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howard (cah3459) – hw17 – turner – (56705)
1
This printout should have 17 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001 (part 1 of 2) 10.0 points
A circular coil oF 693 turns and area 0
.
62 m
2
is in a uniForm magnetic feld oF 0
.
89 T. The
maximum torque exerted on the coil by the
feld is 0
.
0027 N
·
m.
Calculate the current in the coil.
Correct answer: 7
.
06072
×
10

6
A.
Explanation:
Let :
τ
=0
.
0027 N
·
m
,
N
= 693
,
A
.
62 m
2
,
and
B
.
89 T
.
The maximum torque is
τ
=
N I A B ,
so the current in the coil is
I
=
τ
N A B
=
(0
.
0027 N
·
m)
(693) (0
.
62 m
2
) (0
.
89 T)
=
7
.
06072
×
10

6
A
.
002 (part 2 of 2) 10.0 points
Assume the 693 turns oF wire are used to Form
a singleturn coil with the same shape but
much larger area.
What is the current iF the maximum torque
exerted on the coil by the feld is 0
.
0027 N
·
m?
Correct answer: 1
.
01886
×
10

8
A.
Explanation:
Let :
N
= 693
.
The radius oF the coil is
r
=
±
A
π
.
and the total length oF the coil is
C
=
N
2
π r .
The new radius oF the singleturn coil is
R
=
C
2
π
=
N r ,
so the new area oF the coil is
A
new
=
π R
2
=
N
2
A.
Using the equation For the current, we fnd
the new value oF the current:
I
new
=
τ
N
2
AB
=
I
N
=
7
.
06072
×
10

6
A
693
=
1
.
01886
×
10

8
A
.
003 (part 1 of 3) 10.0 points
A circular current loop oF radius
R
is placed
in a horizontal plane and maintains a current
I
. There is a constant magnetic feld
$
B
in the
xy
plane, with the angle
α
(
α <
90
◦
) defned
with respect to
y
axis. The current in the loop
±ows counterclockwise as seen From above.
I
z
ˆ
k
x
ˆ
ı
y
,ˆ
B
α
What is the direction oF the magnetic mo
ment
$μ
?
1.

ˆ
ı
2.
+ˆ
ı
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View Full Document howard (cah3459) – hw17 – turner – (56705)
2
3.
+ˆ
correct
4.

sin
α
ˆ

cos
α
ˆ
ı
5.

ˆ
6.
+
ˆ
k
7.

ˆ
k
8.
cos
α
ˆ

sin
α
ˆ
ı
9.
sin
α
ˆ
+ cos
α
ˆ
ı
10.

cos
α
ˆ
+ sin
α
ˆ
ı
Explanation:
I
z
ˆ
k
τ
x
ˆ
ı
μ
loop
B
α
According to the righthand rule, the mag
netic moment is upward.
004 (part 2 of 3) 10.0 points
If the current in the loop is 0
.
202 A and its
radius is 3
.
29 cm, what is the magnitude of
the magnetic moment of the loop?
Correct answer: 0
.
000686899 A
·
m
2
.
Explanation:
Let :
I
=0
.
202 A
,
and
R
=3
.
29 cm
.
The magnetic dipole moment is
μ
=
IA
=
I π R
2
= (0
.
202 A) (3
.
15) (3
.
29 cm)
2
=
0
.
000686899 A
·
m
2
005 (part 3 of 3) 10.0 points
What is the direction of the torque vector
$τ
?
1.

sin
α
ˆ

cos
α
ˆ
ı
2.

ˆ
k
3.
sin
α
ˆ
+ cos
α
ˆ
ı
4.

ˆ
ı
5.

ˆ
6.

cos
α
ˆ
+ sin
α
ˆ
ı
7.
8.
cos
α
ˆ

sin
α
ˆ
ı
9.
+
ˆ
k
correct
10.
ı
Explanation:
The torque is
=
$μ
×
$
B
=
μ
(ˆ
)
×
[
B
x
(

ˆ
ı
)+
B
y
(+ˆ
)]
and ˆ
×
ˆ
= 0 and +ˆ
×
(

ˆ
ı
)=
ˆ
k
, so
=
μB
x
ˆ
k
so the direction of the torque is +
ˆ
k
, which
agrees with the answer based on the right
hand rule.
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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