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I'm poking around Faraday's law regarding induction and I'm trying to solidify my understanding. In my figure below the light blue shaded area is a region of uniform magnetic field directed into the screen (signified by the one red X). If this magnetic field is increasing in magnitude at a constant rate, $dB/dt$, then it will induce an electric field that drives a current flowing counter clockwise around the purple conductor loop (nature reacts to change).

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This results in an emf being produced around the loop as per the equation, $$\oint \vec E \cdot \vec ds = - \frac{d\phi_B }{dt}$$ And since, $$\phi_B = BA$$ the flux linking the purple ring depends on the area inside the ring. So, here is my question, what if I have a "hole" in the middle of the ring within which is no $B$ field (white area in figure below)? All other things equal, will this configuration induce the same $E$ field and resulting current $i$ as the above case?

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I know that the total magnetic flux linking the coil is smaller now ($BA$), but i think the rate-of-change of that flux linkage is the same as the first scenario...making me think the coil will not know the difference and the emf and induced current $i$, will be same as first case.

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    $\begingroup$ The flux is B.A . So, the change in flux, say in one first case will be larger as the net the flux is larger. The d phi / dt is not same in both cases. $\endgroup$
    – Solid - NMR
    Commented Aug 31, 2020 at 4:06

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As you say, Maxwell's equations states that the emf is:

$$\begin{align} Emf &= - \frac{d\phi}{dt} \\ &= - \frac{dB.A}{dt}\\ &= - A\frac{dB}{dt} \end{align}$$

Where we subbed in the formula for flux, and since A is a constant (doesn't change with time), we can bring A out of the derivative.

We can now compare your two cases.
Assuming that $\frac{dB}{dt}$ is the same between the two cases and that A is smaller in Case 2 vs Case 1 ($A_2 < A_1)$.
Then by the above formula, we see that in Case 2, the Emf induced (and hence current induced) is smaller in magnitude than in Case 1 where we had a larger area.

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$$\phi=B. da$$Thus where ever magnetic field strength is zero flux becomes zero.

The flux linked with ring is given by $\int B. da$ i.e it is the sum of $ B. da$ all over the area of the ring and where $B=0$ contribution to flux is zero

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  • $\begingroup$ So what will be the electric field? $\endgroup$
    – Solid - NMR
    Commented Aug 31, 2020 at 3:58

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