Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
1
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beFore answering.
001
10.0 points
A copper strip (8
.
47
×
10
22
electrons per cu
bic centimeter) 7
.
4 cm wide and 0
.
02 cm thick
is used to measure the magnitudes oF un
known magnetic felds that are perpendicular
to the strip.
The charge on the electron is 1
.
6
×
10
−
19
C.
±ind the magnitude oF
B
when the current
is 24 A and the Hall voltage is 1
.
8
μ
V.
Correct answer: 0
.
20328 T.
Explanation:
Let :
n
= 8
.
47
×
10
22
cm
−
3
,
= 8
.
47
×
10
28
m
−
3
,
q
= 1
.
6
×
10
−
19
C
,
t
= 0
.
02 cm = 0
.
0002 m
,
w
= 7
.
4 cm = 0
.
074 m
,
I
= 24 A
,
and
V
H
= 1
.
8
μ
V = 1
.
8
×
10
−
6
V
.
The current in the metal strip is
I
=
n q v
d
A
=
n q v
d
(
w t
)
v
d
w
=
I
n q t
The Hall voltage is
V
H
=
v
d
w B
B
=
V
H
v
d
w
B
=
n q tV
H
I
=
(8
.
47
×
10
28
m
−
3
) (1
.
6
×
10
−
19
C)
24 A
×
(0
.
0002 m) (1
.
8
×
10
−
6
V)
=
0
.
20328 T
.
002
10.0 points
A small rectangular coil composed oF 49 turns
oF wire has an area oF 41 cm
2
and carries a
current oF 1
.
2 A. When the plane oF the coil
makes an angle oF 36
◦
with a uniForm magnetic
feld, the torque on the coil is 0
.
05 N m.
What is the magnitude oF the magnetic
feld?
Correct answer: 0
.
256361 T.
Explanation:
Let :
N
= 49 turns
,
I
= 1
.
2 A
,
θ
= 36
◦
,
A
= 41 cm
2
= 0
.
0041 m
2
,
and
τ
= 0
.
05 N m
.
The magnetic Force on the current is
±
F
=
I
±
ℓ
×
±
B
and the torque is
±
τ
=
±
r
×
±
F
,
so the torque on the loop due to the magnetic
feld is
τ
= 2
F r
cos
θ
= (
N I ℓ B
)
w
cos
θ
=
N I B
(
ℓ w
) cos
θ
=
N I B A
cos
θ ,
where
A
is the area oF the loop and
θ
is the
angle between the plane oF the loop and the
magnetic feld.
The magnetic feld From above is
B
=
τ
N I A
cos
θ
=
0
.
05 N m
(49 turns) (1
.
2 A) (0
.
0041 m
2
) cos(36
◦
)
=
0
.
256361 T
.
003
10.0 points
Given:
Assume the bar and rails have neg
ligible resistance and Friction.
In the arrangement shown in the fgure,
the resistor is 6 Ω and a 7 T magnetic feld
is directed into the paper.
The separation
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2
between the rails is 5 m
.
Neglect the mass of
the bar.
An applied force moves the bar to the left
at a constant speed of 5 m
/
s
.
m
≪
1 g
5 m
/
s
6 Ω
7 T
7 T
I
5 m
At what rate is energy dissipated in the
resistor?
Correct answer: 5104
.
17 W.
Explanation:
Basic Concept:
Motional
E
E
=
B ℓ v .
Ohm’s Law
I
=
V
R
.
Solution:
The motional
E
induced in the
circuit is
E
=
B ℓ v
= (7 T) (5 m) (5 m
/
s)
= 175 V
.
From Ohm’s law, the current ±owing through
the resistor is
I
=
E
R
=
B ℓ v
R
=
(7 T) (5 m) (5 m
/
s)
R
= 29
.
1667 A
.
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 Fall '08
 Turner
 Physics, Magnetic Field, Kapoor

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