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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

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\frac{2 π m}{qB}

\frac{2 π mq}{B}

\frac{2 π qB}{m}

\frac{2 π q^{2} B}{m}

answer is B.

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Detailed Solution

T = \frac{2 π r}{v}

$(∵ r = \frac{mv}{Bq})$

$T = \frac{2 π (mv)}{v Bq}$

$T = \frac{2 π m}{Bq}$

$∴ T = \frac{2 π m}{Bq}$

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