### 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.