Question

The set-up, shown in the figure, is present in a uniform magnetic field, B, directed perpendicular to the plane of this ‘set-up’. The rod, PQ, of length L, is allowed to slide down vertically, under its own weight, on the (shaded) conducting rods. The terminal velocity, acquired by the sliding rod (of mass m), would equal

(1) \((\frac{m^2g^2R}{Bl})\)

(2) \((\frac{Bl}{m^2g^2R})\)

(3) \((\frac{B^2l^2}{mgR})\)

(4) \((\frac{mgR}{B^2l^2})\)

Answer 1

 (4) \((\frac{mgR}{B^2l^2})\)

As the rod PQ slides down, the magnitude of the induced emf, in it, at an instant when its velocity is v, equals Blv . Due to this an induced current i(= Blv/R) flows in it. Because of this induced current, the sliding rod experiences an (mg). This (upward) force equals Bil = B²l²v/R. When this upward force becomes equal to the weight of the sliding rod, the net force on the rod becomes zero and the rod, subsequently, slides down with a constant velocity, v_T , its terminal velocity. We thus have