A copper rod of mass m slides under gravity on two smooth parallel rails l distance apart and set an angle θ to the horizontal. At the bottom, the rails are joined by a resistance R in figure. There is a uniform

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Correct answer: C. mgRsinθ B^ 2 l^ 2

Terminal velocity of the rod is attained when magnetic force on the rod (Bil)& balances the component of weight of the rod $( mgsin\theta ) ,$ as in figure.

A copper rod of mass m slides under gravity on two smooth parallel rails l distance apart and set an angle θ to the

So, $Bil = mgsin\theta$

$B\left(\dfrac{e}{R}\right)l = mgsin\theta$

$\dfrac{Ble}{R} = mgsin\theta$

$\dfrac{Bl\left(B / {V}_{T}\right)}{R} = mgsin\theta$

${v}_{T} = \dfrac{mgsin\thetaR}{B^{2}l^{2}}$

A copper rod of mass m slides under gravity on two smooth parallel...

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