Inductance

Homework Solutions

Chapter 32

Questions: 1, 4, 5, 6, 7, 9, 14, 16, 17

Problems: 32.2, 12, 19,22,32,53,65

Q1,Why is the induced emf that appears in an inductor called a "counter" or "back" emf?By

Lenz's Law, we expect the induced emf toopposethe change -- to "counter" the change.

Q4,How can a long piece of wire be wound on a spool so that the wire has a negligible self-inductance?If the wire is doubled, so half of it spirals one way and the other half spirals the other way, then the magnetic field created by one half of the wire will just cancel the magnetic field created by the other half.

Q5,A long, fine wire is wound as a solenoid with a self-inductance L. If it is connected across the terminals of a battery, how does the maximum current depend upon L?The maximum current is the steady-state current and that is

notaffected by the self-inductance.

Q6,For the series RL circuit shown here, can the back emf ever be greater than the battery emf?No, the back emf can never be greater than the battery emf.

Q7,Suppose the switch in the diagram above has been closed for a long time and is suddenly opened. Does the current instantaneously drop to zero? Why does a spark appear at the switch contacts at the instant the switch is thrown open?The faster the switch is opened -- that is, the faster we try to stop the current -- the greater will be the

rateat which the magnetic field collapses. As thatrateincreases, the voltage increases (the "back emf") and that can be great enough to cause the arc.

Q9,Discuss the similarities between the enrgy store in the electric field of a charged capacitor and the energy stored in the magnetic field of a current-carrying coil.They are quite similar. Earlier we talked about the energy stored in a capacitor in great detail. We found the total energy -- the potential energy U -- to be

U = (1/2) C V

^{2}In this chapter we found the energy stored in a coil -- the potential energy U -- to be

U = (1/2) L I

^{2}

Q14,In the LC circuit shown in Figre 32.12, the charge on the capacitor is sometime zero, even though there is current in the circuit. How is this possible?When the charge on the capacitor is zero, there is no energy stored in the capacitor. At those times, there is maximum energy stored in the inductor which means maximum current.

Q16,How can you tell whether an RLC circuit is overdamped or underdamped?If it is

underdamped, it goes through several oscillations.If it is

overdamped,it doesnotoscillate -- it just makes one long, partial oscillation.

Q17,What is the significance of critical damping in an RLC circuit?

Critical dampingdescribes the smallest resistanceRthat the circuit can have andnot oscillate. The current will die out to zeromore quicklythan with a greater R. For a greater R -- for more damping -- the current dies out more slowly. For a smaller R -- for less damping -- the current oscillates.

32.2,A spring has a radius of 4.00 cm and an inductance of 125 microHenries when extended to a length of 2.00 m. Find an approximate value for the total number of turns in the spring?A "spring" is but a solenoid, so this is very much like Example 32.1 on page 941. There we found

L =

_{o}N^{2}A / lN

^{2}= L l/_{o}AA = r

^{2}= (0.04 m)^{2}= 0.005 026 m^{2}N

^{2}= L l/_{o}A = (125 x 10^{ - 6})(2.0)/[(4 x 10^{ - 7})(0.005)]N

^{2}= 39,583N = 199

32.12,A solenoid inductor is 30.0 cm long and has a cross-sectional area of 4.00 cm^{2}. When the current through the solenoid decreases at a rate of 0.625 A/s, the induced emf is 200 microVolts. Find the number of turns per unit length of the solenoid.On page 940, we have Equation 32.3,

L = [200 x 10

^{ - 6}V]/[0.625 A/s]L = 0.000 32 H = 0.32 mH

L = 3.2 x 10

^{ - 4}HOn the next page, p 941, we have Equation 32.5 which gives the inductance of a solenoid,

