The induced e.m.f in a coil rotating in a uniform magnetic field will…

The correct answer is Option (2) → B and D Only

A coil of \( N \) turns and area \( A \) rotating with angular speed \( \omega \) in a field \( B \) works as an ac generator. By Faraday's law the induced emf is

    \( \displaystyle e = N B A \omega \sin\omega t, \qquad e_0 = N B A \omega \)

So the emf increases with each of \( N \), \( B \), \( A \) and \( \omega \).

(A) Total number of turns in the coil decreases — Incorrect. Fewer turns means a smaller \( N \), so the emf decreases.

(B) Magnetic field increases — Correct. A larger \( B \) gives a larger flux through the coil and a larger emf.

(C) Speed of rotation of coil decreases — Incorrect. A smaller \( \omega \) means the flux changes more slowly, so the emf decreases.

(D) Area of the coil increases — Correct. A larger \( A \) gives a larger flux and a larger emf.

NCERT (2026-27):
AC Generator – “… From Faraday’s law, the induced emf for the rotating coil of N turns is then,

    \( \displaystyle \varepsilon = -N\frac{\mathrm{d}\Phi_B}{\mathrm{d}t} = -NBA\frac{\mathrm{d}}{\mathrm{d}t}(\cos\omega t) \)

Thus, the instantaneous value of the emf is

    \( \displaystyle \varepsilon = NBA\omega\sin\omega t \qquad (6.19) \)

where NBA\( \omega \) is the maximum value of the emf, which occurs when \( \sin\omega t = \pm 1 \).”