# Video: Mutual Induction

A transformer with an iron core has a primary coil with 25 turns and a secondary coil that also has 25 turns. The current in the primary coil increases the magnetic flux through the core by 0.15 Wb/s. The current in the secondary coil increases at 0.075 A/s. What is the mutual inductance of the coils?

04:24

### Video Transcript

A transformer with an iron core has a primary coil with 25 turns and a secondary coil that also has 25 turns. The current in the primary coil increases the magnetic flux through the core by 0.15 webers per second. The current in the secondary coil increases at 0.075 amperes per second. What is the mutual inductance of the coils?

To get started here, let’s draw a sketch of this transformer with the primary and secondary coils. Okay, so here’s our transformer core. And this is the primary coil; we’ll call this the secondary. And even though we can see that neither of these has 25 turns to it, like our problem statement tells us, we can pretend they do. The idea with the transformer is that as current passes through the primary coil and specifically as it changes and passes through this coil, then this creates a change in magnetic field that the core directs through the loops of the secondary coil. When the secondary coil experiences this change in magnetic flux, current is induced in it.

In our scenario, the current in the primary coil is increasing. We don’t know the rate it’s increasing by, but we do know that it’s affecting the magnetic flux in the core and that that flux is increasing at a steady rate, given as 0.15 webers per second, where a weber is the unit of magnetic flux. Now, just as a reminder, if we have some area 𝐴 and there’s a magnetic field 𝐵 passing through that area, then we can say there’s a magnetic flux Φ sub 𝑚 through that loop. And the units of magnetic flux, as we’ve seen, are webers.

So going back to our transformer, in the core, this iron material that connects our primary and secondary coils, the magnetic flux is increasing at this given rate. And that, our problems statement tells us, drives an increasing current in the secondary coil. We have then a change in current in one coil, creating a change in magnetic flux, which induces a change in current in another coil. This means we have mutual inductance going on. And we want to calculate the value of that inductance.

To begin to do this, we can recall Faraday’s law. The reason we’re thinking of this law is because we have a magnetic flux that changes in time. Faraday’s law tells us that this change in flux, ΔΦ sub 𝑚, over some change in time, Δ𝑡, multiplied by the number of turns in whatever coil we’re considering is equal to the magnitude of the emf induced in that coil. Now, in terms of mutual inductance, induced emf is equal to something else as well. We can recall that induced emf is equal to the mutual inductance between two conducting loops multiplied by the time rate of change of current and what we can call the secondary loop or, in the example of our transformer, the secondary coil.

So it’s 𝑀, the mutual inductance, that we want to solve for. And we see that that value multiplied by Δ𝐼 divided by Δ𝑡 is equal to the emf induced in a secondary coil, which by Faraday’s law is also equal to negative the number of turns or loops in the coil multiplied by the change in magnetic flux divided by the change in time. Since we’re given this time rate of change of magnetic flux, let’s combine these two equations for induced emf by equating the right sides with one another.

When we do this though, we’ll leave off this minus sign here because in our case we’re only concerned with the magnitude of the emf induced. So then we get this. 𝑀, the mutual inductance of the coils, times Δ𝐼 divided by Δ𝑡, the time rate of change of current in the secondary coil, is equal to the number of loops in that coil multiplied by the time rate of change of magnetic flux that each loop in the coil experiences.

What we want to do is isolate 𝑀 since we’re solving for the mutual inductance. And to do that, we’ll divide both sides of the equation by Δ𝐼 divided by Δ𝑡. When we do, on the right-hand side of the resulting equation, we have 𝑁 — the number of turns in the secondary coil, that’s 25 — multiplied by the time rate of change of magnetic flux in the core, that’s given as 0.15 webers per second, divided by the time rate of change of current in the secondary coil. This is the induced current, and its rate of change is 0.075 amperes per second. When we calculate this fraction and keep two significant figures, we find a result of 50 henrys. That’s the mutual inductance of the two coils.