What self-induced emf appears in that coil?

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SUMMARY

The discussion centers on calculating self-induced electromotive force (emf) in a coil using the formula for mutual inductance. The user calculates the induced emf as 198.795 nWb by applying the formula \((6.1 \, \text{mH} \times 7.3 \, \text{mA}) / 224 \, \text{turns}\). The confusion arises from the division by the number of turns, which is clarified through the definition of mutual inductance, \(M = \frac{\Phi_2}{I_1}\). This highlights the importance of understanding the relationship between inductance, current, and turns in coil calculations.

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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?
 
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McAfee said:
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?

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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