per unit length is
F
l
=
I B
= (1567
.
22 A) (4
.
51 T)
= (7068
.
18 N
/
m)
radially outward
.
Although the force per unit length is non
zero (in an outward direction) the net force
per unit length will be zero.
005
(part 1 of 1) 10 points
A sinusoidal voltage is applied directly across
a 11
.
4
μ
F capacitor.
The frequency of the
source is 4
.
81 kHz, and the voltage amplitude
is 51
.
2 V.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Work, Magnetic Field, Correct Answer, Husain, Sonia Paban

Click to edit the document details