# Simple question about definition of Mutual Inductance

1. Apr 9, 2005

### Lomion

I'm getting confused over the formal definition of mutual inductance. I've googled and searched a lot of sites, but I still don't understand, so I'm hoping someone here can explain it to me in plain English!

In our lecture, this was the formula given for mutual inductance:

$$L_{12} = \frac{N_2}{I_1} \oint_{s_2} B_1 dS_2$$

According to my own interpretation, this means that the inductance at the second coil due to the first coil (L 1 on 2) is essentially:

The flux caused by $$B_1$$ from the first coil passing through the area of the second coil. $$\Phi_{12} = B_1 A_2$$ (assuming B constant to make life simpler). And this flux is linked to the N2 turns of the second coil, and divided by the current from the first coil.

Does this make any sense? My textbook doesn't explain this at all (and in fact, doesn't even have this formula! Argh!) so I used Schaum's outline, but it seems to give me something different.

In example 4 from Schaum's Outline (Ch. 11) in case anyone here has it, it has two concentric coils. First, it calculated the B in the first coil. Then it calculated $$\Phi = B_1 A_1$$. (See how it differs from the definition above?)

And then it said $$M_{12} = N_2 \frac{\Phi}{I_1}$$.

Can someone please clarify why this is? And if I'm using the correct formulas at all?

A better explanation of what exactly is going on with mutual inductance would be appreciated as well!

2. Apr 10, 2005

### maverick280857

You are right when you say that mutual inductance is due to change in flux linking circuit A when the flux lines are from circuit B (and vice versa).

If I read correctly you have written a relationship involving the net flux linking a circuit due to the emf induced by the change in current in the circuit itself. This is not mutual inductance. It is self inductance. The self inductance coefficient is normally denoted by L. For mutual inductance we use M. Of course you mean L12 so I guess you mean the same thing.

The Flux linking circuit 1 is due to the B field set up by circuit 2. So the flux linking ckt 1 is

$$\oint_{ckt 1} \vec{B_{2}}dS$$

This must equal the net flux linkage of circuit 1, i.e. $N_{1}\phi_{B,1} = M_{12}I_{1}$.

There is an important theorem governing Mutual Inductance: it is called the Reciprocity Theorem. It simply states that $M_{12} = M_{21}$. The implications of this thorem are quite interesting. As an example if you are given a long solenoid and a finite length solenoid such that the short solenoid is inside the long solenoid and there is a current I running through it: you have to find M for the system. Can you see how the reciprocity theorem helps? The above relationships completely answer this question.

If you want to read more (I am sorry are you referring to Edminster Schaum?) you might want to get hold of Classical Electrodynamics by Griffiths.

Cheers
vivek

Last edited: Apr 10, 2005