Question

An electron is moving along $+x$ direction with a velocity of $6\times 10^{6}m{s}^{-1}$. It enters a region of uniform electric field of $300V/cm$ pointing along $+y$ 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. $5\times 10^{-3}T$, along $+z$ direction
• B. $3\times 10^{-4}T$, along $-z$ direction
• C. $3\times 10^{-4}T$, along $+z$ direction
• D. $5\times 10^{-3}T$, along $-z$ direction

Answer 1

Answer: (A)

$\vec{V}=6\times 10^6\hat{i}$
$\vec{E}=300\hat{j}V/cm=3\times 10^{4}V/m$
$q\vec{E}+q\vec{V}\times \vec{B}=0$
$E=VB$
$B=\frac{E}{V}=\frac{3\times 10^4}{6\times 10^6}=5\times 10^{-3}T$