+2 votes
in JEE by kratos

A thin superconducting (zero resistance) ring is held above a vertical long solenoid, as shown in the figure. The axis of symmetry of the ring is same to that of the solenoid. The cylindrically symmetric magnetic field around the ring can be described approximately in terms of the vertical and radial component of the magnetic field vector as Bz = B0(1 - αz) and Br = B0βr, where B0 , α and β are positive constants, and z & r are vertical and radial position coordinates, respectively. Initially plane of the ring is horizontal, has no current flowing in it. When released, it starts to move downwards with its axis still vertical. Initial coordinates of the centre of the ring 'O' is z = 0 and r = 0.

In the given diagram point O is on the axis and slightly above the solenoid having vertical and radial position coordinates as (0, 0). Ring has mass m, radius r0 and self inductance L. Assume the acceleration due to gravity as g.

1. Find the vertical coordinates z for equilibrium position of the ring.

2. Find the time **** of SHM (for small displacement along z-axis) of the ring.**

1 Answer

+2 votes
by kratos
 
Best answer

Correct option 1. (B) 2. (A)

Explanation:

The lorentz force acting on the ring (which can only be vertical, because of the symmetry of the assembly) can be expressed as

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