A metal rod of length 0.5 m moves at \(5\ \mathrm{m\,s^{-1}}\) in a uniform magnetic field of 0.4 T. The rod, its velocity and the field are perpendicular to one another. The magnitude of the magnetic force on each free electron in the rod due to this motion is (charge of an electron \(=1.6\times10^{-19}\ \mathrm{C}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3.2\times10^{-19}\ \mathrm{N}\)
Each free electron moves along with the rod at speed \( v \). A charge moving normal to a magnetic field feels the Lorentz force \( qvB \). This force is the same for every electron in the rod, so the length of the rod is not needed.
\( \begin{aligned} F &= evB \\ &= 1.6\times10^{-19}\times5\times0.4 \\ &= 3.2\times10^{-19}\ \mathrm{N} \end{aligned} \)
NCERT (2026-27):
Motional Electromotive Force – “Consider any arbitrary charge q in the conductor PQ. When the rod moves with speed v, the charge will also be moving with speed v in the magnetic field B. The Lorentz force on this charge is qvB in magnitude, and its direction is towards Q. All charges experience the same force, in magnitude and direction, irrespective of their position in the rod PQ.”