A metal rod rotates about one of its ends in a plane perpendicular to a uniform magnetic field. The emf between its ends is \(\varepsilon\). A rod twice as long is now rotated in the same way in the same field at half the angular speed. The emf between the ends of this rod is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2\varepsilon\)
Each small piece of a rotating rod moves with speed \( \omega r \), so the pieces far from the pivot give more emf. Adding the emfs of all the pieces gives \( \varepsilon=\frac{1}{2}B\omega l^{2} \) between the ends. So the emf varies as \( \omega l^{2} \).
\( \varepsilon=\frac{1}{2}B\omega l^{2} \)
\( \begin{aligned} \varepsilon^{\prime} &= \frac{1}{2}B\left(\frac{\omega}{2}\right)(2l)^{2} \\ &= 2\times\frac{1}{2}B\omega l^{2} \\ &= 2\varepsilon \end{aligned} \)
NCERT (2026-27):
Example 6.6 – “… 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 dε = Bv dr. Hence,”
\( \displaystyle \varepsilon=\int d\varepsilon=\int_0^R Bv\,dr=\int_0^R B\omega r\,dr= \)\( \displaystyle \frac{B\omega R^{2}}{2} \)
“Note that we have used v = ω r. …”