A device is turned on and 3.00 A flows through it 0.100 ms later. What is the self-inductance of the device if an induced 150 V emf opposes this?
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Final Answer

5.00 mH5.00\textrm{ mH}

Solution video

OpenStax College Physics for AP® Courses, Chapter 23, Problem 58 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A device starts off with no current flowing through it and then 0.100 milliseconds later, 3.00 amps are flowing through it so this change in current occurs over a time of 0.100 times 10 to the minus 3 seconds and the change in current is the final current because it starts off at zero so the change is 3.00 amps and the induced emf in the device is 150 volts and we are asked to figure out what is its self-inductance? So I have an absolute value symbol around here just because the formula is emf induced is negative times self-inductance times the rate of change of current but the negative sign is from Lenz's law to indicate the direction of the emf and that direction is not important for our purposes and L is just always a positive number and so we want L to work out to be positive so we put absolute value around this emf. So the magnitude of the emf then is the self-inductance times the change in current divided by change in time and we can solve this for L by multiplying both sides by Δt over ΔI and then we get that the self-inductance is 150 volts times 0.100 times 10 to the minus 3 seconds divided by 3.00 amps, which is 5.00 millihenries.