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)

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

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

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

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

answer is A.

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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.

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