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A metallic ring of mass m and radius l (ring being horizontal) is

falling under gravity in a region having a magnetic field. If z is the

vertical direction, the z-component of magnetic field is B_{z} = B_{o}

(1+λ z). If R is the resistance of the ring and if the ring falls with a

velocity v, find the energy lost in the resistance. If the ring has

reached a constant velocity, use the conservation of energy to

determine v in terms of m, B, λ and acceleration due to gravity g.

**Rate of change of magnetic flux:**

$\frac{d\varphi}{dt}=\frac{d}{dt}\left(AB\right)\phantom{\rule{0ex}{0ex}}=A\frac{dB}{dt}\phantom{\rule{0ex}{0ex}}=A\frac{d}{dt}{B}_{o}\left(1+\lambda z\right)\phantom{\rule{0ex}{0ex}}=\left(\pi {L}^{2}\right){B}_{o}\lambda \frac{dz}{dt}\phantom{\rule{0ex}{0ex}}=\left(\pi {L}^{2}\right){B}_{o}\lambda v\phantom{\rule{0ex}{0ex}}But,\frac{d\varphi}{dt}=IR\phantom{\rule{0ex}{0ex}}Hence,\phantom{\rule{0ex}{0ex}}IR=\left(\pi {L}^{2}\right){B}_{o}\lambda v\phantom{\rule{0ex}{0ex}}I=\left[\frac{\left(\pi {L}^{2}\right){B}_{o}\lambda}{R}\right]v$

Energy lost per second:

$\frac{d\varphi}{dt}=\frac{d}{dt}\left(AB\right)\phantom{\rule{0ex}{0ex}}=A\frac{dB}{dt}\phantom{\rule{0ex}{0ex}}=A\frac{d}{dt}{B}_{o}\left(1+\lambda z\right)\phantom{\rule{0ex}{0ex}}=\left(\pi {L}^{2}\right){B}_{o}\lambda \frac{dz}{dt}\phantom{\rule{0ex}{0ex}}=\left(\pi {L}^{2}\right){B}_{o}\lambda v\phantom{\rule{0ex}{0ex}}But,\frac{d\varphi}{dt}=IR\phantom{\rule{0ex}{0ex}}Hence,\phantom{\rule{0ex}{0ex}}IR=\left(\pi {L}^{2}\right){B}_{o}\lambda v\phantom{\rule{0ex}{0ex}}I=\left[\frac{\left(\pi {L}^{2}\right){B}_{o}\lambda}{R}\right]v$

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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