Practice HW 16 solution - song(shs546 Practice HW 16...

song (shs546) – Practice HW 16 Solutions – weathers – (22202) 1
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001 0.0points
A coil is wrapped with 224 turns of wire on the
perimeter of a square frame of sides 21 . 6 cm.
Each turn has the same area, equal to that of
the frame, and the total resistance of the coil
is 2 . 42 Ω. A uniform magnetic field is turned
on perpendicular to the plane of the coil.
If the field changes linearly from 0 to
− 0 . 237 Wb / m
2
in a time of 1 . 16 s, find the
magnitude of the induced emf in the coil while
the field is changing.
Correct answer: 2 . 13524 V.
Explanation:
BasicConcept: 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 =
1 . 16 s , the magnetic flux through the loop is
Φ
B
= B A = − 0 . 0110575 Wb . Therefore the
magnitude of the induced emf is
E =
N · ΔΦ
B
Δ t
=
(224 turns) [( − 0 . 0110575 Wb) − 0]
(1 . 16 s)
= − 2 . 13524 V
|E| = 2 . 13524 V .
002 0.0points
The two-loop wire circuit is 43 . 7559 cm wide
and 29 . 1706 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 . 113 Ω / m.
B
B
P
Q
14 . 5853 cm 29 . 1706 cm
29 . 1706 cm
When the magnetic field is 0 . 5 T, find the
magnitude of the current through middle leg
PQ of the circuit.
Correct answer: 215 . 122 μ A.
Explanation:
Let : ℓ = 0 . 291706 m ,
A
l
=
1
2
ℓ
2
= 0 . 0850924 m
2
/ s ,
A
r
= ℓ
2
= 0 . 170185 m
2
/ s ,
δ = 0 . 113 Ω / m , and
d B
dt
=
d
dt
α t = α = 0 . 001 T / s .
BasicConcept:
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.

song (shs546) – 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
29 . 1706 cm, the resistance is
R = δ ℓ = (0 . 113 Ω / m) (0 . 291706 m) = 0 . 0329628 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
right-hand side and the left-hand side. From
Faraday’s law, the magnitude of the induced
emf in the loops equals
E
r
= A
r
d B
dt
=