Sol. Consider a point on the circumference of a circle of radius r(r < R). Let E be the electric field along the tangents to the circle. Then E (electromotive force) = = E. 2 π r.

Also E = – = – ( π r 2 B) = – π r 2 .

∴ E × 2 π r = – π r 2 ⇒ E = – r .

The minus sign suggests that the induced electric field acts to oppose the change of the magnetic field.

Now consider a point P in the rod and a small distance dl = PQ along AB. The electric field at P is E = r along tangent to the circle as shown in the figure.

Elementary work done

= (E cos θ ) dl = E dl cos θ

Draw perpendicular PN from P on OQ.

Then PN = r d θ = dl cos θ

∴ Elementary work done = Erd θ = rd θ

= .

But r 2 d θ = area of the triangle OPQ

∴ Total work done in taking unit charge from A to B

= × summation of the areas of elementary triangle

= × area OAB.

= P.D. between the ends of AB.

∴ V = . 2. l = l .