A circular insulated copper wire loop is twisted to form two loops of area A and 2 A 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 magn

Step-by-step solution

When a conducting loop rotating in a magnetic field B has an angular velocity ω , the flux through the loop of area A at an angle θ = ω t , ϕ = B A cos ⁡ θ = B A cos ⁡ ω t The induced emf, ε = - d ϕ d t = B A ω sin ⁡ ω t So, ε a n d d ϕ d t ∝ sin ⁡ ω t So, the emf is maximum when, ω t = θ = π 2 . Since the emf in the smaller loop will act opposite to that of the larger loop, ε N e t = ε 2 A - ε A = B 2 A ω sin ⁡ ω t - B A ω sin ⁡ ω t = B 2 A - A ω sin ⁡ ω t = B A ω sin ⁡ ω t So, the amplitude of the maximum net emf