A wheel with 10 metallic spokes each 1.0 m long is rotated with a speed of 240 rev/min in a plane normal to magnetic field of 0.4 G at the place, the induced emf between the axle and the rim of the wheel is;
Given that \( 1\,\mathrm{G} = 10^{-4}\,\mathrm{T} \).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 5 \times 10^{-4}\,\mathrm{V} \)
A spoke of length \( R \) rotating with angular speed \( \omega \) in a field \( B \) normal to its plane of rotation has a motional emf between its ends of \( \varepsilon = \tfrac{1}{2}B\omega R^{2} \). All ten spokes join the same axle and rim, so they are in parallel, and the emf between the axle and the rim is that of one spoke; the number of spokes does not matter.
\( f = 240\,\mathrm{rev/min} = 4\,\mathrm{rev/s} \)
\( \omega = 2\pi f = 8\pi \approx 25.1\,\mathrm{rad\,s^{-1}} \)
\( B = 0.4\,\mathrm{G} = 4\times10^{-5}\,\mathrm{T} \)
\( R = 1.0\,\mathrm{m} \)
\( \displaystyle \varepsilon = \frac{1}{2} \times 4\times10^{-5} \times 25.1 \times (1.0)^{2} = 5.03\times10^{-4}\,\mathrm{V} \approx 5\times10^{-4}\,\mathrm{V} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Motional Electromotive Force – …
