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Page 1: Physics 42 Chapter 29 Homework - Santa Rosa …srjcstaff.santarosa.edu/~lwillia2/42/42ch32hw.pdf · Physics 42 Chapter 32 Homework ... An emf of 24.0 mV is induced in a 500-turn coil

Physics 42 Chapter 32 Homework

Problems: 6, 11, 13, 16, 17, 32, 46, 51, 55

6. An emf of 24.0 mV is induced in a 500-turn coil at an instant when the current is 4.00 A

and is changing at the rate of 10.0 A/s. What is the magnetic flux through each turn of the coil?

P32.6 From I

Lt

, we have

3324.0 10 V

2.40 10 H10.0 A s

LI t

.

From BNL

I

, we have

3

22.40 10 H 4.00 A

19.2 T m500

B

LI

N

.

11. A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is

wound onto a plastic tube to form a long solenoid. This coil has a circular cross section

and consists of tightly wound turns in one layer. If the current in the solenoid drops

linearly from 1.80 A to zero in 0.120 seconds, an emf of 80.0 mV is induced in the coil.

What is the length of the solenoid, measured along its axis?

*P32.11 We can directly find the self inductance of the solenoid:

dI

Ldt

0 1.8 A

0.08 V0.12 s

L

2

3 05.33 10 Vs A

N AL

.

Here 2A r , 200 m 2N r , and 3

10 mN . Eliminating extra unknowns step by step,

we have

2 7 22 2 223 0 00

3

3

10 40000 m Tm40000 m200 m5.33 10 Vs A

2 4 A

4 10 W bm A0.750 m

5.33 10 AVs

N r N

N

13. A self-induced emf in a solenoid of inductance L changes in time as ε = ε0e – kt. Find the

total charge that passes through the solenoid, assuming the charge is finite.

P32.13 0kt dI

e Ldt

0 ktdI e dt

L

If we require 0I as t , the solution is 0 kt dqI e

kL dt

0 0

2

0

ktQ Idt e dt

kL k L

0

2Q

k L

.

Page 2: Physics 42 Chapter 29 Homework - Santa Rosa …srjcstaff.santarosa.edu/~lwillia2/42/42ch32hw.pdf · Physics 42 Chapter 32 Homework ... An emf of 24.0 mV is induced in a 500-turn coil

16. Show that I = I0 e – t/τ is a solution of the differential equation

0dt

dILIR

where τ = L/R and I0 is the current at t = 0.

P32.16 Taking L

R , 0

tI Ie

: 0

1tdIIe

dt

0dI

IR Ldt

will be true if 0 0

10

t tI L Ie

Re .

Because L

R , we have agreement with 0 0 .

17. Consider the circuit in Figure P32.17, taking ε =

6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the

inductive time constant of the circuit? (b) Calculate the

current in the circuit 250 μs after the switch is closed. (c)

What is the value of the final steady-state current? (d)

How long does it take the current to reach 80.0% of its

maximum value?

P32.17 (a) 32.00 10 s 2.00 m s

L

R

(b) 0.250 2.00

m ax

6.00 V1 1 0.176 A

4.00

tI I e e

(c) m ax

6.00 V1.50 A

4.00 I

R

(d) 2.00 m s0.800 1 2.00 m s ln 0.200 3.22 m s

te t

FIG. P32.17

32. At t = 0, an emf of 500 V is applied to a coil that has an inductance of 0.800 H and a

resistance of 30.0 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?

P32.32 (a)

22 22

2 2

0.800 5001 127.8 J

2 2 2 8 8 30.0

LU LI L

R R

(b) 1

R L tI e

R

so

11

2 2

R L t R L te e

R R

ln2Rt

L so

0.800

ln2 ln2 18.5 m s30.0

Lt

R

Page 3: Physics 42 Chapter 29 Homework - Santa Rosa …srjcstaff.santarosa.edu/~lwillia2/42/42ch32hw.pdf · Physics 42 Chapter 32 Homework ... An emf of 24.0 mV is induced in a 500-turn coil

46. A 1.00-μF capacitor is charged by a 40.0-V power supply. The fully charged capacitor is

then discharged through a 10.0-mH inductor. Find the maximum current in the resulting

oscillations.

P32.46 At different times, m axm axC LU U so 2 2

m ax m ax

1 1

2 2C V LI

6

m ax m ax 3

1.00 10 F40.0 V 0.400 A

10.0 10 H

CI V

L

.

51. An LC circuit like the one in Figure 32.16 contains an 82.0-

mH inductor and a 17.0-μF capacitor that initially carries a

180-μC charge. The switch is open for t < 0 and then closed

at t = 0. (a) Find the frequency (in hertz) of the resulting

oscillations. At t = 1.00 ms, find (b) the charge on the

capacitor and (c) the current in the circuit.

P32.51 (a) 6

1 1135 H z

2 2 0.0820 H 17.0 10 F

fLC

(b) m axcos 180 C cos 847 0.00100 119 CQ Q t

(c) m ax sin 847 180 sin 0.847 114 m AdQ

I Q tdt

55. Consider an LC circuit in which L = 500 mH and C = 0.100 μF. (a) What is the resonance

frequency ω0? (b) If a resistance of 1.00 kΩ is introduced into this circuit, what is the frequency of

the (damped) oscillations? (c) What is the percent difference between the two frequencies?

P32.55 (a)

06

1 14.47 krad s

0.500 0.100 10LC

(b) 2

14.36 krad s

2d

R

LC L

(c) 0

2.53% low er