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An electron in a circular orbit of radiu...

An electron in a circular orbit of radius 0.05nm performs `10^(16)` revolutions per second. The magnetic moment due to this rotation of electron is (in A . `m^(2)` )

A

`(BR^(3))/(2pimu_(0))`

B

`(2pi^(2)BR^(3))/(mu_(0))`

C

`(BR^(2))/(2pimu_(0))`

D

`(2pi BR^(2))/(mu_(0))`

Text Solution

Verified by Experts

The correct Answer is:
C
r=0.05 nm =`5xx10^(-2) xx10^(-9)=5xx10^(-11)m`
`e=1.6 xx10^(-19)` and i=ne
Magnetic moment of the electron
IA
`= ne.pi^(2)`
`=125xx10^(-25)`
`=1.26 xx10^(-23) Am^(2)`
[`i=(q )/(t)` for the electron q =e and t= T and `(1)/(T )=n`]
`therefore i=(e )/(T)=ne`
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