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A conducting rod AB of mass ‘m’ slides 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 starts moving to the right with an initial velocity $V_{0}$ . Neglect resistance of rails and self induction of the circuit

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The rod covers a distance

\frac{{mRV}_{0}}{2 B^{2} l^{2}}

before stopping

The rod covers a distance

\frac{{mRV}_{0}}{B^{2} l^{2}}

before stopping

The heat liberated in the resistor during motion of the rod is

\frac{1}{2} {mV}_{0}^{2}

The heat liberated in the resistor during motion of the rod is

\frac{1}{4} {mV}_{0}^{2}

answer is C, B.

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Detailed Solution

The initial Kinetic Energy

= \frac{1}{2} {mv}_{0}^{2}

Final K.E = 0

Change in K.E $= \frac{1}{2} {mv}_{0}^{2}$ = work done

Force on the conductor $dF = i Bd l$

Where $i = \frac{{BV}_{0} l}{R}$

$dF = \frac{B^{2} V_{0} l d l}{R}$

$F = \int_{0}^{l} \frac{B^{2} V_{0} l d l}{R} = \frac{B^{2} V_{0} l^{2}}{R}$

Now work done = average force × distance

$\frac{1}{2} {mv}_{0}^{2} = \frac{B^{2} V_{0} l^{2}}{2R} \times S$

$S = \frac{{mV}_{0} R}{B^{2} l^{2}}$

Also Heat generated = loss of K.E $= \frac{1}{2} {mv}_{0}^{2}$

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