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