Question

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

• A. $F_E=4.8\times {10}^{-17}N$ and $F_B=3.2\times {10}^{-18}N$
• B. $F_E=8.4\times {10}^{-18}N$ and $F_B=3.2\times {10}^{-18}N$
• C. $F_E=3.2\times {10}^{-17}N$ and $F_B=4.8\times {10}^{-18}N$
• D. $F_E=4.8\times {10}^{-17}N$ and $F_B=6.4\times {10}^{-17}N$

Answer 1

The correct option is A $F_E=4.8\times {10}^{-17}N$ and $F_B=3.2\times {10}^{-18}N$

We are given that
$E=E_0\sin \left(kx-\omega t\right)$
We get
$E_0=300{\text{Vm}}^{-1}$
For maximum force ${}^{\prime }{\sin }^{\prime }$ will acquire its maximum value as 1.
$F_E=q{E}_0\Rightarrow F_E=1.6\times {10}^{-19}\times 300=4.8\times {10}^{-17}N$
Also magnetic field $B_0$ can be given as
$B_0=\frac{E_0}{c}$
where $c$ is the speed of light
$B_0=\frac{300}{3\times {10}^8}={10}^{-6}\text{T}$
$F_B=|q.\overrightarrow{v}\times \overrightarrow{B}|$
$F_B=qv{B}_0=1.6\times {10}^{-19}\times 2\times {10}^7\times {10}^{-6}$
$F_B=3.2\times {10}^{-18}\text{N}$