L =

_{o}n^{2}A ln

^{2}= L / (_{o}A l)A = 4.0 cm

^{2}[m/100 cm]^{2}= 4 x 10^{ - 4}m^{2}

Be carefulwith unit conversion. It's straightforward -- butbe carefulanyway! While it is "easier" or "more convenient" to talk about area in square-centimeters and length in centimeters, by the time we do a calculation, we need area is square-meters and length in meters. Don't eventhinkof feet, yards, furlongs, or miles.Thatis how NASA crashed one of the Mars probes!n

^{2}= (3.2 x 10^{ - 4})/[(4 x 10^{ - 7})(4 x 10^{ - 4})(0.30)]n

^{2}= 2.12 x 10^{6}n = 1,460

32.19,Calculate the resistance in an RL circuit in which L = 2.5 H and the current increases to 90% of its final value in 3.0 s.We know about the exponential behavior of the current, from Equation 32.7 on page 942,

where the time constant is

R = (2.5/1.3)

R = 1.92

32.22,An inductor that has an inductance of 15.0 H and a resistance of 30.0 ohms is connected across a 100-V battery. What is the rate of increase of the current(a) at t = 0 and

(b) at t = 1.50 s?

= (15/30) s = 0.5 s

We have already used Equation 32.7, from page 942,

For this circuit, we know the final current,

I

_{f}= V/R = 100 V/30 = 3.33AThis describes the

currentbut we want therateof change of the current,dI/dt]

_{(t = 0.0s)}= 16.7 A/sdI/dt]

_{(t = 3.0s)}= 3.73 A/s

32.32,Calculate the energy associated with the magnetic field of a 200-turn solenoid in which a current of 1.75 A produces a flux of 3.70 x 10^{ - 4}Wb in each turn.U = (1/2) L I

^{2}L = N

_{B}/IL = (200)(3.70 x 10

^{ - 4})/1.75L = 4.23 x 10

^{ - 2}HU = (1/2) L I

^{2}= (0.5)(4.23 x10^{ - 2})(1.75)^{2}U = 6.5 x 10

^{ - 2 }JU = 0.065 J

32.53,A 1.00 microFarad 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.Q

_{max}= C V = (1 x 10^{ - 6}F)(40 V)Q

_{max}= 40 x 10^{ - 6}CFrom Equation 32.23 or 32.25, we find that

I

_{max}= Q_{max}and, from Equation 32.22, we know the resonant frequency

= 1/SQRT[(0.010 H)(1 x 10

^{ - 6}F)]= 1 x 10

^{4}HzI

_{max}= Q_{max}= (1 x 10^{4})(40 x 10^{ - 6}) AI

_{max}= 0.4 A

32. 65Consider an LC circuit in which L = 500 mH and C = 0.100 microFarads.(a) What is the resonant frequency (

_{o})?

_{o}= 1/SQRT[(0.500 H)(0.1 x 10^{ - 6}F)]

_{o}= 4.47 x 10^{3}Hz

_{o}= 4,470 Hz = 4.47 kHz(b) If a resistance of 1.00 k is introduced into this circuit, what is the frequency of the (damped) oscillations?

From Eq 32.32, on page 953, we have the frequency of this new, damped oscillator,

_{d}= SQRT[ 1/LC - (R/2L)^{2}]

_{d}= SQRT[ 1/(0.500 H)(0.1 x 10^{ - 6}F) - (1000 /2(0.500 H))^{2}]

_{d}= SQRT[ 1/(0.500)(0.1 x 10^{ - 6}) - (1000/2(0.500))^{2}]

_{d}= SQRT[ 2.0 x 10^{7}- 1.0 x 10^{6}]

_{d}= SQRT[ 1.9 x 10^{7}]

_{d}= 4.36 x 10^{3}Hz

_{d}= 4,360 Hz = 4.36 kHz(c) What is the percent difference between the two frequencies?

=

_{o}-_{d}= 4.47 kHz - 4.36 kHz = 0.11 kHz

%diff = /

_{o}= 0.11/4.47 = 2.5%

Ch33 ToC

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