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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:

mgRcosθB²l²

mgRcotθB²l²

mgRtanθB²l²

mgRsinθB²l²

Solution:

ε = dΦ/dt = d(BA)/dt = d(Blx)/dt = Bldx/dt = BlV
F = ilB = (B2lV/R)(lB) = B²l²/R V
At equilibrium ⇒ mgsinθ = B²l²/R V ⇒ V = mgRsinθ/B²l²