Question

A metallic ring of mass $m$ and radius $l$ (ring being horizontal) is falling under gravity in a region having field is $B_z=B_0(1+\lambda z)$. 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 of determine $v$ in terms of $m,B,\lambda$ and acceleration due to gravity $g$.

Answer 1

Magnetic flux across the ring of mass ${}^{\prime }{m}^{\prime }$

and radius ${}^{\prime }{l}^{\prime }$ falling under gravity in a region of having magnetic field

$B_z=B_0(1+\lambda z)$ is

$\varphi ={\overrightarrow{B}}_z.\overrightarrow{A}=B_0(1+\lambda z).\pi l^2$

The angle between

$\overrightarrow{B}$ and $\overrightarrow{A}$ is $0^{\circ }$

$ϵ=\frac{d}{dt}\left[B_0\right(1+\lambda z\left)\right]\pi l^2$

$IR=(B_0\pi l{)}^2[0+\lambda \frac{dz}{dt}]$

$I=\frac{B_0\pi \lambda l^2}{R}\frac{dz}{dt}=\frac{B_0\pi \lambda l^2}{R}v$

Energy lost $=I^{2}R=\frac{B_0^2{\pi }^2{\lambda }^2{l}^4}{R^2}{v}^2.R$

Energy lost $=\frac{B_0^2{\pi }^2{\lambda }^2{l}^4{v}^2}{R}$

The energy must come from decrease in

$P.E.=mg\frac{dz}{dt}=mgv$

$\therefore mgv=\frac{B_0^2{\pi }^2{\lambda }^2{v}^2{l}^4}{R}$

$v=\frac{mgR}{B_0^2{\pi }^2{\lambda }^2{l}^4}$ or

$v=\frac{mgR}{(\pi l^2\lambda B_0{)}^2}$

It is the required relation.