Question Video: Self-Induction | Nagwa Question Video: Self-Induction | Nagwa

# Question Video: Self-Induction Physics • Third Year of Secondary School

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A loop of wire carries a current of 180 mA. The magnetic flux produced by the current is 0.77 Wb. What is the self-inductance of the loop?

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### Video Transcript

A loop of wire carries a current of 180 milliamperes. The magnetic flux produced by the current is 0.77 webers. What is the self-inductance of the loop?

So then, in this example, we have a loop of wire. Let’s say this is it. And it’s carrying a current of 180 milliamperes. Because of this, a magnetic field is produced that passes through the loop. Let’s just say the magnetic field points like this. And because we have this field passing through the interior area of the loop, that means we have a magnetic flux through the loop. And that’s the value we have given to us here in our problem statement, 0.77 webers. We can refer to that symbolically as Φ sub 𝑚, magnetic flux.

Knowing all this, we want to solve for the self-inductance of the loop. And we can recall that the self-inductance of a loop, capital 𝐿, is equal to the ratio of the magnetic flux Φ sub 𝑚 produced by a current 𝐼 in the loop to that current 𝐼. For our particular loop then, we have exactly that information about it. Our current 𝐼 is a current in the loop. And the magnetic flux is produced thanks to a magnetic field generated by that current. So to solve for the self-inductance of our loop, we only need to divide Φ sub 𝑚 by 𝐼.

With these values plugged in, the one change we’ll want to make before dividing is to convert the units of our current from milliamperes to amperes. Now, since 1000 milliamperes is equal to one amp, we can say that 180 milliamperes is equal to 0.180 amps. And now we’re ready to divide. And when we do, to two significant figures, we find a result of 4.3 henrys. That’s the self-inductance of this loop.

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