# Video: Finding the Mutual Inductance of Two Solenoids

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5 × 10⁻³ m². The solenoid is 0.50 m long and has 500 turns. What is the mutual inductance of this system? The outer coil is replaced by a coil of 40 turns whose radius is three times that of the solenoid. What is the mutual inductance of this configuration?

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### Video Transcript

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5 times 10 to the negative third meters squared. The solenoid is 0.50 meters long and has 500 turns. What is the mutual inductance of this system? The outer coil is replaced by a coil of 40 turns whose radius is three times that of the solenoid. What is the mutual inductance of this configuration?

We can call the number of turns in the coil, 40, 𝑁 sub one and the cross-sectional area of that coil, 7.5 times 10 to the negative third meters squared, capital 𝐴. The coil is wrapped around a solenoid, which itself is 0.50 meters long. We’ll call that length 𝐿. And the solenoid has 500 turns, a number we’ll call 𝑁 sub two.

In part one, we want to solve for the mutual inductance of the system. We’ll call this capital 𝑀. In part two, we want to solve for the same quantity, but this time with a modified radius of the outer coil. We’ll call this value 𝑀 sub 𝑟.

We can start off by drawing a sketch of the coil and the solenoid. In this exercise, we have a coil wrapped around a solenoid, where the solenoid per unit length has many more turns. If we were to look at the coil end on, we would see it has a cross-sectional area 𝐴. And based on the coil and the solenoid being wrapped together, we want to solve for the mutual inductance between them, capital 𝑀.

We can recall that mutual inductance is the capacity of one circuit to induce a new current and electromotive force in a second circuit. Mathematically, it equals the permeability of free space 𝜇 naught times the number of turns in the first coil times the number of turns in the second coil times the cross-sectional area that both coils overlap, all divided by the length 𝑙 over which the two coils overlap.

We’ll treat the constant 𝜇 naught as having an exact value of 1.26 times 10 to the negative sixth tesla meters per amp. Knowing 𝜇 naught, we also know the rest of the parameters in this equation because they’re given in the problem statement. So we’re ready to plug in and solve for mutual inductance.

When we do and enter these values on our calculator, we find that 𝑀, to two significant figures, is 3.8 times 10 to the negative fourth henries. That’s the mutual inductance of the coil and the solenoid.

Now we’re ready to move on to part B where we solve for 𝑀 sub 𝑟, which is the new mutual inductance between the coil and the solenoid when the coil’s radius has been expanded to three times its original value. Now as we consider this setup, we see that indeed the area of the coil has changed. But the mutual overlapping area of the coil and the solenoid both remains constant because that’s limited by the constant radius of the solenoid.

So our value for 𝐴, which is the mutual overlapping of the cross-sectional areas of these two objects, remains the same. So when we solve for 𝑀 sub 𝑟, we use the same equation as we did for 𝑀, plug in the same values and find the same result, 3.8 times 10 to the negative fourth henries. Even with an enlarged coil, this continues to be the mutual inductance of these two objects.