Question

An electron is moving along positive $x$ direction with a velocity of $6\times {10}^{6}m{s}^{-1}$. It enters a region of the uniform electric field of $300V/cm$ pointing along positive $y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$-direction will be

• A. $3\times {10}^{-4}T$, along $–z$ direction
• B. $5\times {10}^{-3}T$, along $–z$ direction
• C. $5\times {10}^{-3}T$, along $+z$ direction
• D. $3\times {10}^{-4}T$, along $+z$ direction

Answer 1

The correct option is C

$5\times {10}^{-3}T$, along $+z$ direction

Step 1: Given data:

Velocity, $\overrightarrow{v}=6\times {10}^6\left(\hat{i}\right)m/s$

Electric field, $$\begin{gathered} E=300Vc{m}^{-1} \\ E=3\times {10}^{4}V{m}^{-1} \end{gathered}$$

Step 2: Calculating magnetic field

Magnetic force, ${\overrightarrow{F}}_B=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$ And, [$q$ is charge, $B$ is magnetic field]

Electric force, $\overrightarrow{F_E}=q\overrightarrow{E}$

Now, equating magnitudes of electric field and magnetic filed.

Now, $\overrightarrow{F_B}=\overrightarrow{F_E}$

$\begin{array}{rcl}& \Rightarrow & q\left(v\times B\right)=qE\\ & \Rightarrow & 6\times {10}^6\times B=3\times {10}^4\\ \therefore B& =& 5\times {10}^{-3}T\end{array}$

Therefore, the value of the magnetic field, $\begin{array}{rcl}B& =& 5\times {10}^{-3}T\end{array}$ along $\mathbf{+}\mathit{z}$ direction.

Hence, correct option is $\left(C\right)$.