Physics 213 — Problem Set 10 — Solutions
Spring 1998
1. Reading Assignment
Serway Sections 31.7, 32.5, 34.1–4.
2. Serway 32.15
A solenoid has 120 turns uniformly wrapped around a wooden core, which has a diameter of 10 mm and a
length of 9 cm. (a) Calculate the inductance of the solenoid. (b) The wooden core is replaced with a soft
iron rod that has the same dimensions but with a magnetic permeability µm = 800µ0 (ie, 1 + χ = 800
where χ is the magnetic susceptibility).
SOLUTION:
a)Assume µm = µ0 for the wooden core, then :
L
=
=
=
µ0 N 2 A
NΦB
=
I
l
(4π × 10−7 T m/A)(120)2 (π(.01/2 m)2 )
0.09 m
1.58 × 10−5 H.
b)For the iron core µm = µ0 (1 + χ) = 800µ0 , so
B = µm H
where H = magnetic field strength = nI (see Serway pgs. 883-885)
B = µm nI = 800µ0 nI
thus the B field inside the solenoid is 800 times larger than for the wooden core.
Since L = (NΦB /I) ∝ B,
µm
Lµ0 = 800(1.58 × 10−5 H),
Lµm =
µ0
so,
Lµm = 1.26 × 10−2 H.
3. Serway 32.20
In an RL circuit (see Fig. 32.2 in your text), the maximum current is 0.5 A. After the switch is thrown,
a current of 0.25 A is reached in 0.15 s, and then the voltage across the resistor is 20 V. Calculate values
of R, L, and ε for this circuit.
SOLUTION: In an RL circuit, such as the one shown in Figure 32.2 of your textbook, Imax = E/R. We
are given that Imax = 0.5 A, and at t = 0.15 s, I = 0.25 A, and VR = 20 V. From this information, we find
R and E:
At t = 0.15 s
⇒
From Imax = E/R
⇒
(20 V)
= 80 Ω.
(0.25 A)
E = Imax R = (0.50 A)(80 Ω) = 40 V.
R = VR /I =
We can calculate L from τ = L/R. After the switch is closed
I(t) =
RI(t)
E
1 − e−t/τ ⇒ τ = −t ln 1 −
R
E
Plugging in values at t = 0.15 s gives
80 · (.25)
40
.
−1
τ
=
− (0.15 s) ln 1 −
L
=
τ R = (0.216 s)(80 Ω) = 17.3H.
1
−1
= 0.216 s
4. Serway 32.36
At t = 0, an emf of 500 V is applied to a coil that has an inductance of 0.8 H and a resistance of
30 Ω. (a) Find the energy stored in the magnetic field when the current reaches half its maximum value.
(b) After the emf is connected, how long does it take the current to reach this value?
SOLUTION: A soleniod with an inductance L and resistance R can be treated as a series RL combination,
so we can apply what we know about RL circuits.
a)Imax = E/R = (500 V)/(30 Ω) = 16.7 A, so the energy stored at I = Imax /2 = 8.3 A is
UB =
1 2
1
LI = (0.8 H)(8.3 A)2 = 27.8 J.
2
2
b)What is t when I(t) = Imax /2? Since I(t) = Imax (1 − e−t/τ ), we have
Imax /2
=
Imax (1 − e−tτ )
1/2
=
1 − e−t/τ
e−t/τ
=
1/2
t
=
t
=
− log(1/2)τ = − log(1/2)L/R
0.8 H
= 18.5 ms.
− log(1/2)
30 Ω
5. Serway 32.60A
The switch in figure P32.60 is connected to point a for a long time. After the switch is thrown to point b,
find (in terms of E, R,L, and C) (a) the frequency of oscillation in the LC ciruit, (b) the maximum charge
in the capacitor, (c) the maximum current in the inductor, and (d) the total energy stored in the circuit
at time t.
SOLUTION: For the circuit in Figure P32.60, with the switch in the position a for a long time, the capacitor
is fully charged and has a voltage equal to the emf E. The switch is then thrown to the position b. After
this time, the circuit is an LC oscillator, with R = 0. (See Section 32.5 of Serway).
a)Then the frequency is
1 1
ω
√
=
,
ν=
2π
2π LC
b)and
Qmax = EC.
c)Since there are no sources of energy loss in this circuit (R = 0), the maximum energy stored in C is equal
to the maxumum energy stored in L, so
Q2max
1 2
= LImax
.
2C
2
Solving for Imax and using part (b) gives
2
Imax
=
Q2max
E 2C 2
E 2C
=
=
,
LC
LC
L
Imax
=
E
r
C
.
L
d)The total energy in the circuit is constant (since R = 0), which we can take to be equal to the total energy
stored in C:
1
Q2
U = max = CE 2 .
2C
2
6. Practice Problems – Don’t Hand In
In Serway Chapt. 32: 11, 25, 35, 55/55A.
2