A square shaped coil of area 70 cm ^ 2 having 600 turns rotates in...

Given here, $N = 600$, $A = 70 \times {10}^{- 4}{m}^{2}$, and $B = 0 . 4T$

Angular speed of coil is $\omega = \dfrac{2\pin}{60} = \dfrac{500 \times 2\pi}{60} = \dfrac{100\pi}{6}rad{s}^{- 1}$

Instantaneous emf in coil is given by $ε = - N\dfrac{d ϕ}{d t}$, where, magnetic flux $ϕ = BAcos\omegat$ and $\omegat = 30 °$ is angle between $A$ and $B$.

So, $ε = - N\dfrac{d \left(BAcos\omegat\right)}{d t} = - N\left(- BA\omega sin\omegat\right) = NBA\omega sin\omegat$

Putting the values, we get

$ε = 600 \times 70 \times {10}^{- 4} \times 0 . 4 \times \dfrac{100\pi}{6} \times \dfrac{1}{2}$

$= 44V$