Q. As shown in the figure a metal rod makes contact and completes the circuit. The circuit is perpendicular to the magnetic field with $B = 0.15\ tesla$. If the resistance is $3\Omega$, force needed to move the rod as indicated with a constant speed of $2m/sec$ is
A) $3.75 \times 10^{- 3}\ N$
B) $3.75 \times 10^{- 2}\ N$
C) $3.75 \times 10^{2}\ N$
D) $3.75 \times 10^{- 4}\ N$
Explanation
(a) Induced current in the circuit $i = \frac{Bvl}{R}$
Magnetic force acting on the wire $F_{m} = Bil = B\left( \frac{Bvl}{R} \right)l$
$\Rightarrow F_{m} = \frac{B^{2}vl^{2}}{R}$. External force needed to move the rod with constant velocity
$\left( F_{m} \right) = \frac{B^{2}vl^{2}}{R} = \frac{(0.15)^{2} \times (2) \times (0.5)^{2}}{3} = 3.75 \times 10^{- 3}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.