A straight conducting rod of length $0.25\text{m}$ moves with a speed $3.0{\text{m s}}^{-1}$ perpendicular to a uniform magnetic field of $0.40\text{T}$ directed into the page. Find the magnitude of the induced emf across the ends of the rod.

Step-by-step solution

1. The motional emf is given by Faraday’s law for a moving conductor: $\epsilon =B\ell v$.
2. Substitute the given values: $B=0.40\text{T}$, $\ell =0.25\text{m}$, $v=3.0{\text{m s}}^{-1}$.
3. Calculate: $\epsilon =0.40\times 0.25\times 3.0=0.30\text{V}$.
4. Thus the induced emf is $0.30\text{V}$.

Concept tested: Motional emf (Faraday’s law)

Published 21 September 2026 · How this solution was prepared and checked