JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 18)
In an ac generator, a rectangular coil of 100 turns each having area $14 \times 10^{-2} \mathrm{~m}^{2}$ is rotated at $360 ~\mathrm{rev} / \mathrm{min}$ about an axis perpendicular to a uniform magnetic field of magnitude $3.0 \mathrm{~T}$. The maximum value of the emf produced will be ________ $V$.
$\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$
$\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$
Answer
1584
Explanation
$$\phi=B.A$$
$$\phi=\mathrm{BNA}\cos\omega t$$
So, $$Emf = {{ - d\phi } \over {dt}} = NBA\omega \sin \omega t$$
So maximum value of emf is
$${E_{\max }} = NBA\omega $$
$$ = 100 \times 3 \times 14 \times {10^{ - 2}} \times {{360 \times 2\pi } \over {60}} = 1584$$
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