Question

A light beam travelling in the $x-$ direction is described by the electric field $E_y=\left(300{\text{Vm}}^{-1}\right)\sin \omega (t-x/c)$. An electron is constrained to move along the $y-$direction with a speed of $2.0\times {10}^7{\text{ms}}^{-1}$. If the maximum electric force acting on the electron is $F_E$ and the maximum magnetic force acting on the electron is $F_B$, then

• A. $F_E=1.60\times {10}^{-17}\text{N}$
• B. $F_E=4.8\times {10}^{-17}\text{N}$
• C. $F_B=1.6\times {10}^{-18}\text{N}$
• D. $F_E>F_B$

Answer 1

Answer: (B), (D)

$F_E>F_B$

The maximum value of electric field is
$E_0=300{\text{Vm}}^{-1}$.

The maximum electric force on the electron is

$F_E=e{E}_0$

$F_E=(1.6\times {10}^{-19}\text{C})\left(300{\text{Vm}}^{-1}\right)$

$\Rightarrow F_E=4.8\times {10}^{-17}\text{N}$

The maximum value of magnetic field is
$B_0=\frac{E_0}{c}=\frac{300}{3\times {10}^8}={10}^{-6}\text{T}$
along the $z-$direction.

The maximum magnetic force on the electron is

$F_B=\left|e\right(\overrightarrow{v}\times \overrightarrow{B}\left)\right|=ev{B}_0$

$=(1.6\times {10}^{-19}\text{C})\times (2.0\times {10}^7{\text{ms}}^{-1})\times \left({10}^{-6}\text{T}\right)$

$=3.2\times {10}^{-18}\text{N}$

Therefore, $F_E>F_B$.

Hence, options $\left(B\right)$ and $\left(D\right)$ are the correct answers.