JEE Main · PhysicsHard

$Two long parallel conducting rails are separated by a distance L and connected by a resistor R . A conducting bar of mass m slides without friction along the rails. A uniform magnetic field B is perpendicular to the plane of the rails. If the bar is given an initial velocity v_{0}, find the distance it travels before coming to rest.$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{m v _{0} R}{B ^{2} L ^{2}}$

Solution

$The induced EMF is E = B Lv, so the current is I = B Lv / R .$
$The magnetic force is F = I L B = B^{2} L^{2} v / R .$
$Using Newton's second law, m (d v / d t) = - B^{2} L^{2} v / R .$
$Since d v / d t = v (d v / d x), we have m v (d v / d x) = - B^{2} L^{2} v / R, which simplifies to d v / d x = - B^{2} L^{2} / (m R) .$
$Integrating from v_{0} to 0 gives x = m v_{0} R / (B^{2} L^{2}) . Hence the answer is (A).$