# What self-induced emf appears in that coil?

1. Apr 6, 2012

### McAfee

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I now know the units are nWb. I'm just getting a little confused by answer.

((6.1mH)(7.3mA))/224turnes=.198795) (mH*mA)/turnes

So would that equal= 198.795 nWb?

2. Apr 6, 2012

### Staff: Mentor

Why are you dividing by the number of turns? From the definition of mutual inductance,
$$M = \frac{\Phi_2}{I_1}$$