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What role does the induced emf in Faraday's law play in generating current in a ring in which the magnetic flux is changing ?

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Can someone help me ? –  Carpediem Mar 27 '13 at 15:30

2 Answers 2

The induced EMF is really just like a voltage provided by a battery. Consider, for example, a situation in which an EMF of magnitude $\mathcal E$ were induced in a loop. Then this would be equivalent to a putting a battery with voltage $V=\mathcal E$ in the loop, and this could cause current to flow. If the resistance of the loop were $R$ for example, then an amount of current $I = V/R$ would flow.

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To understand the subtle difference between a battery emf and that produced by a coil we need to write down the basic equations for two circuits:

(i) An ideal battery(no internal resistance) with emf = $e(t)$ in series with a resistor, R:

emf=$e(t) = i(t)R$ from which $i(t)=\frac{e(t)}{R}$

(ii) An ideal coil (with negligible resistance) with variable current flowing through it

emf=$e(t)=L\frac{di(t)}{dt}$ from which $i(t)=\frac{1}{L}\int_0^t e(\tau)d\tau$

These two equations show a subtle difference between the two emfs. Note that the current that develops due to the emf of the coil is bult up through all past moments of the emf, unlike the case of the battery where the current takes the value $e(t)/R$ regardless what the value of $e(t)$ has been a short while ago.

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