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(D) The magnetic force acting on the conductor is given by $F = I l B$,where $I$ is the current.Since $I = \frac{dq}{dt}$,the impulse provided to the

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$\frac{m\sqrt{2gh}}{Bl}$

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(D) The magnetic force acting on the conductor is given by $F = I l B$,where $I$ is the current.Since $I = \frac{dq}{dt}$,the impulse provided to the conductor is $J = \int F dt = \int (I l B) dt = l B \int I dt = l B q$,where $q$ is the total charge passed.This impulse provides an initial velocity $v_0$ to the conductor such that $J = m v_0$.Equating the two expressions for impulse: $m v_0 = q l B$,which gives $v_0 = \frac{q l B}{m}$.Using the kinematic equation for vertical motion,$v^2 = u^2 - 2gh$. At the maximum height $h$,the final velocity is $0$,so $0 = v_0^2 - 2gh$,which implies $v_0 = \sqrt{2gh}$.Substituting $v_0$ into the impulse equation: $\sqrt{2gh} = \frac{q l B}{m}$.Solving for $q$,we get $q = \frac{m \sqrt{2gh}}{l B}$.

$A$ conductor of length $l$ and mass $m$ is placed along the east-west line on a table. Suddenly,a certain amount of charge is passed through it and it is found to jump to a height $h$. The earth's magnetic induction is $B$. The charge passed through the conductor is:

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