Question

A circular insulated copper wire loop is twisted to form two loops of area $A$ and $2A$ as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field $B$ point into the plane of the paper. At $t=0$, the loop starts rotating about the common diameter as axis with a constant angular velocity $\omega$ in the magnetic field. Which of the following options is/are correct?

• A. The amplitude of the maximum net $emf$ induced due to both the loops is equal to the amplitude of maximum $emf$ induced in the smaller loop alone
• B. The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper
• C. The $emf$ induced in the loop is proportional to the sum of the areas of the two loops
• D. The net $emf$ induced due to both the loops is proportional to $\cos \omega t$

Answer 1

Answer: (A), (B)

The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper
The amplitude of the maximum net $emf$ induced due to both the loops is equal to the amplitude of maximum $emf$ induced in the smaller loop alone

Initial condition

${\varphi }_1=AB,{\varphi }_2=2AB$

General form

${\varphi }_1=ABcos\omega t$

${\varphi }_2=2ABcos\omega t$

$\therefore E_1=\frac{-d{\varphi }_1}{dt}=AB\omega sin\omega t$ ------ (1)

$E_2=2AB\omega sin\omega t$-------- (2)

Now, the emf induced in the two loops have opposite polarising because of the given orientation of the wires.

$\therefore |E_{net}|=|E_1-E_2|=AB\omega sin\omega t$

Option A is correct.

Rate of change of flux = emf induced is proportional to $sin\omega t$ from (1) and (2).

Therefore induced emf in the loop = $AB\omega sin\omega t$

$\Rightarrow$ proportional to sum of areas.

It is proportional to difference of the areas.

Rate of change of the flux is maximum which implies emf is maximum.

$AB\omega sin\omega t$ is maximum after one fourth and three-fourth of the revolution, when the loop is perpendicular.

Hence A and B are the correct answer.