A square shaped coil of area $70\ \text{cm}^2$ having $600$ turns rotates in a magnetic field of $0.4\ \text{Wb m}^{-2}$, about an axis which is parallel to one of the sides of the coil and perpendicular to the direction of field. If the coil completes $500$ revolutions in a minute, the instantaneous emf when the plane of the coil is inclined at $60^\circ$ with the field, will be ______ V. (Take $\pi=\frac{22}{7}$)
Numerical value type. Enter your answer.
Answer: 44
$\omega=2\pi\times\dfrac{500}{60}=\dfrac{1100}{21}\ \text{rad s}^{-1}$, so $NBA\omega=600\times0.4\times7\times10^{-3}\times\dfrac{1100}{21}=88$ V.
With the plane at $60^\circ$ to the field, the normal is at $30^\circ$ to it: $\varepsilon=NBA\omega\sin30^\circ=44$ V.
Solution by Sreeraj P, M.Sc Physics