A metallic rod of length L is rotated about one of its ends at an ang…

The correct answer is Option (3) → \( \displaystyle\frac{\omega BL^{2}}{2} \)

A rotating rod cuts magnetic field lines; the motional emf in an element \( dr \) at distance \( r \) from the pivot is \( d\varepsilon = Bv\,dr \) with \( v = \omega r \). Different parts of the rod move at different speeds, so the elements are added by integration.

    \( \displaystyle \varepsilon = \int_{0}^{L} B\omega r\,dr = B\omega\left[\frac{r^{2}}{2}\right]_{0}^{L} = \frac{\omega BL^{2}}{2} \)

Shortcut: Use the average speed: the speed rises linearly from 0 at the pivot to \( \omega L \) at the free end, so \( v_{\mathrm{avg}} = \omega L/2 \) and \( \varepsilon = BLv_{\mathrm{avg}} = B\omega L^{2}/2 \).

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27):
Motional Electromotive Force – “… Example 6.6 A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (Fig. 6.11). A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?”

Motional Electromotive Force – “… Using Eq. (6.5), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by

    \( \displaystyle d\varepsilon = Bv\,dr \)

Hence,

    \( \displaystyle \varepsilon = \int d\varepsilon = \int_0^R Bv\,dr = \int_0^R B\omega r\,dr = \frac{B\omega R^2}{2} \)

”