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A rod of length $l$ , negligible resistance and mass $m$ slides on two horizontal frictionless rails of negligible resistance by hanging a block of mass $m_{1}$ by the help of insulating mass less string passing through fixed massless pulley, (as shown). If a constant magnetic field $B$ acts upward perpendicular to the plane of the figure, find steady state velocity of hanging mass -

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$\frac{m_{1} g R}{B^{2} l^{2}}$ upward

$\frac{m g R}{B^{2} l^{2}}$ downward

$\frac{m_{1} g R}{B^{2} l^{2}}$ downward

$\frac{m g R}{B^{2} l^{2}}$ upward

answer is B.

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Detailed Solution

m1g=T,T=ilB

ϵ=Blv

i=BlvR

v=m1gRB2l2 downward.

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