Question

An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6} ms ^{-1}$. It enters a region of uniform electric field of $300 V / cm$ 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)

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