Q43 / 54 · JEE Advanced 2017 (Paper 1) · Physics

 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 in the magnetic field. Which of the following options is/are correct?
Figure

  • A
    The net emf induced due to both the loops is proportional to cos⁡ωt
  • Correct answer
    B
    The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of the maximum emf induced in the smaller loop alone
  • Correct answer
    C
    The rate of change of flux is maximum when plane of loop is perpendicular to the plane of paper
  • D
    The emf induced in the loop is proportional to sum of areas of two loops

Solution

Figure

When a conducting loop rotating in a magnetic field Bhas an angular velocity ω, the flux through the loop of area A at an angle θ = ωt, 

ϕ=BAcos⁡θ

=BA cos⁡ωt

The induced emf,

ε= -dϕdt=BAωsin⁡ωt

So,  ε anddϕdt ∝sin⁡ωt

So, the emf is maximum when, ωt=θ=π2.

Since the emf in the smaller loop will act opposite to that of the larger loop,

εNet= ε2A- εA=B2Aωsin⁡ωt-BAωsin⁡ωt

=B2A-Aωsin⁡ωt=BA ωsin⁡ωt

So, the amplitude of the maximum net emf is equal to BA which is same as that in the smaller loop.

Hence, the correct answer is options: 'B', 'C'.

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