A light beam travelling in the x- direction is described by the electric field `E_y=(300V (m^-1) sin omega (t-(x//c))`. An electron is constrained to move along the y-direction with a speed `(2.0 xx (10^7) m (s^-1))`. Find the maximum electric force and the maximum magnetic force on the electron.

Explanation

Solution:
The maximum electric field is `E_0=300V(m^-1)`. The
maximum magnetic field is
` B_0= ((E_0)/c) = (300V(m^-1)/(3 xx (10^8)m(s^-1))) = (10^-6)T.`
along the z-direction.
The maximum electric force on the electron is
` F_e=(q(E_0))=(1.6 xx (10^-19)C) xx (300 V (m^-1)).`
` =4.8 xx (10^-17) N.`
The maximum magnetic force on the electron is
` F_b=(|(vec qv) xx (vec B)|^max) = qv B_0`
` =(1.6 xx (10^-19)C) xx (2.0 xx (10^7)m (s^-1)) xx ((10^-8)T)`
` 3.2 xx (10^-18)N.`