JEE MAIN - Physics (2005 - No. 14)
A charged particle of mass $$m$$ and charge $$q$$ travels on a circular path of radius $$r$$ that is perpendicular to a magnetic field $$B.$$ The time taken by the particle to complete one revolution is
$${{2\pi {q^2}B} \over m}$$
$${{2\pi mq} \over B}$$
$${{2\pi m} \over {qB}}$$
$${{2\pi qB} \over m}$$
Explicació
Equating magnetic force to centripetal force.
$${{m{V^2}} \over r} = qvB\,\sin \,{90^ \circ }$$
Time to complete one revolution.
$$T = {{2\pi r} \over v} = {{2\pi m} \over {qB}}$$
$${{m{V^2}} \over r} = qvB\,\sin \,{90^ \circ }$$
Time to complete one revolution.
$$T = {{2\pi r} \over v} = {{2\pi m} \over {qB}}$$