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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
$5 \times 10^{-3} T ,$ along $+ z$ direction
$3 \times 10^{-4} T ,$ along $- z$ direction
$3 \times 10^{-4} T ,$ along $+ z$ direction
$5 \times 10^{-3} T ,$ along $- z$ direction
Solution
$\overrightarrow{ B }$ must be in $+ z$ axis.
$\overrightarrow{ V }=6 \times 10^{6} \hat{ i }$
$\overrightarrow{ E }=300 \hat{ j } V / cm =3 \times 10^{4} V / m$
$q \vec{ E }+q \vec{ V } \times \vec{ B }=0$
$E = VB$
$B =\frac{ E }{ V }=\frac{3 \times 10^{4}}{6 \times 10^{6}}=5 \times 10^{-3} T$
