nguyen (jmn727) – homework 25 – Turner – (59070)
1
This
printout
should
have
11
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
In the arrangement shown in the figure,
the resistor is
R
and a
B
magnetic field is
directed into the paper.
The separation be
tween the rails is
ℓ .
Neglect the mass
m
of the
bar. Assume the bar and rails have negligible
resistance and friction.
An applied force moves the bar to the left
at a constant speed of
v .
m
v
R
B
B
ℓ
a
z
What is the relationship between the elec
tric potential at the ends of the resistor (
a
or
z
) while the bar is moving on the “right” side
of the resistor (towards the resistor) and mov
ing on the “left” side of the resistor (after the
bar moves past the resistor)?
1.
V
z
> V
a
(right)
and
V
a
> V
z
(left)
2.
V
a
> V
z
(right)
and
V
z
> V
a
(left)
3.
V
a
> V
z
(right)
and
V
a
> V
z
(left)
4.
V
a
=
V
z
(right)
and
V
a
=
V
z
5.
V
z
> V
a
(right)
and
V
z
> V
a
(left)
correct
Explanation:
As the bar moves toward the resistor, the
area of the current loop decreases, so the in
duced
vector
B
ind
is downward with
I
ind
clockwise
from above.
Lenz’s law dictates that before moving past
the resistor, current flows from
z
to
a
, so
z
is
at a higher potential.
After going past the resistor, Lenz’s law
dictates that the induced
vector
B
ind
is now upward.
This requires the current to reverse its rota
tional direction to be counterclockwire from
above. However, the direction
z
to
a
(through
the resistor
R
) also reverses its rotational di
rection.
The
emf
across the bar does not
change sign;
i.e.
, the current through the re
sistor
R
remains in the same direction.
002
10.0 points
A rectangular coil of 37 turns, 0
.
24 m by
0
.
16 m, is rotated at 55 rad
/
s in a magnetic
field so that the axis of rotation is perpendicu
lar to the direction of the field. The maximum
emf
induced in the coil is 0
.
3 V.
What is the magnitude of the field?
Correct answer: 3
.
83907 mT.
Explanation:
Let :
N
= 37 turns
,
ω
= 55 rad
/
s
,
ǫ
max
= 0
.
3 V
,
x
= 0
.
24 m
,
and
y
= 0
.
16 m
.
From
ǫ
max
=
N A B ω
, where the area
A
is
A
= (0
.
24 m)(0
.
16 m) = 0
.
0384 m
2
,
so
B
=
ǫ
max
N A ω
=
0
.
3 V
(37 turns)(0
.
0384 m
2
)(55 rad
/
s)
= 0
.
00383907 T =
3
.
83907 mT
.
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 Spring '08
 Turner
 Work, Magnetic Field, Electric charge

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