Gujarati
Hindi
4.Moving Charges and Magnetism
hard

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.

A

$\frac{{m\,\cos \,\theta }}{{qB}}$

B

$\frac{{m\,\cos ec\,\theta }}{{qB}}$

C

$\frac{{m\,\cot \,\theta }}{{qB}}$

D

none

Solution

$F=q V B$

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 $t$

$v=g \sin \theta t$

$\mathrm{t}=\frac{\mathrm{v}}{\mathrm{g} \sin \theta}=\frac{\mathrm{mg} \cot \theta}{\mathrm{qgB}}=\frac{\mathrm{mcot} \theta}{\mathrm{qB}}$

Standard 12
Physics

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