A proton goes undeflected in a crossed electric 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

`m_(p) = 1.6 xx 10^(-27) kg `,
`v = 2 xx 10^(5) m/s`
`r = 4 cm = 4 xx 10^(-2) m`
Since the proton is not deflected in the combine magnetic and electric field , hence force a circle due to both field must be same.
i. e. `qE = qvB rArr E = vB`
Now know `r = (mu)/(qB)`
`rArr B (mu)/(qr)`
`B = 0.5 xx 10^(-1) = 0.05 T`
`E = uB = 2 xx 10^(5) xx 0.05`
`= 1 xx 10^(6) N/c` .