A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicul

Question

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T . If the resistance of the total circuit is $2 \Omega$ then the force needed to move the rod towards right with constant speed $(v)$ of $1.5 \mathrm{~m} / \mathrm{s}$ is $\_\_\_\_$ N.

Professor PC Thomas Classes · Answer

Motional emf induced in the rod is $e = Bvl$. The induced current is $I = \frac{e}{R} = \frac{Bvl}{R}$. The magnetic force on the rod is $F = IlB = \left(\frac{Bvl}{R}ight)lB = \frac{B^2 v l^2}{R}$. Substituting the values: $F = \frac{(0.10)^2 	imes 1.5 	imes (1)^2}{2} = \frac{0.01 	imes 1.5}{2} = 0.0075 	ext{ N} = 7.5 	imes 10^{-3} 	ext{ N}$.