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). 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? 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.

• 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. – Aditya Roychowdhury Aug 31 '20 at 4:06

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.

$$\phi=B. da$$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

• So what will be the electric field? – Aditya Roychowdhury Aug 31 '20 at 3:58