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An electron (=9xx10^(-31)kg,e=1.6xx10^(-...

An electron `(=9xx10^(-31)kg,e=1.6xx10^(-19))` is moving in a circular in a magnetic field of `1.0xx10^(-7)Wb//m^(2)`. Its period of revolution is

A

`7.0xx10^(-7)s`

B

`3.5xx10^(-7)s`

C

`1.05xx10^(-7)s`

D

`2.1xx10^(-7)s`

Text Solution

Verified by Experts

The correct Answer is:
B
`T=(2pim)/(qB)=(2xx3.14xx9xx10^(-31))/(1.6xx10^(-19)xx1xx10^(-4))=35xx10^(-7)s`
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