Question 364: Electromagnetic Induction

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 $$\overrightarrow B $$ 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?

Accepted Answer

Consider the coordinate axes shown in the figure. The magnetic field is into the paper, i.e., $$\overrightarrow B  = B\widehat k$$. Let loop 1 be the loop of area A and loop 2 be the loop of area 2A. At t = 0, both loops are in the plane of the paper. They start rotating about the x-axis with a constant angular velocity $$\overrightarrow \omega   = \omega \widehat i$$. Let the area vectors of loop 1 and 2 be $${\overrightarrow A _1}$$ and $${\overrightarrow A _2}$$, respectively. At t = 0, $${\overrightarrow A _1}$$ = $$A\widehat k$$ and $${\overrightarrow A _2} = 2A\widehat k$$ (we chose area into the plane of the paper as positive). In time t, the loops rotate by an angle $$	heta$$ = $$\omega$$t. Their area vectors make an angle $$	heta$$ = $$\omega$$t with z-axis (direction of the magnetic field). Thus, the magnetic fluxes passing through the loop 1 and the loop 2 at time t are given by
$$\phi$$
1
= $$\overrightarrow B .{\overrightarrow A _1}$$ = B A
1
cos$$	heta$$ = B A cos $$\omega$$t,
$$\phi$$
2
= $$\overrightarrow B .{\overrightarrow A _2}$$ = B A
2
cos$$	heta$$ = 2B A cos $$\omega$$t.
By Faraday's law of electromagnetic induction, induced emfs in the loop 1 and in the loop 2 are given by
$${\varepsilon _1} =  - {{d{\phi _1}} \over {dt}} = B\,A\omega \sin \omega t$$,
$${\varepsilon _2} =  - {{d{\phi _2}} \over {dt}} = 2B\,A\omega \sin \omega t$$.
The rate of change of flux is maximum when $$\omega$$t = $$\pi$$/2 i.e., when the plane of the loops is perpendicular to the plane to the paper.
Let T = 2$$\pi$$/$$\omega$$ be the time period to complete one revolution. The fluxes $$\phi$$
1
and $$\phi$$
2
decrease with time if t $$\in$$ (0, T/4). By Lenz's law, the induced current should oppose reductions in $$\phi$$
1
and $$\phi$$
2
. Thus, induced current in both the loops should be clockwise. Hence, the polarity of 'equivalent batteries' are as shown in the figure. These batteries are connected with reverse polarity. Thus, net emf induced due to both the loop is
$$\varepsilon  = {\varepsilon _2} - {\varepsilon _1} = B\,A\omega \sin \omega t$$.
Note that the net emf $$\varepsilon $$ is proportional to the difference in areas of two loops and $$\left| \varepsilon  ight| = \left| {{\varepsilon _1}} ight|$$.

Answer

The emf induced in the loop is proportional to the sum of the areas of the two loops

Answer

The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude if maximum emf induced in the smaller loop alone

Answer

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

Answer

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

Electromagnetic Induction

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