Version PREVIEW – HW9 – feng – (58230)
1
This
printout
should
have
25
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
Changing Magnetic Field
001
10.0 points
A coil is wrapped with 193 turns of wire on
the perimeter of a circular frame (of radius
42 cm). Each turn has the same area, equal
to that of the frame.
A uniform magnetic
field is directed perpendicular to the plane of
the coil. This field changes at a constant rate
from 30 mT to 49 mT in 80 ms.
What is the magnitude of the induced
E
in
the coil at the instant the magnetic field has
a magnitude of 46 mT?
Correct answer: 25
.
4021 V.
Explanation:
Basic Concepts:
E
=

N
d
Φ
B
dt
Φ
B
≡
integraldisplay
vector
B
·
d
vector
A
=
B
·
A
Solution:
E
=

N
d
Φ
B
dt
=

N A
Δ
B
Δ
t
=

N π r
2
(
B
2

B
1
)
Δ
t
=

(193)
π
(42 cm)
2
×
(49 mT)

(30 mT)
80 ms
=

25
.
4021 V
E
= 25
.
4021 V
.
Bar Moving in a Field
002
(part 1 of 2) 10.0 points
A straight rod moves along parallel conduct
ing rails, as shown below. The rails are con
nected at the left side through a resistor so
that the rod and rails form a closed rectangu
lar loop. A uniform field perpendicular to the
movement of the rod exists throughout the
region.
Assume the rod remains in contact with
the rails as it moves. The rod experiences no
friction or air drag.
The rails and rod have
negligible resistance.
4
.
6 g
6
.
8 Ω
2
.
3 T
2
.
3 T
0
.
47 A
1
.
3 m
At what speed should the rod be moving to
produce the downward current in the resistor?
Correct answer: 1
.
0689 m
/
s.
Explanation:
Let :
I
= 0
.
47 A
,
ℓ
= 1
.
3 m
,
m
= 4
.
6 g
,
R
= 6
.
8 Ω
,
and
B
= 2
.
3 T
.
m
R
B
B
I
ℓ
From Ohm’s Law, the
emf
inside the loop is
E
=
I R .
and the motional
emf
is
E
=
B ℓ v .
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Version PREVIEW – HW9 – feng – (58230)
2
Thus the velocity of the rod is
v
=
E
B ℓ
=
I R
B ℓ
=
(0
.
47 A) (6
.
8 Ω)
(2
.
3 T) (1
.
3 m)
=
1
.
0689 m
/
s
.
003
(part 2 of 2) 10.0 points
The rod is
1.
moving to the left.
2.
stationary.
3.
moving to the right.
correct
4.
moving either left or right, since both
directions produce the same result.
Explanation:
The downward current through the resistor
reduces the existing flux Φ
B
in the loop;
i.e.
the current produces magnetic flux in the op
posite direction as the existing flux, into the
paper.
Such a response counteracts the re
duction of enclosed flux within the rail and
rod loop that arises if the rod moves to the
right, as would follow from Lenz’ law.
Therefore the rod must be moving to the
right.
Circuit in a Magnetic Field 02
004
10.0 points
The twoloop wire circuit is 123
.
523 cm wide
and 82
.
3486 cm high.
The wire circuit in
the figure is located in a magnetic field whose
magnitude varies with time according to the
expression
B
= (0
.
001 T
/
s)
t
and its direction
is into the page.
Assume The resistance per length of the
wire is 0
.
134 Ω
/
m.
B
B
P
Q
41
.
1743 cm
82
.
3486 cm
82
.
3486 cm
When the magnetic field is 0
.
2 T, find the
magnitude of the current through middle leg
PQ
of the circuit.
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 Spring '08
 Turner
 Physics, Inductance, Magnetic Field, Inductor, iL, Faraday's law of induction

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