Maximum voltage produced in an AC generator completing $60$ cycles in $30$ seconds is $250V$. What is the maximum e.m.f. produced when the armature completes $1800$ rotation?

Hint: An a.c. A generator is a device which converts mechanical energy into a.c. energy. It consists of a rectangular coil of several turns rotating in a uniform magnetic field. Consequently, it produces alternating e.m.f.

Complete step by step answer:
An a.c. the generator is based on the principle of electromagnetic induction i.e., whenever the amount of magnetic flux linking a coil changes, an e.m.f. is induced in the coil. The direction of induced current in the coil is given by Fleming’s right-hand rule.Let’s calculate the e.m.f. induced in the rotating coil.

Let the strength of the magnetic field be $B$. $N$ is the number of turns of the coil. $A$ is the area of each turn. $\theta $ is the angle between the normal to the surface of the coil and the magnetic field at any time $t$.Then,
$\theta = \omega t$
$\omega $ is the angular frequency
The magnetic flux $\phi $ through each turn of the coil is given by
$\phi = AB\cos \theta = AB\cos \omega t$

The e.m.f. induced in the coil, $E = - \dfrac{d}{{dt}}\left( {N\phi } \right)$
$ \Rightarrow E = - \dfrac{d}{{dt}}\left( {NAB\cos \omega t} \right)$
Further calculating
$ \Rightarrow E = - NAB\left( { - \omega \sin \omega t} \right)$
$ \Rightarrow E = NAB\omega \sin \omega t$
The maximum e.m.f. ${E_0}$ induced in the coil when $\sin \omega t = 1$
${E_0} = NAB\omega $
The maximum e.m.f. induced in the coil depends on the number of turns of the coil, the area of the coil, the strength of the magnetic field and the angular frequency.

Hence, the maximum e.m.f. induced in the coil is equal to the maximum voltage produced i.e., maximum e.m.f. produced is $250\,V$.

Note: The e.m.f. induced in the coil doesn’t depend on the number of cycles (rotation) completed by the coil. As long as the angular frequency $\omega $ of the coil is uniform, the maximum e.m.f. induced in the coil is constant $\left( {{E_0} = NAB\omega } \right)$.