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← JEE Advanced 2017 · Paper 1
PhysicsMultiple correct+4 / -2Question 5 of 54

JEE Advanced 2017 · Paper 1, Physics question 5

A circular insulated copper wire loop is twisted to form two loops of area AA and 2A2A 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 B⃗\vec{B} points into the plane of the paper. At t=0t = 0, the loop starts rotating about the common diameter as axis with a constant angular velocity ω\omega in the magnetic field. Which of the following options is/are correct?

A figure-of-eight wire loop in the plane of the paper: an upper circle of area A and a lower circle of area 2A, crossing (insulated) at the joint. A uniform field B points into the paper (crosses). The loop rotates with angular velocity ω about the vertical dashed line through both centres (the common diameter).
Choose one or more options
Answer and worked solution

Official answer: B, D

Twisted loop. Twisting a loop into a figure of eight makes the current circulate in opposite senses in the two lobes. Their emfs therefore subtract. The effective area is
2A−A=A.2A-A=A.

  • (A) The emf is proportional to the difference of the areas, not the sum. False.
  • (B) The net amplitude equals that of the smaller loop alone. True.
  • (C) Φ=BAcos⁡ωt\Phi=BA\cos\omega t gives ε=BAωsin⁡ωt\varepsilon=BA\omega\sin\omega t, not cos⁡ωt\cos\omega t. False.
  • (D) ∣dΦ/dt∣|d\Phi/dt| is largest when Φ=0\Phi=0, i.e. when the loops' plane is perpendicular to the paper (parallel to B⃗\vec B). True.

(B), (D)