choe (dec865) – Practice HW 16 Solutions – Weathers – (22202)
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A coil is wrapped with 207 turns of wire on
the perimeter of a square frame of sides 20 cm.
Each turn has the same area, equal to that of
the frame, and the total resistance of the coil
is 4
.
66 Ω. A uniform magnetic field is turned
on perpendicular to the plane of the coil.
If
the
field
changes
linearly
from
0
to
0
.
737 Wb
/
m
2
in a time of 0
.
798 s, find the
magnitude of the induced emf in the coil while
the field is changing.
Correct answer: 7
.
64707 V.
Explanation:
Basic Concept:
Faraday’s Law is
E
=
−
d
Φ
B
dt
.
Solution:
The magnetic flux through the
loop at
t
= 0 is zero since
B
= 0.
At
t
=
0
.
798 s
,
the magnetic flux through the loop
is Φ
B
=
B A
= 0
.
02948 Wb
.
Therefore the
magnitude of the induced emf is
E
=
N
·
ΔΦ
B
Δ
t
=
(207 turns) [(0
.
02948 Wb)
−
0]
(0
.
798 s)
= 7
.
64707 V
E
= 7
.
64707 V
.
002
10.0 points
The twoloop wire circuit is 117
.
543 cm wide
and 78
.
362 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
.
0519 Ω
/
m.
B
B
P
Q
39
.
181 cm
78
.
362 cm
78
.
362 cm
When the magnetic field is 0
.
5 T, find the
magnitude of the current through middle leg
PQ
of the circuit.
Correct answer: 1258
.
22
μ
A.
Explanation:
Let :
ℓ
= 0
.
78362 m
,
A
l
=
1
2
ℓ
2
= 0
.
61406 m
2
/
s
,
A
r
=
ℓ
2
= 1
.
22812 m
2
/
s
,
δ
= 0
.
0519 Ω
/
m
,
and
d B
dt
=
d
dt
α t
=
α
= 0
.
001 T
/
s
.
Basic Concept:
Faraday’s Law is
E
=
−
d
Φ
B
dt
,
where
Φ
B
=
A
B .
Ohm’s Law is
V
=
I R .
Solution:
The instantaneous value of the
magnetic field (
B
= 0
.
5 T) is not germane to
this problem.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
choe (dec865) – Practice HW 16 Solutions – Weathers – (22202)
2
B
B
P
Q
ℓ
2
ℓ
ℓ
I
r
I
l
I
PQ
The resistance for the wire is proportional
to the length of the wire.
For a length of
78
.
362 cm, the resistance is
R
=
δ ℓ
= (0
.
0519 Ω
/
m) (0
.
78362 m) = 0
.
0406699 Ω
.
Using Ohm’s law for the right perimeter and
left perimeter of the left loops in the circuit,
we have
E
r
= 4
ℓ δ I
r
= 4
R
I
r
,
and
(1)
E
l
= 3
ℓ δ I
l
= 3
R
I
l
.
(2)
Note:
When the magnetic field changes
with time, there is an induced emf in both the
righthand side and the lefthand side. From
Faraday’s law, the magnitude of the induced
emf in the loops equals
E
r
=
A
r
d B
dt
=
ℓ
2
d B
dt
,
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Work, Magnetic Field, Faraday's law of induction, Choe

Click to edit the document details