A metal conductor of length 1 m rotates in vertical plane about an axis passing through one of its ends at angular velocity \( 4\,\mathrm{rad\,s^{-1}} \). What is the emf developed between the two ends of the conductor, if the horizontal component of the earth's magnetic field is 0.2 G?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 40\,\mu\mathrm{V} \)
A rod rotating in a magnetic field cuts the field lines, and a motional emf appears between its ends. Here the rod turns in a vertical plane, and the horizontal component of the earth's field is taken perpendicular to that plane, as the question expects.
In one rotation the rod sweeps a circle of area \( \pi l^{2} \) in time \( 2\pi/\omega \), so the rate at which it sweeps area is \( \frac{1}{2}\omega l^{2} \). By Faraday's law the emf is the rate of flux cut:
\( \displaystyle \varepsilon = \frac{1}{2}\,B\,\omega\,l^{2} \)
\( B = 0.2\,\mathrm{G} = 0.2\times10^{-4}\,\mathrm{T} = 2\times10^{-5}\,\mathrm{T} \)
\( \omega = 4\,\mathrm{rad\,s^{-1}} \)
\( l = 1\,\mathrm{m} \)
\( \displaystyle \varepsilon = \frac{1}{2} \times 2\times10^{-5} \times 4 \times (1)^{2} = 4\times10^{-5}\,\mathrm{V} \)
\( \varepsilon = 40\times10^{-6}\,\mathrm{V} = 40\,\mu\mathrm{V} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
