Discussion Forum

A circular insulated copper wire loop is twisted to form two loops of areas 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 an axis with a constant angular velocity ω in the magnetic field. Which of the following options is/are correct?

Answer:

Explanation:

The net magnetic flux through that loops at time t is

$\phi = B(2A-A)\cos\omega t=B\omega A \cos\omega t$

So,  $\mid\frac{\text{d}\phi}{\text{d}t}\mid =BA \cos\omega t$

  $\therefore$     $\mid\frac{\text{d}\phi}{\text{d}t}\mid$ is maximum when $\phi=\omega t$ = $\frac{\pi}{2}$

The emf induced in the smaller loop

ε_smaller = $-\frac{\text{d}}{\text{d}t}(BA \cos\omega t)=B\omega A\sin\omega t$

$\therefore$    Amplitude of maximum net induced in  both the loops= Amplitudeof maximum emf induced the smaller loop alone.