$A$ 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 the magnetic field is $B_z = B_0(1 + \lambda z)$. If $R$ is the resistance of the ring and the ring falls with a velocity $v$,find the energy lost in the resistance per unit time. If the ring has reached a constant velocity,use the conservation of energy to determine $v$ in terms of $m, B_0, l, \lambda, R$ and acceleration due to gravity $g$.
(D) Magnetic flux linked with the ring is given by $\phi = B_z A = B_0(1 + \lambda z) \pi l^2$.
According to Faraday's law,the induced emf is $\varepsilon = |\frac{d\phi}{dt}| = |\frac{d}{dt} [B_0(1 + \lambda z) \pi l^2]| = B_0 \pi l^2 \lambda \frac{dz}{dt} = B_0 \pi l^2 \lambda v$.
The induced current is $I = \frac{\varepsilon}{R} = \frac{B_0 \pi l^2 \lambda v}{R}$.
The energy lost in the resistance per unit time (power dissipated) is $H = I^2 R = (\frac{B_0 \pi l^2 \lambda v}{R})^2 R = \frac{B_0^2 \pi^2 l^4 \lambda^2 v^2}{R}$.
When the ring reaches a constant terminal velocity $v$,the rate of loss of gravitational potential energy equals the rate of heat dissipation: $mgv = H$.
Substituting $H$,we get $mgv = \frac{B_0^2 \pi^2 l^4 \lambda^2 v^2}{R}$.
Solving for $v$,we get $v = \frac{mgR}{B_0^2 \pi^2 l^4 \lambda^2}$.
According to Faraday's law,the induced emf is $\varepsilon = |\frac{d\phi}{dt}| = |\frac{d}{dt} [B_0(1 + \lambda z) \pi l^2]| = B_0 \pi l^2 \lambda \frac{dz}{dt} = B_0 \pi l^2 \lambda v$.
The induced current is $I = \frac{\varepsilon}{R} = \frac{B_0 \pi l^2 \lambda v}{R}$.
The energy lost in the resistance per unit time (power dissipated) is $H = I^2 R = (\frac{B_0 \pi l^2 \lambda v}{R})^2 R = \frac{B_0^2 \pi^2 l^4 \lambda^2 v^2}{R}$.
When the ring reaches a constant terminal velocity $v$,the rate of loss of gravitational potential energy equals the rate of heat dissipation: $mgv = H$.
Substituting $H$,we get $mgv = \frac{B_0^2 \pi^2 l^4 \lambda^2 v^2}{R}$.
Solving for $v$,we get $v = \frac{mgR}{B_0^2 \pi^2 l^4 \lambda^2}$.
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