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 {arrow}{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 ω in the magnetic field. Which of the following options is/are correct? [JEE Adv 2017 P1]

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

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

The net emf induced due to both the loops is proportional to ω t

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

Test Index

JEE Advanced 2017 Paper 1

Section: Physics

Question : 7 of 54

Marks: +1, -0

\overset{\rightarrow}{B}

\omega

The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude ofmaximum emf induced in the smaller loop alone
The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of thepaper
The net emf induced due to both the loops is proportional to $\cos\omega t$
The emf induced in the loop is proportional to the sum of the areas of the two loops

Solution:

The angle that area vector makes with

\overset{\rightarrow}{B}

at time t is

\theta = \omega t

|\phi_1|=BA\cos\omega t\Rightarrow|\epsilon_1|=BA\omega\sin(\omega t)

|\phi_2|=2BA\cos(\omega t)\Rightarrow|\epsilon_2|=2BA\omega\sin(\omega t)

Due to orientation of loops, the two EMFs will work against each other.

So,

\epsilon(\text{net})|=BA\omega\sin(\omega t)

So, (A) is correct .

We can see that

\epsilon_{\text{net}}

is maximum when

\theta = \frac{\pi}{2}

.So,(B) is correct.Obviously,(C) is incorrect.(D) is incorrect as the EMF is proportional to difference in areas.

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