A thin superconducting ring is held above a vertical long solenoid, as shown in the figure, having the same axis. The cylindrically symmetric magnetic field around the ring can be described approximately in terms of the vertical and radial components of the magnetic field vector as $B_{z} = B_{0} (1 - α z)$ and $B_{r} = B_{0} β r$ , (r= radial distance measured from the center of the ring)
Where $B_{0}, α$ and $β$ are positive constants and z and r are vertical and radial position coordinates, respectively. Initially plane of the ring is horizontal and the ring has no current flowing in it. When released, it starts to move downwards with its axis still in vertical direction. 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 $r_{0}$ and self-inductance L. Assume, the acceleration due to gravity as g.

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The magnitude of current in the ring is $\frac{1}{L} B_{0} α π {r_{0}}^{2} z .$

The time period of ring’s motion is $\frac{1}{B_{0} {r_{0}}^{2}} \sqrt{\frac{2 m L}{α β}} .$

The force(magnitude) acting on the ring is $\frac{2 {B_{0}}^{2} α β π^{2} {r_{0}}^{4} z}{L} .$

Vertical coordinate z for equilibrium position of the ring is $- \frac{m g L}{2 {B_{0}}^{2} α β π^{2} {r_{0}}^{4}} .$

answer is A, C, D.

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Detailed Solution

Total magnetic flux at any position, $ϕ = B_{z} π r_{0}^{2} - L I =$ constant
From initial condition (z=0, I=0), the value of constant is $ϕ = B_{0} π r_{0}^{2}$ .
Using the above equation, the current in the ring, $I = \frac{1}{L} B_{0} α π r_{0}^{2} z$
The Lorentz force acting on the ring (which can only be vertical, because of the symmetry of the assembly) can be expressed as $F_{z} = - B_{r} I (z) 2 π r_{0}$
$= - \frac{2 {B_{0}}^{2} α β π^{2} r_{0}^{4} z}{L} = - k z$
Equation of motion of the ring is $m a_{z} = F_{z} + m g = - k z + m g$
At equilibrium position,
$z_{0} = - \frac{m g}{k} = - \frac{m g L}{2 B_{0}^{2} α β π^{2} r_{0}^{4}}$

Now, $ω_{0} = \sqrt{\frac{k}{m}} = \sqrt{\frac{2 B_{0}^{2} α β π^{2} r_{0}^{4}}{L m}}$ $= B_{0} π r_{0}^{2} \sqrt{\frac{2 α β}{m L}}$ $∴ T = \frac{2 π}{ω} = \frac{2 π}{B_{0} π r_{0}^{2}} \sqrt{\frac{m L}{2 α β}} = \frac{1}{B_{0} r_{0}^{2}} \sqrt{\frac{2 m L}{α β}}$