A very small circular loop of radius 'a' is initially (at t = 0 ) coplanar and concentric with a much larger fixed circular loop of radius 'b'. A constant current I flows in the larger loop. The smaller loop is rotated

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Quantrex Academy · Accepted Answer

Correct answer: C. ((π a^2μ_0 I)/(2b))ω ω t

The magnetic field produced by the larger loop at its centre is given by

B = \dfrac{\mu_0 I}{2b}

Since the smaller loop is very small (a \ll b

), the magnetic field over its area can be assumed to be uniform and equal to

B

.<br><br>The magnetic flux linked with the smaller loop at any time

t

is<br><br>\Phi = \vec{B} \cdot \vec{A} = B A \cos(\omega t)

Substituting

A = \pi a^2

and

B = \dfrac{\mu_0 I}{2b}

, we get

\Phi = \left(\dfrac{\mu_0 I}{2b}\right) (\pi a^2) \cos(\omega t)

According to Faraday's law of induction, the induced emf is

e = -\dfrac{d\Phi}{dt} = -\dfrac{d}{dt}

\left[\left(\dfrac{\mu_0 I \pi a^2}{2b}\right) \cos(\omegat) \right]

e =

\left(

\dfrac{\pi a^2 \mu_0 I}{2b}\right) \omega \sin

(\omegat)

<br><br>Answer: \left(

\dfrac{\pi a²\mu_0 I}{2b}\right)\omega\sin\omegat

A very small circular loop of radius 'a' is initially (at t = 0 )...

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