A proton goes undflected in a crossed and magnetic field (the fields are perpendicular to each other) at a speed of `2.0X10^5 ms^(-1).` The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and the magnetic fields. take the mass of the proton `=1.6 X10^(-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`