An electron beam travels with a velocity of `1.6xx10^(7)ms^(-1)` perpendicular to magnetic field of intensity 0.1 T. The radius of the path of the electron beam `(m_(e)=9xx10^(-31)kg)`

Options

A. `9xx10^(-5)m`
B. `9xx10^(-2)m`
C. `9xx10^(-4)m`
D. `9xx10^(-3)m`

Answer

C

Explanation

The radius of circular path travelled by electron beam in a perpendicular magnetic field is,
`r=(mv)/(eB)`
`m=9xx10^(-31)kg, v=1.6xx10^(7)ms^(-1), e=1.6xx10^(-19),
B=0.1` (Given)
`r=(9xx10^(-31)xx1.6xx10^(7))/(1.6xx10^(-19)xx0.1)=9xx10^(-4)m.`