An electron (mass $9 \times 10^{-31}$ kg and charge $1.6 \times 10^{-19}$ C) moving with speed $c/100$ ($c$ = speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4}$ T perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^8\ \text{m s}^{-1}$)
Answer: (B) $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^2\ \text{V m}^{-1}$
For no deflection, the electric force must cancel the magnetic force: $q\vec{E} = -q\vec{v} \times \vec{B}$. The magnetic force is perpendicular to $\vec{B}$, so $\vec{E}$ must be perpendicular to $\vec{B}$ (and to $\vec{v}$).
Magnitude: $eE = evB$, so
$$E = vB = \frac{3 \times 10^8}{100} \times 9 \times 10^{-4} = 3 \times 10^6 \times 9 \times 10^{-4} = 27 \times 10^2\ \text{V m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics