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 $\vec{B}$ 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 $\omega$ 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 the areas of the two loops
• B. The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone
• C. The net emf induced due to both the loops is proportional to $\cos \omega t$
• D. The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper

Answer 1

Answer: (B), (D)

(A) ${\varphi }_1=-BA\cos \omega t$
$\Rightarrow {\epsilon }_1=\frac{d\varphi }{dt}=-BA\omega \sin \omega t$
${\varphi }_2=2BA\cos \omega t$
${\varphi }_{\text{net }}={\varphi }_2+{\varphi }_1=BA\cos \omega t$
$\epsilon =-\frac{d\varphi }{dt}=+BA\omega \sin \omega t$
$|\epsilon |=\left|{\epsilon }_1\right|$
(B) $\varphi =BA\cos \omega t$
$\frac{d\varphi }{dt}=BA\omega \sin \omega t$
$\frac{d\varphi }{dt}$ is maximum when $\sin \theta =1$
$\theta =90^{\circ }$
$\vec{B}\perp \vec{A}$
(C) $\epsilon \propto \sin \omega t$
(D) $\epsilon \propto A_{\text{net }}$
when $A_{\text{net }}=A_1-A_2$