Question

A particle of charge $q=16\times {10}^{-18}$ C moving with $10m{s}^{-1}$ along $x-$axis enter a magnetic field of induction $B$ along the $y-$axis and an electric field ${10}^4$ Vm${}^{-1}$ along negative $z-$direction. If the particle continues to move along $x-$axis then the strength of magnetic field is

• A. ${10}^3$ Wbm${}^{-2}$
• B. ${10}^5$ Wbm${}^{-2}$
• C. ${10}^{-3}$ Wbm${}^{-2}$
• D. ${10}^{16}$ Wbm${}^{-2}$

Answer 1

Answer: (A)

The particle moves along $x-$axis undeflected.
Magnetic force: $\overrightarrow{F_{mag}}=q(\overrightarrow{v}\times \overrightarrow{B})=q(10\hat{i}\times \overrightarrow{B})$
Electric force: $\overrightarrow{F_e}=q\overrightarrow{E}=-q\times {10}^4\hat{k}$
$\overrightarrow{F_{mag}}+\overrightarrow{F_e}=0$
$\Rightarrow q(10\hat{i}\times \overrightarrow{B})=q\times {10}^4\hat{k}$
$\Rightarrow \overrightarrow{B}={10}^3\hat{j}Wb{m}^{-2}$