Question

(a) Calculate the self-inductance of a 50.0 cm long, 10.0 cm diameter solenoid having 1000 loops. (b) How much energy is stored in this inductor when 20.0 A of current flows through it? (c) How fast can it be turned off if the induced emf cannot exceed 3.00 V?

Final Answer

- $19.7\textrm{ mH}$
- $3.95\textrm{ J}$
- $0.132\textrm{ s}$

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 23, Problem 62 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A solenoid is 50.0 centimeters long, which is 50.0 times 10 to the minus 2 meters and it has a diameter of 10.0 centimeters with 1000 loops of wire and a current going through it initially of 20.0 amps and the question in part (a) is what is the self-inductance of this solenoid? And the formula for that is permeability of free space times the number of turns squared times the cross-sectional area of the solenoid divided by its length. Its area is gonna be

*π*times the radius squared but we are given the diameter so we have to divide that by 2 in place of*r*so we substitute*d*over 2 and this works out to*π*times*d squared*over 4. So we substitute that in for area in our formula for self-inductance and so the self-inductance then is 4*π*times 10 to the minus 7 tesla meters per amp times 1000 turns squared times*π*times 10.0 times 10 to the minus 2 meters squared divided by 4 times 50.0 times 10 to the minus 2 meters, this works out to 19.7 millihenries. Part (b) asks for the amount of energy stored in the solenoid so that's one-half times its self-inductance times the current squared. So that's one-half times—the answer for part (a)—0.0197392 henries times 20.0 amps squared and that is 3.95 joules. Part (c) says suppose there's a limit on what the induced*emf*should be when the current is turned off and the limit is 3.00 volts... how quickly can the current be turned off in that case? So the induced*emf*will be the self-inductance times the change in current divided by this change in time and we want to solve for*Δt*so we multiply both sides by*Δt*over induced*emf*. So the time then is the self-inductance times the change in current divided by the*emf*that is intended, which is 3.00 volts. So that's the inductance there times 20.0 amp current because it goes from 20.0 to 0 so the change is 20.0 amps divided by 3.00 and this is 0.132 seconds. So if the current was turned off quicker than this then the*emf*would be higher than 3.00 volts; in other words, if the time was reduced the denominator of this fraction would be smaller and the quotient then would be bigger and the induced*emf*would be bigger if the time was smaller. So anyway... that's the shortest time possible.