homework 09 – KIM, JI – Due: Mar 23 2008, 4:00 am
1
Question 1, chap 31, sect 2.
part 1 of 1
10 points
A square loop of wire of resistance
R
and
side
a
is oriented with its plane perpendicular
to a magnetic field
B
, as shown in the figure.
B
B
a
I
What must be the rate of change of the
magnetic field in order to produce a current
I
in the loop?
1.
d B
dt
=
I a
2
R
2.
d B
dt
=
R a
I
3.
d B
dt
=
I R a
4.
d B
dt
=
I R
a
2
correct
5.
d B
dt
=
I a
R
Explanation:
The
emf
produced is given by
E
=
d
φ
dt
=
a
2
d B
dt
.
Using Ohm’s law we can solve for B
I R
=
a
2
d B
dt
d B
dt
=
I R
a
2
.
Question 2, chap 31, sect 2.
part 1 of 1
10 points
A toroid having a rectangular cross section
(
a
= 2
.
62 cm by
b
= 4
.
11 cm) and inner ra
dius 2
.
34 cm consists of
N
= 210 turns of
wire that carries a current
I
=
I
0
sin
ω
t
, with
I
0
= 31
.
6 A and a frequency
f
= 42
.
9 Hz.
A
loop
that
consists
of
N
=
29
turns
of wire links the toroid,
as in the figure.
b
a
N
R
N
l
Determine the maximum
E
induced in the
loop by the changing current
I
.
Correct answer: 0
.
275601
V (tolerance
±
1
%).
Explanation:
Basic Concept:
Faraday’s Law
E
=

d
Φ
B
dt
.
Magnetic field in a toroid
B
=
μ
0
N I
2
π
r
.
Solution:
In a toroid, all the flux is confined
to the inside of the toroid
B
=
μ
0
N I
2
π
r
.
So, the flux through the loop of wire is
Φ
B
1
=
B dA
=
μ
0
N I
0
2
π
sin(
ω
t
)
b
+
R
R
a dr
r
=
μ
0
N I
0
2
π
a
sin(
ω
t
) ln
b
+
R
R
.
Applying Faraday’s law, the induced emf can
be calculated as follows
E
=

N
d
Φ
B
1
dt
=

N
μ
0
N I
0
2
π
ω
a
ln
b
+
R
R
cos(
ω
t
)
=

E
0
cos(
ω
t
)
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homework 09 – KIM, JI – Due: Mar 23 2008, 4:00 am
2
where
ω
= 2
π
f
was used.
The maximum magnitude of the induced
emf
,
E
0
, is the coe
ffi
cient in front of cos(
ω
t
).
E
0
=

N
d
Φ
B
1
dt
=

N μ
0
N I
0
ω
2
π
a
ln
b
+
R
R
=

(29 turns)
μ
0
(210 turns)
×
(31
.
6 A) (42
.
9 Hz) (2
.
62 cm)
×
ln
(4
.
11 cm) + (2
.
34 cm)
(2
.
34 cm)
=

0
.
275601 V
E
= 0
.
275601 V
.
Question 3, chap 31, sect 1.
part 1 of 2
10 points
In the arrangement shown in the figure, the
resistor is 2
Ω
and a 2 T magnetic field is
directed out of the paper.
The separation
between the rails is 4 m
.
Neglect the mass of
the bar.
An applied force moves the bar to the left
at a constant speed of 4 m
/
s
.
Assume the
bar and rails have negligible resistance and
friction.
m
1 g
4 m
/
s
2
Ω
2 T
2 T
I
4 m
Calculate
the
applied
force
required
to
move the bar to the left at a constant speed
of 4 m
/
s.
Correct answer: 128 N (tolerance
±
1 %).
Explanation:
Motional
emf
is
E
=
B
v .
Magnetic force on current is
F
=
I
×
B .
Ohm’s Law is
I
=
V
R
.
The motional
emf
induced in the circuit is
E
=
B
v
= (2 T) (4 m) (4 m
/
s)
= 32 V
.
From Ohm’s law, the current flowing through
the resistor is
I
=
E
R
=
32 V
2
Ω
= 16 A
.
Thus, the magnitude of the force exerted on
the bar due to the magnetic field is
F
B
=
I
B
= (16 A)(4 m)(2 T)
= 128 N
.
To maintain the motion of the bar, a force
must be applied on the bar to balance the
magnetic force
F
=
F
B
=
128 N
.
Question 4, chap 31, sect 1.
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