NTA Abhyas JEE Main2020PhysicsElectromagnetic InductionPractice
A non-conducting ring of mass m = 4 kg and radius R = 10 cm has a charge Q = 2 C uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that the plane of the ring is parallel to the surface. A vertical magnetic field B = 4 t 3 T is switched on at t = 0 . At t = 5 s ring starts to rotate about the vertical axis through the centre. The coefficient of friction between the rin
Correct answer
18
Step-by-step solution
The induced electric field at the periphery of the ring is E = R 2 d B d t ⇒ E = R 2 12 t 2 = 6 R t 2 When the ring is about to slip, q E R = μ m g R ⇒ 6 q R t 2 = μ m g ⇒ μ = 6 q R t 2 m g = 3 4 = 18 24 k = 18