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001 10.0 points
In an RL series circuit, an inductor of 3.2 H
and a re
...
This print-out should have 14 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
In an RL series circuit, an inductor of 3.2 H
and a resistor of 7.08 Ω are connected to a
24.6 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: 19.3163 J.
Explanation:
Let : L = 3.2 H ,
R = 7.08 Ω , and
E = 24.6 V .
The current in an RL circuit is
I = E
R 1 − e−Rt/L .
The final equilibrium value of the current,
which occurs as t → ∞ , is
I0 = E
R
=
24.6 V
7.08 Ω = 3.47458 A .
The energy stored in the inductor carrying a
current 3.47458 A is
U = 1
2
L I2
=
1 2
(3.2 H) (3.47458 A)2
= 19.3163 J .
002 (part 1 of 3) 10.0 points
A long solenoid carries a current I2 . Another
coil (of larger diameter than the solenoid) is
coaxial with the center of the solenoid, as in
the figure below.
ℓ2
ℓ1
Outside solenoid has N1 turns
Inside solenoid has N2 turns
A1 A2
The current I2 is held constant.
The energy stored in the solenoid is given
by
1. U = µ0
2
N2 A2
ℓ2 I22
2. U = µ0
2
N2
2
A2
ℓ2 I22 correct
3. U = µ0
2
N2
2
ℓ2
A2 I22
4. U = µ0
2
N2
2
A2
ℓ2 I2
5. U = µ0
2
A2 ℓ2 I2
6. U = µ0
2
N2
2 A2 I22
7. U = 1
2 µ0
N2
2 A2 I22
8. U = µ0
2
N2
2
A1
ℓ2 I22
Explanation:
The magnetic energy density is given by
B2
2 µ0
. Inside the solenoid the magnetic field
is B = µ0 N2 I2
ℓ2 , and the volume enclosed by
the solenoid is A2 ℓ2, so
U = 1
2 µ0 µ0 Nℓ22 I22 A2 ℓ2
=
µ0
2
N2
2
A2
ℓ2 I22 .
003 (part 2 of 3) 10.0 points
The mutual inductance M12 between the coil
and the solenoid is given by
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