# Mutual Inductance of a conducting loop and a solenoid

## Homework Statement

A 0.100m long solenoid has a radius of 0.05m and 15000 turns. The current in the solenoid changes at a rate of 6.0 A/s. A conducting loop of radius 0.0200m is placed at the center of the solenoid with its axis the same as that of the solenoid. Determine the mutual inductance of this combination.

## Homework Equations

Es=-M($$\Delta$$Ip/$$\Delta$$t)

## The Attempt at a Solution

The answer has to be the induced emf divided by the rate of current change(6 A/s), but I just don't understand where and how we can glean the emf from the information given. I'm having a hard time on this emf stuff. I thought I understood very well how and when it is induced, but this is the second instance that I've had to solve for two variables and I just don't know how my teacher expects us to do it.

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gneill
Mentor
The emf induced in a loop due to changing flux inside the loop is

$$\left| E \right| = \left| \frac{d \Phi}{dt} \right|$$

The flux in side the loop depends upon the magnetic field strength inside the loop and the area of the loop.

So, I suggest you start with the equation for the field strength inside the solenoid. Differentiate to find rate of change.

Ok, I see that I need the change in magnetic flux to calculate the Emf of the inner conducting loop, but I still can't calculate that without a single hint about magnetic field strength. I thought I could figure out the change in field strength by doing two calculations of
B=uo(N/L)I.
I made I=6 in one equation and I=12 in another, to assume a one second time variable in every other equation to come. So I got Bo=1.131T and B=2.262T. I multiplies each with the cross sectional area of the primary solenoid ($$\pi$$.052) and I took the difference to get .00888 for the change in magnetic flux over 1 second inside the solenoid. From there, I thought I could set that number equal to the negative mutual inductance times the rate of current change. But that would mean that .00888/-6 would equal my answer, and I know the answer isnt -0.00148 H. I'm expecting a positive number to the 10-4 power

gneill
Mentor
You have

B=uo(N/L)I

and the flux in an area A is just

Φ = B*A

So what's dΦ/dt for the area of the loop? (Hint: the only variable that changes in B is I).

ok, ok. so I didn't need to do two magnetic field strength equations? If I let I=6, I get B=1.131. The flux inside the conducting loop then is 0.00142, and Es would be -.00142 for one second. Then I can finally solve for M to get 2.369*10-4. I hope I've done it right this time. It makes a little more sense now. I don't know why I thought the magnetic field strength would change throughout the system

gneill
Mentor
ok, ok. so I didn't need to do two magnetic field strength equations? If I let I=6, I get B=1.131. The flux inside the conducting loop then is 0.00142, and Es would be -.00142 for one second. Then I can finally solve for M to get 2.369*10-4. I hope I've done it right this time. It makes a little more sense now. I don't know why I thought the magnetic field strength would change throughout the system
You got to a correct answer, but you should be able to get there without plugging in arbitrary values for the current and assuming a particular time interval (sort of a "poor man's derivative").

I think that the intention of the problem writer was to get you to recognize that the given rate of change of current is in fact dI/dt, which you could use directly to find dΦ/dt for the loop, hence the induced emf, E. Then E/(dI/dt) yields the mutual inductance.