in this case would be in the counterclockwise
direction.
005
10.0 points
A 39
.
2 turns circular coil with radius 6
.
92 cm
and resistance 3
.
14 Ω is placed in a magnetic
field directed perpendicular to the plane of
the coil. The magnitude of the magnetic field
varies in time according to the expression
B
=
a
1
t
+
a
2
t
2
,
where
a
1
= 0
.
0141 T
/
s,
a
2
= 0
.
0893 T
/
s
2
are
constants, time
t
is in seconds and field
B
is
in Tesla.
Find the magnitude of the induced
emf
in
the coil at
t
= 7
.
76 s.
Correct answer: 0
.
825634 V.
Explanation:
Let :
n
= 39
.
2 turns
,
r
= 6
.
92 cm = 0
.
0692 m
,
R
= 3
.
14 Ω
,
a
1
= 0
.
0141 T
/
s
,
a
2
= 0
.
0893 T
/
s
2
,
and
t
= 7
.
76 s
.
The area of the circular coil is
A
=
π r
2
=
π
(0
.
0692 m)
2
= 0
.
015044 m
2
,
so from Faraday’s law,
E
=

n
d
Φ
B
dt
=

n
d A B
dt
=

n A
d B
dt
=

n A
(
a
1
+ 2
a
2
t
)
=

(39
.
2 turns) (0
.
015044 m
2
)
×
[0
.
0141 T
/
s + 2 (0
.
0893 T
/
s
2
) (7
.
76 s)]
=

0
.
825634 V
,
johnson (rj6247) – hw 11 – Opyrchal – (121014)
3
which has a magnitude of
0
.
825634 V
.
006
10.0 points
Given:
μ
0
= 1
.
25664
×
10

6
N
/
A
2
.
A 3 m long large coil with a radius of
15
.
1 cm and 280 turns surrounds a 7
.
1 m
long solenoid with a radius of 5
.
1 cm and 6300
turns, see figure below.
The current in the solenoid changes as
I
=
I
0
sin(2
π f t
)
,
where
I
0
= 30 A and
f
= 60 Hz.
7
.
1 m
3 m
15
.
1 cm
5
.
1 cm
16 Ω
E
=
E
0
sin
ω t
ℓ
2
ℓ
1
A
1
A
2
R
E
=
E
0
sin
ω t
Basic Concepts:
Faraday’s law
E
=

N
d
Φ
dt
and Ohm’s law
E
=
I R .
Solution:
The angular velocity is
ω
= 2
π f
= 2
π
(60 Hz) = 376
.
991 rad
/
s
.
Find the maximum induced current
I
max
in
the large coil.
The
solenoid
carries
a
current
I
=
johnson (rj6247) – hw 11 – Opyrchal – (121014)
4
where
A
2
=
π
(0
.
051 m)
2
= 0
.
00817128 m
2
.
The induced
emf
is
E
1
=
M
12
d I
dt
=
M
12
I
0
d
dt
sin
ω t
=
M
12
I
0
ω
cos
ω t
= (0
.
00255118 H) (30 A)
×
(376
.
991 rad
/
s) cos
ω t
= (28
.
8531 V) cos
ω t ,
where
the
maximum
emf
is
E
max
1
=
28
.
8531 V
.
We can then determine the maximum cur
rent in the resistor using Ohm’s law.
I
max
=
E
max
1
R
=
28
.
8531 V
16 Ω
=
1
.
80332 A
007
10.0 points
A powerful electromagnet has a field of 1
.
39 T
and a cross sectional area of 0
.
863 m
2
. Now we
place a coil of 171 turns with a total resistance
of 2
.
13 Ω around the electromagnet and turn
off the power to the electromagnet in 0
.
0359 s.
What will be the induced current in the
coil?
For coil of many turns, the emf is the sum
of induced emf in each turn.
Besides, when
=
ΔΦ
B
Δ
t
=
Δ
B
·
A
Δ
t
= 33
.
4142 V
So, the emf in the whole coil is:
E
=
n
· E
1
= 171 turns
×
33
.
4142 V
= 5713
.
83 V
Apply Ohm’s law, we get the current as fol
lowing:
I
=
V
R
=
E
R
=
5713
.
83 V
2
.
13 Ω
= 2682
.
55 a
008
(part 1 of 2) 10.0 points
calculating the induced emf in the coil, we
actually are try to get a average emf. For one
turn of coil, the induced emf is: