A thin conducting rod M N of mass 20 gm , length 25 cm and resistance 10 Ω is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field B 0 = 4 T directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time t = 0 and it moves down along the rails. Assume air drag is negligible. Match each quantity in List- I with

Step-by-step solution

Induced emf ε = B l v ⇒ Induced current i = ε R = B l v R Direction of induced current would be from N to M . Now magnetic force acting on the rod due to this current would be in the upward direction. Therefore, we can write ⇒ m g - i l B = m a [Applying 2 nd law] ⇒ m g - B 2 l 2 v R = m d v d t ⇒ ∫ 0 v d v m g - B 2 l 2 v R = ∫ 0 t d t m ⇒ ln m g - B 2 l 2 v R 0 v - B 2 l 2 R = t m ⇒ m g - B 2 l 2 v R m g = e - B 2 l 2 m R t Now, m g = 20 × 10 - 3 × 10 = 0 . 2 , B 2 l 2 R = 4 2 × 0 . 25 2 10 = 0 . 1 and B 2 l 2 m