Anderson, Matthew – Homework 11 – Due: Apr 15 2008, 9:00 am – Inst: Vitaly
1
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11
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before answering.
The due time is Central
time.
001
(part 1 of 1) 15 points
A plane loop of wire of area A is placed in a
region where the magnetic field is perpendicu
lar to the plane. The magnitude of B varies in
time according to the expression
B
=
B
0
e

at
.
That is, at
t
= 0 the field is
B
0
, and for
t >
0,
the field decreases exponentially in time.
Find the induced emf,
E
, in the loop as a
function of time.
1.
E
=
a B
0
t
2.
E
=
a A B
0
e

2
at
3.
E
=
a A B
0
e

at
correct
4.
E
=
A B
0
e

at
5.
E
=
a B
0
e

at
6.
E
=
a A B
0
Explanation:
Basic Concepts:
Faraday’s Law:
E ≡
I
E
·
ds
=

d
Φ
B
dt
Solution:
Since B is perpendicular to the
plane of the loop, the magnetic flux through
the loop at time
t >
0 is
Φ
B
=
B A
=
A B
0
e

a t
Also, since the coefficient
AB
0
and the pa
rameter a are constants, and Faraday’s Law
says
E
=

d
Φ
B
dt
the induced emf can be calculated the from
Equation above:
E
=

d
Φ
B
dt
=

A B
0
d
dt
e

a t
=
a A B
0
e

a t
That is, the induced emf decays exponentially
in time.
Note:
The maximum emf occurs at
t
= 0
,
where
E
=
a A B
0
.
B
=
B
0
e

at
B
0
0
0
~
t
The plot of
E
versus
t
is similar to the
B
versus
t
curve shown in the figure above.
keywords:
002
(part 1 of 1) 15 points
The plane of a rectangular coil, 7
.
2 cm by
11
.
4 cm, is perpendicular to the direction of a
uniform magnetic field
B
.
If the coil has 38 turns and a total resistance
of 11
.
1 Ω, at what rate must the magnitude of
B
change to induce a current of 0
.
14 A in the
windings of the coil?
Correct answer: 4
.
9823 T
/
s.
Explanation:
Given :
x
= 7
.
2 cm = 0
.
072 m
,
y
= 11
.
4 cm = 0
.
114 m
,
N
= 38 turns
,
r
= 11
.
1 Ω
,
and
I
= 0
.
14 A
.
The induced emf is
E
=
I R
=
N
d
Φ
dt
=
N
d
(
B A
)
dt
=
N A
d B
dt
,
so
d B
dt
=
I R
N
(
x y
)
=
(0
.
14 A) (11
.
1 Ω)
38 (0
.
072 m) (0
.
114 m)
= 4
.
9823 T
/
s
.
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Anderson, Matthew – Homework 11 – Due: Apr 15 2008, 9:00 am – Inst: Vitaly
2
keywords:
003
(part 1 of 1) 15 points
A conducting bar moves as shown near a long
wire carrying a constant
I
= 30 A current.
I
a
v
L
A
B
If
a
= 2
.
4 mm,
L
= 140 cm, and
v
=
26
m
/
s,
what
is
the
potential
difference,
Δ
V
≡
V
A

V
B
?
Correct answer: 91 mV.
Explanation:
Given;
E
=

d
Φ
B
dt
=

d
dt
(
B ‘ x
)
=

B ‘
dx
dt
=

B ‘ v .
From
Ampere’s
law,
the
strength
of
the
magnetic field created by the long current
carrying wire at a distance
a
from the wire
is
B
=
μ
0
I
2
π a
.
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 Spring '08
 Opyrchal
 Work, Magnetic Field

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