Question

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

• A. $4.8\times {10}^{-17}\text{N},3.2\times {10}^{-18}\text{N}$
• B. $9.6\times {10}^{-17}\text{N},6.4\times {10}^{-18}\text{N}$
• C. $2.4\times {10}^{-17}\text{N},1.6\times {10}^{-18}\text{N}$
• D. $3.6\times {10}^{-17}\text{N},2.5\times {10}^{-18}\text{N}$

Answer 1

The correct option is A $4.8\times {10}^{-17}\text{N},3.2\times {10}^{-18}\text{N}$

Given,
$E_y=300\sin \omega [t-\frac{x}{c}]$
$E_0=300V/m$
Force due to electric field,
$F_E=qE=eE=1.6\times {10}^{-19}\times 300=4.8\times {10}^{-17}\text{N}$
As we know that,
$B_0=\frac{E_0}{c}$
where,
$B_0$ = Max. magnetic field
$E_0$ = Max. electric field
$c=$ speed of light
It is given that,
$E_0=300$
So, $B_o=\frac{300}{3\times {10}^8}={10}^{-6}T$
Now, maximum magnetic force = $qv{B}_0$
$=1.6\times {10}^{-19}\times 2\times {10}^7\times {10}^{-6}$
$=3.2\times {10}^{-18}\text{N}$