A block of mass $m$ $\&$ charge $q$ is released on a long smooth inclined plane magnetic field $B$ is constant, uniform, horizontal and parallel to surface as shown. Find the time from start when block loses contact with the surface.
Diffcult
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$F=q V B$
Particle will leave the inclined plane when
$\mathrm{F}=\mathrm{mg} \cos \theta$
$\Rightarrow \mathrm{qvB}=\mathrm{mg} \cos \theta$
$v=\frac{m g \cos \theta}{q B}$
Time taken to reach $v$ is $t$
$v=g \sin \theta t$
$\mathrm{t}=\frac{\mathrm{v}}{\mathrm{g} \sin \theta}=\frac{\mathrm{mg} \cot \theta}{\mathrm{qgB}}=\frac{\mathrm{mcot} \theta}{\mathrm{qB}}$
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