# A metallic ring of mass m and radius l (ring being horizontal) isfalling under gravity in a region having a magnetic field. If z is thevertical direction, the z-component of magnetic field is Bz = Bo(1+λ z). If R is the resistance of the ring and if the ring falls with avelocity v, find the energy lost in the resistance. If the ring hasreached a constant velocity, use the conservation of energy todetermine v in terms of m, B, λ and acceleration due to gravity g.

Rate of change of magnetic flux:

Energy lost per second:

${I}^{2}R=\frac{{\left(\pi {L}^{2}\lambda \right)}^{2}{{B}_{o}}^{2}{v}^{2}}{R}$

Rate of change in PE:

Rate of change = $mg\frac{dz}{dt}$
= mgv

​Equate rate of change in PE to energy lost per second:

$mgv=\frac{{\left(\pi {L}^{2}\lambda {B}_{o}\right)}^{2}{v}^{2}}{R}\phantom{\rule{0ex}{0ex}}v=\frac{mgR}{{\left(\pi {L}^{2}\lambda {B}_{o}\right)}^{2}}$

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