Q163 / 200 · WBJEE 2014 · Physics

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 with a constant angular speed $\omega$ about the common diameter. The emf induced in the smaller loop as a function of time $t$ is

  • A
    $\frac{\pi a^{2} \mu_{0} l}{2 b} \omega \cos (\omega t)$
  • B
    $\frac{\operatorname{\pi a}^{2} \mu_{0}l}{2 b} \omega \sin \left(\omega^{2} t^{2}\right)$
  • Correct answer
    C
    $\frac{\pi a^{2} \mu_{0} l}{2 b}$ $\omega sin$ $(\omega t)$
  • D
    $\frac{\pi a^{2} \mu_{0}l}{2 b} \omega \sin ^{2}(\omega t)$

Solution

We know that $\tau=$ NBAosin $\omega t$ where $N=$ number of loops $=1$ $$ \begin{aligned} B &=\frac{\mu_{0} I}{2 b} \text { newton/amp-m } \\ A &=\pi a^{2} \text { metre }^{2} \\ \begin{aligned} \therefore &=\frac{\mu_{0} I}{2 b}\left(\pi a^{2}\right) \omega \sin \omega t \\ &=\frac{\pi a^{2} \mu_{0} I}{2 b} \cdot \omega \sin \omega t \end{aligned} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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