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 takeen by the particle to complete one revolution is
A
`( 2 piq^(2)B)/(m)`
B
`( 2 pi m q)/(B)`
C
`( 2 pi m)/(q ^(B))`
D
`( 2 piqB)/(m)`
Text Solution
Verified by Experts
The correct Answer is:
C
Equating magnetic force to centripetal force, `(mv^(2))/ ( r ) = qvB sin 90^(@)` Time to complete one revolution, `T = ( 2pir)/(v) = ( 2 pi m)/(qB)`
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