A conducting bar of mass m and length l moves on two frictionless parallel conducting rails in the presence of a uniform magnetic field B₀ -directed into the paper. The bar is given an initial velocity v₀ to the right and is released at t=0. The velocity of the bar as a function of time is given by

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$v_{0} e^{- \frac{B_{0}^{2} I^{2} t}{mR}}$

$v_{0} \times e^{\frac{- 2 B_{0}^{2} I^{2} t}{mR}}$

$v_{0} e^{- \frac{B_{0}^{2} I^{2} t}{2 mR}}$

$v_{0} \times \frac{mRt}{B_{0}^{2} l^{2}}$

answer is B.

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Detailed Solution

Let at any time t, velocity of rod be v, then emf developed across its ends is e = B₀vl . Due to this induced emf, a current will establish in circuit given by $I = \frac{B_{0} v /}{R}$

For rod,
$\begin{array}{l} m \frac{dv}{dt} = - {IB}_{0} I = - \frac{B_{0}^{2} I^{2}}{R} v \\ \Rightarrow \int_{v_{0}}^{v} \frac{dv}{v} = - \int_{0}^{t} \frac{B_{0}^{2} I^{2}}{mR} dt \Rightarrow v = v_{0} e^{- \frac{B_{0}^{2} /^{2} t}{mR}} \end{array}$