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An electron ( = 9 xx 10^(-31) kg, e = 1....

An electron `( = 9 xx 10^(-31) kg, e = 1.6 xx 10^(-19))` is moving in a circular in a magnetic field of `1.0 xx 10^(4) Wb// m^(2)` . Its period of revolution is .

A

`7.0 xx 10^(-7) s`

B

`3.5 xx 10^(-7)s`

C

`1.05 xx 10^(-7) s `

D

`2.1 xx 10^(-7) s`

Text Solution

Verified by Experts

The correct Answer is:
B
`T = (2pi m)/(qB) = ( 2 xx 3.14 xx 9 xx 10^(-31))/(1.6 xx 10^(-19) xx 1 xx 10^(-4)) = 3.5 xx 10^(-7)s`
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Knowledge Check

  • 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`

  • An electron (mass =9.0xx10^(-31) kg and charge =1.6xx10^(-19) coulomb) is moving in a circular orbit in a magnetic field of 1.0xx10^(-4)"weber"//m^(2) . Its perido of revolution is

    A
    `3.5xx10^(-7)sec`
    B
    `7.0xx10^(-7)sec`
    C
    `1.05xx10^(-6)sec`
    D
    `2.1xx10^(-6)sec`

  • An electron (mass =9xx10^(-31)kg . Charge =1.6xx10^(-19)C) whose kinetic energy is 7.2xx10^(-18) joule is moving in a circular orbit in a magnetic field of 9xx10^(-5) "weber"//m . The radius of the orbit is

    A
    `1.25cm`
    B
    `2.5cm`
    C
    `12.5cm`
    D
    `25.0cm`

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