In an ac generator, if coil of $\mathrm{N}$ turns and area $\mathrm{A}$ is rotated at $u$ revolutions per…

In an ac generator, if coil of $\mathrm{N}$ turns and area $\mathrm{A}$ is rotated at $u$ revolutions per second in a uniform magnetic field $B$, then the motional emf produced is equal to (At $\mathrm{t}=0 \mathrm{~s}$, the coil is perpendicular to the field)
  1. $\operatorname{NBA}(2 \pi u) \sin (2 \pi u t)$
  2. $\operatorname{NBA}^2(2 \pi v) \sin (2 \pi v t)$
  3. $\mathrm{N}^2 \mathrm{~B}^2 \mathrm{~A}^2(2 \pi v) \sin (2 \pi v t)$
  4. $\mathrm{NBA}(4 \pi v) \sin (2 \pi v t)$

Solution

Flux, $\phi=\mathrm{B}(\mathrm{NA}) \cos \omega \mathrm{t}$ and $\omega=2 \pi \mathrm{v}$ $\mathrm{emf}, \mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{-\mathrm{d}}{\mathrm{dt}} \mathrm{BNA} \cos \omega \mathrm{t}=-\mathrm{NBA} \omega \sin (\omega \mathrm{t})$ $|\mathrm{e}|=\mathrm{NBA}(2 \pi \mathrm{v}) \sin (2 \pi \mathrm{vt})$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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