Discussion Forum



1)

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?


A) the emf induced in the loop is proportional to the sum of area of the two loops,

B) The rate of change of the flux is maximum when the plane of the loops perpendicular to plane of the paper

C) The net emf induced due to both the loops is proportional to $\cos\omega t$

D) The amplitude of the maximum net emf induced due to the loops is equal to the amplitude of maximum emf induced in the smaller loop alone.

Answer:

Option B,D

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

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

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