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

Question

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:

TestHub · Accepted Answer

From Faraday's law of electomagnetic induction, $e = \frac{d ϕ}{d t} = \frac{d ( B A )}{lt} = \frac{d ( Bll )}{d t} = \frac{B d l \times l}{d t} = B V l$ Also, $F = i lB = (\frac{B V}{R}) (l^{2} B) = \frac{B ^{2} l ^{2} V}{R}$ At equilibrium
$m g sin θ = \frac{B ^{2} l V}{R} \Rightarrow V = \frac{m g R sin θ}{B ^{2} l ^{2}}$

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