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.

Explanation

Since proton goes undeflected and `vec(E), vec(B)` and `v` are mutually perpendicular i.e. net force on proton is zero.

`F_(m) = F_(e)`
`Bev = eE rArr v = (E)/(B)` …(i)
When electric field is absent, proton moves in a circle of radius `R = 4.0 cm`
`R = (mv)/(Bq) = (mv)/(Be)`
`B = (mv)/(Re) = (1.67 xx 10^(-27) xx 2 xx 10^(5))/(4 xx 10^(-2) xx 1.6 xx 10^(-19)) ~= 0.05T`
`v = (E)/(B)`
`E = Bv = 0.05 xx 2 xx 10^(5) = 10^(4) N//C`