A coil of area A and N turns is rotating with angular velocity \(\omega\) in a uniform magnetic field…
- \(\varphi=\mathrm{AB}, \varepsilon=0\)
- \(\varphi=0, \varepsilon=0\)
- \(\varphi=0, \varepsilon=\mathrm{NAB} \omega\)
- \(\varphi=\mathrm{AB}, \varepsilon=\mathrm{NAB} \omega\)
Solution

$\begin{aligned}
\phi & =\mathrm{BAN} \cdot \cos (\omega \mathrm{t}) \\ \varepsilon & =\frac{-\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{BA} \omega \mathrm{~N} \cdot \sin (\omega \mathrm{t})
\end{aligned}$
When $B$ is parallel to plane, $\underline{\underline{\omega t}}=\frac{\pi}{2}$
$\Rightarrow \phi=0, \varepsilon=\mathrm{BA} \omega \mathrm{~N}$
Asked in: JEE Main 2025 (29 Jan Shift 1)
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