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What self-induced emf appears in that coil?

  1. Apr 6, 2012 #1
    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. jcsd
  3. Apr 6, 2012 #2


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    Staff: Mentor

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