Question

An electron beam passes through a magnetic field of `4xx10^-3 weber//m^2` and an electric field of `2xx10^4Vm^-1`, both acting simultaneously. The path of electron remaining undeviated, calcualte the speed of the electrons. If the electric field is removed, what will be the radius of the electron path?

Answer 1

Here, `B=4xx10^-3weber//m^2`:
`E=2xx10^4V//m`.
As the path of moving electron is undeviated, so force on moving electron due to electric field is equal and opposite to the force on moving electron due to magnetic field, i.e., `eE=evB`,
or `v=E/B=(2xx10^4)/(4xx10^-3)=5xx10^6m//s`
When electron moves perpendicular to magnetic field, the radius r of circualr path traced by electron is
`r=(mv)/(eB)=((9*1xx10^(-31))xx(5xx10^6))/((1*6xx10^(-19))xx4xx10^(-3))`