A circular coil of 100 turns and area of cross-section \( 10^{-2}\,\mathrm{m^2} \) is rotated about its vertical diameter at 60 rpm in a uniform horizontal magnetic field of \( 3.5 \times 10^{-2}\,\mathrm{T} \). The maximum emf induced in the coil is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 0.22\,\mathrm{V} \)
For a coil rotating with angular speed \( \omega \) in a uniform field, the flux is \( \Phi = NBA\cos\omega t \), so by Faraday's law the induced emf is \( \varepsilon = NBA\omega\sin\omega t \). Its maximum value is
\( \displaystyle \varepsilon_0 = NBA\omega \)
\( N = 100 \)
\( A = 10^{-2}\,\mathrm{m^2} \)
\( B = 3.5\times10^{-2}\,\mathrm{T} \)
The coil makes 60 revolutions per minute, that is 1 revolution per second.
\( \displaystyle \omega = 2\pi\nu = 2\pi \times 1\,\mathrm{s^{-1}} = 2\pi\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle \varepsilon_0 = 100 \times 3.5\times10^{-2} \times 10^{-2} \times 2\pi = 0.07\pi\,\mathrm{V} \approx 0.22\,\mathrm{V} \)