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The question is simple. We know induced emf in a conducting loop due to a changing flux is given by

$$ E = -\frac{d\Phi}{dt} $$

My question is if the flux is changing only in a small part of the loop , will the emf still be induced in the loop?

Example:

pic

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    $\begingroup$ w.r.t means with respect to $\endgroup$
    – Jdeep
    Sep 16, 2020 at 12:50

2 Answers 2

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In Faraday's equation:

$$ E = -\frac{d\Phi}{dt} \tag{1} $$

the flux $\Phi$ is the total magnetic flux passing through the loop. That is, there is some magnetic field $B$ passing through the loop and we integrate this field across the area of the loop to get the flux $\Phi$.

The magnetic field through the loop $B$ does not have to be constant across the loop. Indeed it can vary across the loop in any fashion you want and that makes no difference. All that matters in equation (1) is the total flux $\Phi$ through the loop and not how that flux is distributed across the loop.

So the answer is that yes in the diagram you have drawn an EMF will be induced even when the field $B$ is non-zero only in part of the loop. As long as the total flux $\Phi$ is changing with time an EMF will be induced.

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  • $\begingroup$ When we calculate the flux at any instant , we have to consider only that Area (A) where B is present ,right? $\endgroup$
    – Jdeep
    Sep 16, 2020 at 16:11
  • $\begingroup$ @NoahJ.Standerson yes. You are integrating the field over the whole area, but when $B=0$ that means $\int B dA$ is zero in that region so you can just ignore it. You need only integrate over the regions where $B$ is non-zero. $\endgroup$ Sep 16, 2020 at 16:17
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The answer to your question is yes. The total emf will be determined by the rate of change of total flux, no matter where it is in the loop. If the flux is not symmetrical within the loop, then the induced E field will change from point to point around the loop. Since the current must be the same all round the loop, the charge density must also change from point to point.

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