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 points into the plane of the paper. At t=0, the loop starts rotating about the common diameter as axis with a constant angular velocity ω in the magnetic field. Which of the following options is/are correct?

The rate of change of the flux is maximum when the plane of the loops is perpendicular to the plane of the paper

The net emf induced due to both the loops is proportional to cosωt

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

The emf induced in the loop is proportional to the sum of the areas of the two loops

Solution:

Initial condition φ₁=AB, φ₂=2AB
General form φ₁=ABcosωt
φ₂=2ABcosωt
∴E₁=-dφ₁/dt=ABωsinωt ——— (1)
E₂=2ABωsinωt ——— (2)
Now, the emf induced in the two loops have opposite polarising because of the given orientation of the wires.
∴|Enet|=|E₁-E₂|=ABωsinωt
Option A is correct.
Rate of change of flux = emf induced is proportional to sinωt from (1) and (2).
Therefore induced emf in the loop =ABωsinωt ⇒ 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.
ABωsinωt is maximum after one fourth and three-fourth of the revolution, when the loop is perpendicular.
Hence A and B are the correct answer.