JEE MAIN - Physics (2024 - 27th January Morning Shift - No. 23)
Two coils have mutual inductance $$0.002 \mathrm{~H}$$. The current changes in the first coil according to the relation $$\mathrm{i}=\mathrm{i}_0 \sin \omega \mathrm{t}$$, where $$\mathrm{i}_0=5 \mathrm{~A}$$ and $$\omega=50 \pi$$ rad/s. The maximum value of emf in the second coil is $$\frac{\pi}{\alpha} \mathrm{~V}$$. The value of $$\alpha$$ is _______.
Risposta
2
Spiegazione
$$\begin{aligned}
& \phi=\mathrm{Mi}=\mathrm{Mi}_0 \sin \omega \mathrm{t} \\
& \mathrm{EMF}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=-0.002\left(\mathrm{i}_0 \omega \cos \omega \mathrm{t}\right) \\
& \mathrm{EMF}_{\max }=\mathrm{i}_0 \omega(0.002)=(5)(50 \pi)(0.002) \\
& \mathrm{EMF}_{\max }=\frac{\pi}{2} \mathrm{~V}
\end{aligned}$$