maldonado (om2892) – Homework
4
2010
UPII – Stone – (13750)
1
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printout
should
have
21
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
Two identical parallel sections of metal rods
are connected parallel to a battery as shown.
The two sections of metal rods are free to
move.
5
.
1 m
0
.
24 m
2
.
8 V
Note:
Neglect the resistance of the part of
the circuit not specifically mentioned below.
The battery is 2
.
8 V, each metal rod has a
resistance of 2
.
04 Ω, a length of 5
.
1 m, and a
mass of 0
.
0024 kg
.
The rods are separated by
0
.
24 m
.
Find the magnitude of the acceleration of
each rod.
Correct answer: 0
.
00333606 m
/
s
2
.
Explanation:
Let :
E
= 2
.
8 V
,
R
= 2
.
04 Ω
,
ℓ
= 5
.
1 m
,
m
= 0
.
0024 kg
,
and
Δ
r
= 0
.
24 m
.
ℓ
Δ
r
E
The attractive forces on each metal rod are
equal and opposite to each other, and each
force has its magnitude given by
F
=
I ℓ B
with
I
the current in a rod,
ℓ
the length of
that rod and B the magnetic field at the rod
due to the current in the other rod. Since the
separation Δ
r
of the rods, is much less than
ℓ
we can use Ampere’s law to find
contintegraldisplay
vector
B
·
d
vector
ℓ
= 2
π
Δ
r B
=
μ
0
I
B
=
μ
0
I
2
π
Δ
r
.
Further, the voltage drop through the rod
equals the voltage
E
of the battery;
i.e.
,
E
=
I R ,
and
I
=
E
R
,
and
B
=
μ
0
parenleftbigg
E
R
parenrightbigg
2
π
Δ
r
.
Substituting
I
=
E
R
, and the expression for
B ,
gives for the force,
F
=
I ℓ B
=
parenleftbigg
E
R
parenrightbigg
2
μ
0
ℓ
2
π
Δ
r
With
F
=
m a ,
we find
a
=
parenleftbigg
E
R
parenrightbigg
2
μ
0
ℓ
2
π
Δ
r m
=
parenleftbigg
2
.
8 V
2
.
04 Ω
parenrightbigg
2
(1
.
25664
×
10
−
6
N
/
A
2
)(5
.
1 m)
2
π
(0
.
24 m)(0
.
0024 kg)
=
0
.
00333606 m
/
s
2
.
002
(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 field
vector
B
in the
xy
plane, with the angle
α
(
α <
90
◦
) defined
with respect to
y
axis. The current in the loop
flows counterclockwise as seen from above.
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maldonado (om2892) – Homework
4
2010
UPII – Stone – (13750)
2
I
z
ˆ
k
x
ˆ
ı
y
, ˆ
B
α
What is the direction of the magnetic mo
ment
vectorμ
?
1.
+ˆ
ı
2.

sin
α
ˆ

cos
α
ˆ
ı
3.
sin
α
ˆ
+ cos
α
ˆ
ı
4.
+ˆ
correct
5.

cos
α
ˆ
+ sin
α
ˆ
ı
6.
cos
α
ˆ

sin
α
ˆ
ı
7.

ˆ
k
8.
+
ˆ
k
9.

ˆ
ı
10.

ˆ
Explanation:
I
z
ˆ
k
τ
x
ˆ
ı
μ
loop
B
α
According to the righthand rule, the mag
netic moment is upward.
003
(part 2 of 3) 10.0 points
If the current in the loop is 0
.
235 A and its
radius is 1
.
67 cm, what is the magnitude of
the magnetic moment of the loop?
Correct answer: 0
.
000205897 A
·
m
2
.
Explanation:
Let :
I
= 0
.
235 A
,
and
R
= 1
.
67 cm
.
The magnetic dipole moment is
μ
=
I A
=
I π R
2
= (0
.
235 A) (3
.
15) (1
.
67 cm)
2
=
0
.
000205897 A
·
m
2
004
(part 3 of 3) 10.0 points
What is the direction of the torque vector
vector
τ
?
1.

ˆ
2.

sin
α
ˆ

cos
α
ˆ
ı
3.

cos
α
ˆ
+ sin
α
ˆ
ı
4.
sin
α
ˆ
+ cos
α
ˆ
ı
5.
+ˆ
ı
6.

ˆ
ı
7.

ˆ
k
8.
cos
α
ˆ

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

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

ˆ
ı
) =
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 Summer '10
 Stone
 Work, Magnetic Field, maldonado

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