segovia (cs39966) – Homework 5 – Spurlock – (14441)
1
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001
10.0points
A circular coil consisting of a single loop of
wire has a radius of 19
.
6 cm and carries a
current of 37
.
3 A. It is placed in an external
magnetic field of 0
.
258 T.
Find the magnitude of the torque on the
wire when the plane of the coil makes an
angle of 25
.
3
◦
with the direction of the field.
Correct answer: 1
.
05002 N
·
m.
Explanation:
Let :
r
= 19
.
6 cm = 0
.
196 m
,
B
= 0
.
258 T
,
I
= 37
.
3 A
,
and
θ
= 25
.
3
◦
.
The area of the circular coil is
A
=
π r
2
=
π
(0
.
196 m)
2
= 0
.
120687 m
2
.
The torque is
bardbl
vector
τ
bardbl
=
bardbl
vectorμ
×
vector
B
bardbl
=
μ B
sin
θ ,
where
θ
is the angle that the normal to the
loop makes with the magnetic field, and
μ
=
I A
is the magnetic moment of the loop.
Thus the magnitude of the torque is
τ
=
I A B
sin
θ
= (37
.
3 A) (0
.
120687 m
2
) (0
.
258 T)
×
sin(90
◦
−
25
.
3
◦
)
=
1
.
05002 N
·
m
.
002
10.0points
Two long straight wires carry currents per
pendicular to the
xy
plane. One wire carries a
current of 97 A and passes through the point
x
= 11 cm on the
x
axis. The second wire car
ries a current of 12000 A and passes through
the point
y
= 7
.
1 cm on the
y
axis.
What is the magnitude of the resulting
magnetic field at the origin?
Correct answer: 33803
.
3
μ
T.
Explanation:
Let :
x
= 11 cm = 0
.
11 m
,
I
1
= 97 A
,
y
= 7
.
1 cm = 0
.
071 m
,
I
2
= 12000 A
,
and
μ
0
= 4
π
×
10
−
7
T
·
m
/
A
.
The magnetic field from the first wire will
be
vector
B
1
=
μ
0
I
1
2
π x
(
±
ˆ
)
directed along the
y
axis at the origin.
The
magnetic field from the second wire will be
vector
B
2
=
μ
0
I
2
2
π y
(
±
ˆ
ı
)
directed along the
x
axis at the origin. Since
these contributions are perpendicular to each
other, the magnitude of the total magnetic
field at the origin will be
B
=
radicalBig
B
2
1
+
B
2
2
=
radicalBigg
parenleftbigg
μ
0
I
1
2
π x
parenrightbigg
2
+
parenleftbigg
μ
0
I
2
2
π y
parenrightbigg
2
=
radicalBigg
parenleftBig
μ
0
2
π
parenrightBig
2
parenleftbigg
I
2
1
x
2
+
I
2
2
y
2
parenrightbigg
=
μ
0
2
π
radicalBigg
I
2
1
x
2
+
I
2
2
y
2
=
4
π
×
10
−
7
T
·
m
/
A
2
π
×
radicalBigg
(97 A)
2
(0
.
11 m)
2
+
(12000 A)
2
(0
.
071 m)
2
×
10
6
μ
T
1 T
=
33803
.
3
μ
T
.
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segovia (cs39966) – Homework 5 – Spurlock – (14441)
2
003
10.0points
A long, thin conductor carries a current of
10
.
5 A.
At what distance from the conductor is
the magnitude of the resulting magnetic field
0
.
000114 T?
Correct answer: 1
.
84211 cm.
Explanation:
Let :
μ
0
= 1
.
25664
×
10
−
6
N
/
A
2
,
B
= 0
.
000114 T
,
and
I
= 10
.
5 A
.
The magnetic field is
B
=
μ
0
I
2
π r
r
=
μ
0
I
2
π B
=
(1
.
25664
×
10
−
6
N
/
A
2
) (10
.
5 A)
2
π
(0
.
000114 T)
= 0
.
0184211 m
=
1
.
84211 cm
.
keywords:
004
10.0points
Two long, straight parallel wires 8
.
2 cm apart
carry currents of equal magnitude
I
.
The
parallel wires repel each other with a force
per unit length of 2
.
8 nN
/
m.
The permeability of free space is 4
π
×
10
−
7
T
·
m
/
A.
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 Fall '11
 HOFFMAN
 Work, Magnetic Field, Correct Answer, Segovia

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