A 1.0 m long metallic rod is rotated with an angular frequency of \( 500\,\mathrm{rad\,s^{-1}} \) with one end hinged at the center and the other end at the circumference of a circular metallic ring of radius 1 m about an axis passing through the center and perpendicular to the plane of the ring. A constant and uniform magnetic field of 0.6 T parallel to the axis exists everywhere. The emf developed between the center and the ring is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 150 V
The rotating rod cuts the magnetic field lines, so a motional emf appears across it. A small piece \( dr \) at distance \( r \) from the centre moves with speed \( v = \omega r \), so its emf is \( B\,\omega r\,dr \).
\( \displaystyle \varepsilon = \int_{0}^{l} B\,\omega\,r\,dr = \frac{1}{2}B\,\omega\,l^{2} \)
\( B = 0.6\,\mathrm{T} \)
\( \omega = 500\,\mathrm{rad\,s^{-1}} \)
\( l = 1.0\,\mathrm{m} \)
\( \displaystyle \varepsilon = \frac{1}{2} \times 0.6 \times 500 \times (1.0)^{2} = 150\,\mathrm{V} \)
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 – …
