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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 ω 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

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B

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

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C

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

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D

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

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Solution

The correct option is B

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


Initial condition

ϕ1=AB, ϕ2=2AB

General form

ϕ1=ABcosωt

ϕ2=2ABcosωt

E1=dϕ1dt=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|=|E1E2|=ABωsinωt
Enet will be maximum when sinωt=1
(max)Enet=ABω
Also maximum emf induced in loop 1, (max)E1=ABω for sinωt=1
Hence 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

Option A is correct.

Let area makes an angle θ with magnetic field at time t. Then, θ=ωt
Since emf induced is maximum when sinωt=1 i.e θ=ωt=90. Hence area vector is perpendicular to magnetic field. Since magnetic field points into the plane of paper, this means that area which is perpendicular to magnetic field must lie in the plane of paper.
Hence option B is correct.


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