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 $B$ point 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 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
• B. The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper
• C. The $emf$ induced in the loop is proportional to the sum of the areas of the two loops
• D. The net $emf$ induced due to both the loops is proportional to $\cos \omega t$

Answer 1

Answer: (A), (B)

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

Initial condition

${\varphi }_1=AB,{\varphi }_2=2AB$

General form

${\varphi }_1=ABcos\omega t$

${\varphi }_2=2ABcos\omega t$

$\begin{array}{}\therefore E_1=\frac{-d{\varphi }_1}{dt}=AB\omega sin\omega t& \text{(1)}\end{array}$

$\begin{array}{}{E}_2=2AB\omega sin\omega t& \text{(2)}\end{array}$

Now, the emf induced in the two loops have opposite polarising because of the given orientation of the wires.

$\therefore |E_{net}|=|E_1-E_2|=AB\omega sin\omega t$
$E_{net}$ will be maximum when $sin\omega t=1$
$\left(max\right)E_{net}=AB\omega$
Also maximum $emf$ induced in loop 1, $\left(max\right)E_1=AB\omega$ for $sin\omega t=1$
Hence 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

Option A is correct.

Let area makes an angle $\theta$ with magnetic field at time $t$. Then, $\theta =\omega t$
Since $emf$ induced is maximum when $sin\omega t=1$ i.e $\theta =\omega t={90}^{\circ }$. Hence area vector is perpendicular to magnetic field. Since magnetic field points into the plane of paper, this means that area which is perpendicular to magnetic field must lie in the plane of paper.
Hence option B is correct.