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

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 $B$ 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 ?

TestHub · Accepted Answer

ϕ = B A cos ( ω t ) , so ε = − d t d ϕ ​ = B A ω sin ( ω t ) ε and d t d ϕ ​ are maximum when ω t = 2 π ​ , i.e. plane of loops perpendicular to B (A correct) Polarities of the two loops are opposite: e net ​ = B ( 2 A ) ω sin ω t − B ( A ) ω sin ω t = B A ω sin ω t Net emf ∝ sin ω t (B wrong) and ∝ difference of areas (C wrong) Amplitude of net emf = B A ω , same as the smaller loop's amplitude (D correct) Answers: (A), (D)

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Options:(select one or more)

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