A straight conducting wire of mass 10 g and length 20 cm slides vertically downwards, maintaining contact with two parallel, frictionless, vertical conducting rails. The rails are connected at the top by a resistor of 2 . A uniform horizontal magnetic field of 0.5 T exists perpendicular to the plane of the rails. The terminal velocity acquired by the falling wire is: (Take g = 10 ~m/s^2 and neglect the resistance of

Step-by-step solution

When the wire falls with a velocity v , the motional EMF induced is E = BLv . The induced current in the circuit is I = E R = BLv R . This current experiences an upward magnetic force F_m = ILB . Substituting I , we get F_m = ( BLv R )LB = B^2L^2v R . The wire attains terminal velocity when the upward magnetic force balances the downward gravitational force: mg = B^2L^2v R Rearranging for terminal velocity v : v = mgR B^2L^2 Given values: m = 10 g = 0.01 kg g = 10 m/s ^2 R = 2 B = 0.5 T L = 20 cm = 0.2 m Substitute