JEE Main · 2021 · 26 Aug 2021 · Shift 2 · EM Waves

A light beam is described by

E = 800 sin ω (t - \frac{x}{c})

. An electron is allowed to move normal to the propagation of light beam with a speed of

3 \times 1 0^{7} ms^{- 1}

. What is the maximum magnetic force exerted on the electron?

Answer: (D)

Solution

In an electromagnetic wave the field amplitudes are related by

B_{0} = \frac{E _{0}}{c}

, so with

E_{0} = 800 V m^{- 1}

B_{0} = \frac{800}{3 \times 1 0 ^{8}} T

The electron travels perpendicular to the direction of propagation, so its velocity can be perpendicular to

B

and the magnetic force reaches

F = e v B_{0} = (1.6 \times 1 0^{- 19}) (3 \times 1 0^{7}) \frac{800}{3 \times 1 0 ^{8}} = 12.8 \times 1 0^{- 18} N

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option