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A conducting rod AB of mass ‘m’ without friction over two long conducting horizontal rails separated by a distance $l .$ At the left rails are inter connected by a resistor R. The system is located in uniform magnetic field acting perpendicular to the plane of the loop. At the moment t = 0, the rod AB is projected horizontal on the rails to the right with an initial velocity $V_{0} .$ Neglect resistance of rails and self induction of the circuit

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The heat liberated in the resistor during motion of the rod is $\frac{1}{4} m V_{0}^{2}$

The heat liberated in the resistor during motion of the rod is $\frac{1}{2} m V_{0}^{2}$

The rod covers a distance $\frac{m R V_{0}}{2 B^{2} l^{2}}$ before stopping

The rod covers a distance $\frac{m R V_{0}}{B^{2} l^{2}}$ before stopping

answer is B, C.

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Detailed Solution

$F_{n e t} = m a$

$- B i l = m V (\frac{d V}{d x})$ but induced current $i = B V l / R$

$\Rightarrow \frac{- V B^{2} l^{2}}{m R} = V (\frac{d V}{d x}) \Rightarrow \int_{0}^{x} d x = \frac{- m R}{B^{2} l^{2}} \int_{V_{0}}^{0} d V \Rightarrow x = \frac{m R V_{0} }{B_{^{}}^{2} l^{2}}$

Heat liberated = change in K.E of rod = $\frac{1}{2} m V_{0}^{2}$

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