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An insulated circular copper wire loop is twisted (figure-eight style) into two coplanar loops of areas A and 2A. At the crossing point the wires stay electrically insulated from one another. The whole arrangement lies in the plane of the paper and a uniform magnetic field B is directed into the page. At t = 0 the system begins rotating about the common diameter with constant angular velocity omega. Which statement(s) is/are correct?

1. (A) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper
2. (B) The net emf induced due to both the loops is proportional to cos omega t
3. (C) The emf induced in the loop is proportional to the sum of the areas of the two loops
4. (D) The amplitude of the maximum emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone

Correct answer: (A) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper

Solution

Because the loop is twisted into a figure-eight, the two loops carry the induced emf in opposite directions, so the net emf is the difference, proportional to (2A - A) = A, not the sum. Net flux phi = B(2A - A)cos(omega t) = BA cos(omega t). The induced emf = -dphi/dt = BA omega sin(omega t), which is maximum (dphi/dt max) when sin(omega t)=1, i.e. when the plane of the loops is perpendicular to the plane of the paper. So (A) is correct. (B) is wrong: emf is proportional to sin(omega t), not cos(omega t). (C) is wrong: the emf depends on the difference of the areas (2A - A), not their sum. (D): the net peak emf is B*A*omega, and the smaller loop alone (area A) also gives peak emf B*A*omega, so (D) is also true for these particular areas. This is a multiple-correct question (official correct set A, B, D in the original); the single best/primary choice recorded here is (A).

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