A thin superconducting (zero resistance) ring a 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 `B_(z) = B_(0)(1-alpha z)` and `B_(r) = B_(0) beta r`, where `B_(0),alpha` and `beta` 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 relreased, it starts to move downwards with its axis 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 m, radius `r_(0)` and self inductance L. Assume the acceleration due to gravity as g.

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

Options

A. `(mgL)/(2 B_(0)^(2)alpha beta pi^(2)r_(0)^(4))`
B. `-(mgL)/(2 B_(0)^(2)alpha beta pi^(2)r_(0)^(4))`
C. zero
D. none of these

Answer

B

Explanation

Total magnetic flux at any position
`phi =B_(z)pi r_(0)^(2)-LI`
Since, `R=0`, so `phi=B_(0)(1-alphaz)pi r_(0)^(2)-LI=` constant
From initial condition `(z= 0,I=0)`, the value of constant is `phi = B_(0) pi r_(0)^(2)`
Using the above equation the current in the ring `I = (1)/(L) B_(0) alpha pi r_(0)^(2)z`
The lorentz foce acting on the ring (which can only be vertical because of the symmetry of the assembly) can be expressed as)
`F_(z)=-B_(0)I(z)2pi r_(0)=-(2 B_(0)^(2)alpha beta pi^(2)r_(0)^(4)z)/(L)=-kz`
Equation of motion of the ring is `ma_(z) = F_(z) - mg = -kz - mg`
Equilibrium position
`z_(0) = -mg//k omega_(0) = sqrt((k)/(m))`.