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?

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

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

The net emf induced due to both loops is proportional to cos ωt

The amplitude of the maximum net emf induced due to the both loops is equal to the amplitude of maximum emf induced in the smaller loop alone

answer is B.

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Detailed Solution

emf induced in the loop 1=ddtBAcosωtemf induced in the loop 2=ddtB2AcosωtBut these two are in opposite sense , as Area vectors are in opposite direction.Enet=E2−E1=−ddtB2Acosωt−−ddtBAcosωt⇒Enet=B2Asinωt−BAsinωt⇒Enet=Bsinωt 2A−A⇒ Net emf induced ∝ difference of area, so option A is wrongRate of change of flux ddtBAcosωt or ddtB2Acosωt will be max when ωt=90° so option B is correct.Enet=E1 so option D is correct

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