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. [2017]

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 $ω$ in the magnetic field. Which of the following options is/are correct [2017]

Smart Achievers · Accepted Answer

(2, 4)

The net magnetic flux through the loops at time $t$

$ϕ = B (2 A) \cos ω t - B A \cos ω t = B (2 A - A) \cos ω t = B A \cos ω t$

$∴$ $| \frac{d ϕ}{d t} |$ $= B ω A \sin ω t$

So, $| \frac{d ϕ}{d t} |$ is maximum when $ω t = \frac{π}{2}$

The emf induced in the smaller loop is

$ε_{smaller} = - \frac{d}{d t} (B A \cos ω t) = B ω A \sin ω t$

$∴$ Amplitude of maximum net emf induced in both the loops = Amplitude of maximum emf induced in the smaller loop alone.

Solution

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

2 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.

3 The net emf induced due to both the loops is proportional to $\cos ω t .$

4 The rate of change of the flux is maximum when the plane of the loop is perpendicular to the plane of the paper.

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