A coil of 200 turns and area 0 . 20 m ^ 2 is rotated at half a...

Time taken for one complete revolution = time period $= 2s$.

Therefore, $\omega = \dfrac{2\pi}{T} = \pi rad{s}^{- 1}$

Now flux, $ϕ = NABcos ( \omegat )$

Therefore, induced EMF $ε = \left|- \dfrac{dϕ}{dt}\right| = NAB\omega sin ( \omegat )$

The maximum voltage generated, {ε}_\max = NAB\omega

$= 200 \times 0 . 2 \times 0 . 01 \times \pi$

$= \dfrac{4\pi}{10} = \dfrac{2\pi}{5}$ volt

Hence, $\beta = 5$.