A beam of proton passes undeflected with a horizontal velocity v, through a region of electric and magnetic fields, mutually perpendicular to each other and perpendicular to the direction of the beam. If the magnitudes of the electric and magnetic fields are `100kV//m`, `50mT` respectively, calculate the velocity of the beam v.

Answer

`2xx10^ms^-1`

Explanation

Here, `E=100kV//m=100xx10^3V//m=10^5V//m`,
`B=50mT=50xx10^-3T`
`v=E/B=(10^5)/(50xx10^-3)=2xx10^6ms^-1`