The expression of electric field for a light beam travelling in the X- direction is
`E= 250 sin omega ( t - (x)/(c ) ) V//m`
A proton is travelling along the y-direction with a speed of `2.2 xx 10^7 m//s ` calculate the maximum electric force and maximum magnetic force on the electron .

Explanation

maximum manitude of electric field ,
`E_0 = 250 V//m`
maximum magnitude of magnetic field
`B_0 = (E_0 )/(c ) = ( 250 )/( 3xx 10^(8))`
` = 8.3 xx 10^(-7) t,` along the Z- direction
maximum electric force on the proton is
` F_p = q E_0 `
` = 1.6 xx 10^(-19) xx 250 `
`= 4.0 xx 10^(-17 ) N `
maximum magnetic force on the proton is
` F_m = qv B_0`
`= 1.6xx 10^(-19) xx 2.2 xx 10^(-7) xx 8.3 xx 10^(-7)`
`= 2.92 xx 10^(-18 ) N`