Shie, Gary – Homework 25 – Due: Nov 1 2004, 4:00 am – Inst: Turner
1
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The due time is Central
time.
001
(part 1 of 3) 10 points
A long solenoid carries a current
I
2
.
Another
coil (of larger diameter than the solenoid) is
coaxial with the center of the solenoid, as in
the figure below.
‘
2
‘
1
Outside solenoid has
N
1
turns
Inside solenoid has
N
2
turns
A
1
A
2
The current
I
2
is held constant. The energy
stored in the solenoid is given by
1.
U
=
μ
0
2
N
2
2
A
2
‘
2
I
2
2.
U
=
μ
0
2
N
2
2
‘
2
A
2
I
2
2
3.
U
=
μ
0
2
N
2
2
A
2
I
2
2
4.
U
=
1
2
μ
0
N
2
2
A
2
I
2
2
5.
U
=
μ
0
2
A
2
‘
2
I
2
6.
U
=
μ
0
2
N
2
2
A
1
‘
2
I
2
2
7.
U
=
μ
0
2
N
2
2
A
2
‘
2
I
2
2
correct
8.
U
=
μ
0
2
N
2
A
2
‘
2
I
2
2
Explanation:
Basic concept:
Magnetic flux magnetic
field of a solenoid at the center
B
=
μ
0
N I
‘
,
magnetic energy, mutual induction.
Solution:
The magnetic energy density is
given by
B
2
2
μ
0
.
Inside the solenoid the mag
netic field is
B
=
μ
0
N
2
I
2
‘
2
, and the volume
enclosed by the solenoid is
A
2
‘
2
, so
U
=
1
2
μ
0
μ
0
N
2
‘
2
I
2
¶
2
A
2
‘
2
=
μ
0
2
N
2
2
A
2
‘
2
I
2
2
.
002
(part 2 of 3) 10 points
The mutual inductance
M
12
between the coil
and the solenoid is given by
1.
M
12
=
μ
0
N
1
N
2
A
1
‘
2
2.
M
12
=
μ
0
N
1
N
1
N
2
3.
M
12
=
μ
0
N
1
N
2
A
2
‘
2
correct
4.
M
12
=
μ
0
N
1
N
2
A
1
5.
M
12
=
μ
0
N
2
A
2
‘
2
6.
M
12
=
μ
0
‘
2
N
1
N
2
A
2
7.
M
12
=
μ
0
N
1
N
2
‘
2
Explanation:
The mutual inductance
M
12
of loop 1 with
respect to loop 2 is defined as
M
12
≡
N
1
Φ
12
I
2
,
where Φ
12
is the flux through a single loop 1
due to loop 2. Since the magnetic field inside
the coil is restricted to the part inside the
solenoid, we have Φ
12
=
μ
0
N
2
‘
2
I
2
A
2
,
so
M
12
=
μ
0
N
2
N
1
A
2
‘
2
.
003
(part 3 of 3) 10 points
Let the current
I
2
be dependent on time,
I
=
I
0
exp(

a t
), where
I
0
= 1
.
6 A,
a
= 5 s

1
,
and
t
is measured in seconds.
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Shie, Gary – Homework 25 – Due: Nov 1 2004, 4:00 am – Inst: Turner
2
If the mutual inductance between the coil
and the solenoid is 6 mH, what is the mag
nitude of the
emf
induced in the coil at time
t
1
= 3
.
7 s ?
Correct answer: 4
.
43398
×
10

7
mV.
Explanation:
The magnitude of the induced
emf
on the
circular coil is
E
ind
=
M
12
fl
fl
fl
fl
dI
2
dt
fl
fl
fl
fl
=
M
12
I
0
a e

a t
1
= 4
.
43398
×
10

7
mV
.
004
(part 1 of 2) 10 points
A smalldiameter inductor having a self
inductance of 120 mH and a largediameter
inductor having a selfinductance of 76 mH
are connected in series as shown in the figure.
Consider the two inductors to be far apart.
Thus, the mutual inductance between the two
inductors is
M
= 0 mH.
a
b
Determine the equivalent selfinductance
L
eq
for the system.
Correct answer: 196 mH.
Explanation:
Let :
L
s
= 120 mH
and
L
‘
= 76 mH
.
Let
E
‘
be the
emf
across the largerdiameter
inductor,
E
s
be the
emf
across the smaller
diameter inductor, and
E
be the total
emf
across both inductors.
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 Spring '10
 ERSKINE
 Physics, Inductance, Energy, Work, Inductor, dt dt dI

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