A conducting rod AB (length $l$ )moves parallel to x-axis in the x-y plane. A uniform magnetic field B pointing normally out of the plane exists throughout the region. A force F acts perpendicular to the rod, so that the rod moves with uniform velocity v. The force F is given by (neglect resistance of all the connecting wires)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{v B^{2} l^{2}}{R} e^{- t / R C}$

$\frac{v B^{2} l^{2}}{R} (1 - e^{- 2 t / R C})$

$\frac{v B^{2} l^{2}}{R} (1 - e^{- t / R C})$

$\frac{v B^{2} l^{2}}{R}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

: I. Induced emf in the rod $ε = B I v$
Current in the circuit $I = \frac{ε}{R} e^{- t / R C} = \frac{B I v}{R} e^{- t / R C}$
Since the net force on the rod should be zero, the external force will be equal in magnitude but opposite to the magnetic force.

$\Rightarrow F = I I B = \frac{B^{2} I^{2} V}{R} e^{- t / R C}$