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As an alternator’s coil moves, one side of the coil induces a voltage and thus a charge flow in one direction However, simultaneously the other side of the coil induces a voltage and charge flow in the opposite direction.

This intuitively to me means that the equal and opposite induced electromotive forces should cancel out, but instead an alternating current is generated. Why is this?

For example if a series circuit had two 3V cells applying voltage in opposite directions, then the net voltage in the circuit would be 0. How is this any different to the opposite voltages induced in the alternator?


There are no two voltages existing in the same loop! It is not like having $3V$ batteries making the net voltage zero either.

Considering the setup that you have drawn:

The changing magnetic flux through the loop due to the bar magnet forces the generation of an EMF to counter that change in magnetic flux (this is the Lenz' law).

The important thing here to realise is that this sort of an EMF is not the conservative type.

This is a special case of those non-conservative electric fields: $$\int E dl \not= 0$$

where the integral covers the whole loop. In fact this integral equals the EMF $\epsilon$, which is written as: $\epsilon = -\frac{d\phi}{dt}$, with $\phi$ being the magnetic flux.

In usual cases, for example, Kirchoff's rule, the above integral always used to be zero (over the whole loop).

Although electric fields are in general conservative, induced electric fields are non-conservative.

  • $\begingroup$ +1 for stating how alternators differ from two opposing cells in a series circuit. But drum roll how does an alternator produce AC? $\endgroup$
    – Hisham
    Apr 13 '19 at 15:04
  • 1
    $\begingroup$ AC means a sinusoidal wave. So, for an AC current to flow, the EMF should be sinusoidal in nature. From the definition of $\phi=BAcos(\theta)$, if you use the formula for $\epsilon$ that I have written above, and differentiate $\phi$, you would get a sinusoidal form, thus confirming AC currents! $\endgroup$
    – Karthik
    Apr 13 '19 at 15:10
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    $\begingroup$ I don’t care what the website says. +1 and thanks ALOT $\endgroup$
    – Hisham
    Apr 13 '19 at 15:12
  • $\begingroup$ No problems! :) $\endgroup$
    – Karthik
    Apr 13 '19 at 15:13

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