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