A coil of 400 turns of effective area \( 0.05\,\mathrm{m^{2}} \) is kept perpendicular to a magnetic field of \( 5 \times 10^{-3}\,\mathrm{T} \). When the plane of coil is rotated by \( 90^\circ \) around any of its co-planar axis in 0.2 s in magnetic field, the emf induced in the coil is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.5 V
By Faraday's law the average emf is \( \varepsilon = N\,|\Delta\Phi|/\Delta t \). The flux through one turn is \( \Phi = BA\cos\theta \), where \( \theta \) is the angle between the field and the normal to the coil.
Initially the coil is perpendicular to the field, so \( \theta = 0^\circ \) and \( \Phi_i = BA \).
After a rotation of \( 90^\circ \), \( \theta = 90^\circ \) and \( \Phi_f = 0 \).
\( \displaystyle |\Delta\Phi| = BA = 5\times10^{-3} \times 0.05 = 2.5\times10^{-4}\,\mathrm{Wb} \)
\( \displaystyle \varepsilon = \frac{N\,|\Delta\Phi|}{\Delta t} = \frac{400 \times 2.5\times10^{-4}}{0.2} = 0.5\,\mathrm{V} \)