brown (twb493) – Ch23
h1 – chiu – (56565)
1
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printout
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have
28
questions.
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before answering.
001
10.0points
A magnetic field near the floor points up and
is increasing. Looking down at the floor, does
the nonCoulomb electric field curl clockwise
or counterclockwise?
1.
Not enough information
2.
Counterclockwise
3.
Clockwise
correct
Explanation:
From the Maxwell’s equation, we see that
change in the magnetic field
d
vector
B
dt
in this case is
down so the electric field
vector
E
NC
curls clockwise.
002
10.0points
A magnetic field near the ceiling points down
and is decreasing.
Looking up at the ceil
ing, does the nonCoulomb electric field curl
clockwise or counterclockwise?
1.
Counterclockwise
correct
2.
Not enough information
3.
Clockwise
Explanation:
From the Maxwell’s equation, we see that
change in the magnetic field
d
vector
B
dt
in this case is
toward the floor so the electric field
vector
E
NC
curls
counterclockwise.
003(part1of2)10.0points
On a circular path of radius 8 cm in air around
a solenoid with increasing magnetic field, the
emf is 35 V.
What is the magnitude of the nonCoulomb
electric field on this path?
Correct answer: 69
.
6303 V
/
m.
Explanation:
Here we will use the definition of emf writ
ten in terms of the electric field.
Since the
value of potential is given, we can find out the
magnitude of the electric field. We begin by
writing down the equation for the emf:
emf =
contintegraldisplay
C
vector
E
NC
·
d
vector
ℓ
=
vextendsingle
vextendsingle
vextendsingle
vector
E
NC
vextendsingle
vextendsingle
vextendsingle
contintegraldisplay
C
d
vector
ℓ
= 2
π r
vextendsingle
vextendsingle
vextendsingle
vector
E
NC
vextendsingle
vextendsingle
vextendsingle
= 35 V
.
So, rearranging this expression, we obtain
vextendsingle
vextendsingle
vextendsingle
vector
E
NC
vextendsingle
vextendsingle
vextendsingle
=
35 V
2
π r
=
35 V
2
π
(8 cm)
=
69
.
6303 V
/
m
.
004(part2of2)10.0points
A wire with resistance 5 Ω is placed along the
path. What is the current in the wire?
Correct answer: 7 A.
Explanation:
Now we use Ohm’s Law to figure out the
current flow:
I
=
Δ
V
R
=
emf
R
=
35 V
5 Ω
=
7 A
.
005
10.0points
A uniform magnetic field of 3 T points 28
degrees away from the perpendicular to the
plane of a rectangular loop of wire 0
.
2 m by
0
.
4 m as in the figure below.
ˆ
n
28
◦
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brown (twb493) – Ch23
h1 – chiu – (56565)
2
What is the magnetic flux on this loop?
Correct answer: 0
.
211907 T
·
m
2
.
Explanation:
We can write
Φ
mag
=
integraldisplay
S
vector
B
·
ˆ
n dA
=
integraldisplay
S
vextendsingle
vextendsingle
vextendsingle
vector
B
vextendsingle
vextendsingle
vextendsingle
cos 28
◦
dA
=
vextendsingle
vextendsingle
vextendsingle
vector
B
vextendsingle
vextendsingle
vextendsingle
cos 28
◦
integraldisplay
S
dA
=
vextendsingle
vextendsingle
vextendsingle
vector
B
vextendsingle
vextendsingle
vextendsingle
(cos 28
◦
)
A
= (3 T)(cos 28
◦
)(0
.
2 m)(0
.
4 m)
=
0
.
211907 T
·
m
2
.
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 Fall '08
 Turner
 Physics, Magnetic Field

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