Q. A wire of length $1\ m$ is moving at a speed of $2ms^{- 1}$ perpendicular to its length and in a homogenous magnetic field of $0.5\ T$. The ends of the wire are joined to a circuit of resistance $6\ \Omega$. The rate at which work is being done to keep the wire moving at constant speed is
A) $\frac{1}{12}\ W$
B) $\frac{1}{6}\ W$
C) $\frac{1}{3}\ W$
D) $1W$
Explanation
(b) Rate of work $= \frac{W}{t} = P = Fv;$ also $F = Bil = B\left( \frac{Bvl}{R} \right)l$
$\Rightarrow P = \frac{B^{2}v^{2}l^{2}}{R} = \frac{(0.5)^{2} \times (2)^{2} \times (1)^{2}}{6} = \frac{1}{6}W$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.