Q. 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 $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^8$ ms$^{-1}$)
A) $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^4$ V $m^{-1}$
B) $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^2$ V m$^{-1}$
C) $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^2$ V m$^{-1}$
D) $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^4$ V m$^{-1}$
Explanation
[Solution by Claude - not from source] No deflection needs $qE=qvB$ with $E\perp B$ (and $\perp v$). $v=3\times10^6$ m/s, so $E=vB=3\times10^6\times9\times10^{-4}=2.7\times10^3=27\times10^2$ V/m. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.