### Video Transcript

At the instant a current of 0.50 amps is flowing through a coil of wire, the energy stored in its magnetic field is 8.0 times 10 to the negative third joules. What is the self-inductance of the coil?

So we’ve been told here we have a coil of wire and that a current flows through it. And that current value is 0.50 amps at the instant we’re interested in. Along with that, at that particular moment in time, the energy stored in the magnetic field of the coil is given as 8.0 times 10 to the negative third joules.

Knowing all this, we want to solve for the self-inductance of the coil. We can call that self-inductance capital 𝐿. To start solving for it, let’s go back to our energy in our magnetic field 𝐸 sub 𝐵. Where, we might wonder, does this magnetic field come from?

Looking at our sketch of our coil, we know that the current flowing through the loops of this coil will help to create a magnetic field through those windings. It’s that magnetic field that has the energy we’ve written as 𝐸 sub 𝐵.

But notice how this field is really created by the inductor itself. And therefore, we could say that this energy is not just of the magnetic field. But it’s really energy of the inductor.

That fact can remind us of an equation that would be helpful when moving toward our solution for the self-inductance of the coil. The inductor energy, in this case equivalent to the magnetic field energy since the field is created by the inductor, is equal to one-half the self-inductance of the coil multiplied by the current running through it squared.

This equivalence between magnetic field and inductor energy means we can write 𝐸 sub 𝐵 is equal to one-half 𝐿𝐼 squared. And since it’s the self-inductance 𝐿 that we want to solve for, we’ll rearrange this equation to isolate that term.

The self-inductance of the coil is equal to two times the energy in the magnetic field, all divided by the current in the coil squared. The current in the coil is given as 0.50 amps. And the magnetic field energy is also a given quantity.

We calculate the self-inductance to be 0.064 henrys. That’s the self-inductance of this particular coil at this particular instant.