A metallic ring of field. If z is the radius l (ring being horizontal is falling under gravity in a region haivng a magnetic field. If z is the vertical direction, the z-component of magnetic field is `B_(z) = B_(0) (1 + lambda z)`. If R the resistance of the ring and if the ring falls with a velocity `upsilon`, find the energy lost in the resistance If the ring has reached a constant velocity, use the conservation of energy to determine `upsilon` in terms of m, B, `lambda` and acceleration due to gravity g.

Explanation

The magnetic flux linked with the metallic ring of mass m and radius l falling under gravity in a region having a magnetic field whose z-component of magnetic field is `B_z=B_0(1+lambdaz)` is
`phi-B_z(pil^2)=B_0(1+lambdaz)(pil^2)`
Applying Faraday's law of EMI, we have emf induced given by `(dphi)/(dt)` = rate of change of flux also, by Ohm's law
`B_0(pil^2)lambda (dz)/(dt)=lR`
On rearranging the terms, we have `l=(pil^2B_0lambda)/(R)`
Energy lost/second `l^2R=((pil^2lambda)^2B_(0)^(2)V^(2))/(R)`
This must come from rate of change in `PE=mg(dz)/(dt)=mgv`
Thus , `mgv=(pi^2lambdaB_(0)^2v^(2))/(R)`or `v=(mgR)/(pil^2lambdaB_(0))^2`
This is the required expression of velocity.