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A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle $θ$ with respect to the horizontal. The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the final velocity?

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$\frac{m g R \sin θ}{B^{2} d^{2} \cos^{2} θ}$

$\frac{m g R \sin^{2} θ}{B^{2} d^{2} \cos θ}$

$\frac{m g R}{B^{2} d^{2}}$

$\frac{m g R \sin θ}{B^{2} d^{2}}$

answer is A.

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Detailed Solution

power due to gravity= heat power across resistor

$(m g s i n θ) (v) = {(B c o s θ)}^{2} v^{2} d^{2}$

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