A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of `10^(5)m//s`.The velocity is perpendicular to both the fields.When the electric field is swithed off.the proton moves along a circle of radius `2 cm`.Find the magnitudes of electric and the magnitic fields.Take the mass of the proton`=1.6xx10^(-27) kg`.

Answer

A, B, C

Explanation

`V=E/B`
`r=(mV)/(qB)`
`B=(mV)/(rq)`
`=(1.6xx10^(-27)xx2xx10^(5))/(4xx10^(-2)xx1.6xx10^(-19))=5x10^(-2)T`
`E=VB=5xx10^(3)N//C`