A copper rod of mass m slides under gravity on two smooth parallel rails , with separation $l$ and set at an angle of $θ$ with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic field B normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is

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$\frac{mgRcosθ}{B^{2} l^{2}}$

$\frac{mgRsinθ}{B^{2} l^{2}}$

$\frac{mgR \tan θ}{B^{2} l^{2}}$

$\frac{mgRcotθ}{B^{2} l^{2}}$

answer is A.

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Detailed Solution

Copper rod will aquire terminal velocity when magnetic force = gravitational force
Or $I l B = mgsin θ .$
Also, $I = \frac{induced emf}{R} = \frac{B l v}{R}$
From equations (i) and (ii) we get
$\frac{B^{2} l^{2} v}{R} = mgsin θ; ∴ v = \frac{mgRsin θ}{B^{2} l^{2}}$

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