An electron revolves in a circle of radius 0.4 Å with a speed of `10^(5)m//s`. The magnitude of the magnetic field produced at the centre of the circular path due to the motion of the electron in `W//m^(2)` is

Options

A. 0.01
B. 10
C. 1
D. 0.005

Answer

C

Explanation

Given that
`r=0.4` Å=`0.4xx10^(-10)m`
Speed `v=10^(5)m//s` `l=(ev)/(2pir)`
`thereforeB=(mu_(0)Nl)/(2r)=(mu_(0)N)/(2r)xx(ev)/(2pir)`
or `B=(mu_(0)Nev)/(4pir^(2))`
or `B=(10^(-7)xx1xx1.6xx10^(-19)xx10^(5))/(0.16xx10^(-20))`
or `B=(1.6xx10^(-21))/(0.16xx10^(-20))`
`=(0.16)/(0.16)=1Wb//m^(2)`