terry (ect328) – oldhomework 26 – Turner – (59130)
1
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printout
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have
14
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before answering.
001
10.0 points
In an RL series circuit, an inductor of 4
.
26 H
and a resistor of 6
.
45 Ω are connected to a
28
.
4 V battery.
The switch of the circuit
is initially open.
Next close the switch and
wait for a long time. Eventually the current
reaches its equilibrium value.
At this time, what is the corresponding
energy stored in the inductor?
Correct answer: 41
.
2949 J.
Explanation:
Let :
L
= 4
.
26 H
,
R
= 6
.
45 Ω
,
and
E
= 28
.
4 V
.
The current in an RL circuit is
I
=
E
R
parenleftBig
1

e
−
Rt/L
parenrightBig
.
The final equilibrium value of the current,
which occurs as
t
→ ∞
,
is
I
0
=
E
R
=
28
.
4 V
6
.
45 Ω
= 4
.
4031 A
.
The energy stored in the inductor carrying a
current 4
.
4031 A is
U
=
1
2
L I
2
=
1
2
(4
.
26 H) (4
.
4031 A)
2
=
41
.
2949 J
.
002
(part 1 of 3) 10.0 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.
ℓ
2
ℓ
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
ℓ
2
A
2
I
2
2
2.
U
=
μ
0
2
A
2
ℓ
2
I
2
3.
U
=
μ
0
2
N
2
A
2
ℓ
2
I
2
2
4.
U
=
μ
0
2
N
2
2
A
1
ℓ
2
I
2
2
5.
U
=
μ
0
2
N
2
2
A
2
I
2
2
6.
U
=
μ
0
2
N
2
2
A
2
ℓ
2
I
2
2
correct
7.
U
=
μ
0
2
N
2
2
A
2
ℓ
2
I
2
8.
U
=
1
2
μ
0
N
2
2
A
2
I
2
2
Explanation:
The magnetic energy density is given by
B
2
2
μ
0
.
Inside the solenoid the magnetic field
is
B
=
μ
0
N
2
I
2
ℓ
2
, and the volume enclosed by
the solenoid is
A
2
ℓ
2
, so
U
=
1
2
μ
0
parenleftbigg
μ
0
N
2
ℓ
2
I
2
parenrightbigg
2
A
2
ℓ
2
=
μ
0
2
N
2
2
A
2
ℓ
2
I
2
2
.
003
(part 2 of 3) 10.0 points
The mutual inductance
M
12
between the coil
and the solenoid is given by
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terry (ect328) – oldhomework 26 – Turner – (59130)
2
1.
M
12
=
μ
0
N
1
N
2
A
2
ℓ
2
correct
2.
M
12
=
μ
0
N
1
N
2
A
1
ℓ
2
3.
M
12
=
μ
0
N
1
N
1
N
2
4.
M
12
=
μ
0
N
1
N
2
A
1
5.
M
12
=
μ
0
N
2
A
2
ℓ
2
6.
M
12
=
μ
0
ℓ
2
N
1
N
2
A
2
7.
M
12
=
μ
0
N
1
N
2
ℓ
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
ℓ
2
I
2
A
2
,
so
M
12
=
μ
0
N
2
N
1
A
2
ℓ
2
.
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 Fall '09
 Turner
 Inductance, Inductor, Terry, iron core

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