Question

In a chamber, a uniform magnetic field of 6.5 G `(1G=10^(-4)T)` is maintained. An electron is shot into the field with a speed of `4.8xx10^(6)m s^(-1)` normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
`(e=1.5xx10^(-19)C, m_(e)=9.1xx10^(-31)Kg)`

Answer 1

Given magnetic field `B=6.5 G =6.5 xx10^(-4)T`
Charge `e=-1.6xx10^(-19)C`
Speed of electron `V=4.8xx10^(6) m//s`
Mass of electron `m_(e)=9.1xx10^(-31) kg`
Angle between magnetic field and electron `(theta)=90^(@)`
The force on charge particle entering in the magnetic field
`F=q(V xx B)=e (V xxb)`
The electron attains a ciruclar path and necessary centripetal force is provided by magnetic force.
`e(V xxB)=(mV^(2))/(r )`
`e V B sin 90^(@)=(m V^(2))/(r )`
`r=(m V)/(e Bxx1)=(9.1xx10^(-31)xx4.8xx10^(6))/(1.6xx10^(-19)xx6.5xx10^(-4))=4.2xx10^(-2)m=4.2 cm`