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A charged particle of mass m and charge q moves along 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

A

`(2pimq)/B`

B

`(2piq^(2)B)/m`

C

`(2piqB)/m`

D

`(2pim)/(qB)`

Text Solution

Verified by Experts

The correct Answer is:
D

Magnetic force `F=qvB [ :' theta=90^(@)]`…..i
Centripetal force `F=(mv^(2))/r`………….ii
Fromeqs I and ii we have
`(mv^(2))/r=qvB` or `r=(mv)/(qB)`
The time taken by the particle to complete one revolution,
`T=(2pir)/V=(2pim)/(vqB)=(2pim)/(qB)`
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