A metal rod of length 1.0 m turns about an axis through its midpoint with an angular speed of 20 rad s−1. The rod turns in a plane normal to a uniform magnetic field of 0.5 T. The emf induced between the midpoint and one end of the rod is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.25\ \mathrm{V}\)
Take a rod of length \( R \) that turns about one end with angular speed \( \omega \). If it turns in a plane normal to \( B \), the emf across it is \( \frac{1}{2}B\omega R^{2} \). Each half of this rod acts like such a rod. Its fixed end is the midpoint, and its length is half the rod.
\( B=0.5\ \mathrm{T} \)
\( \omega=20\ \mathrm{rad\,s^{-1}} \)
\( R=0.5\ \mathrm{m} \) (half of 1.0 m)
\( \begin{aligned} \varepsilon &= \frac{1}{2}B\omega R^{2} \\ &= \frac{1}{2}\times0.5\times20\times(0.5)^{2} \\ &= 1.25\ \mathrm{V} \end{aligned} \)
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,”
\( \displaystyle \varepsilon=\int d\varepsilon=\int_0^R Bv\,dr=\int_0^R B\omega r\,dr= \)\( \displaystyle \frac{B\omega R^{2}}{2} \)