Question

A particle of charge $16\times 10^{-16}C$ moving with velocity $10m{s}^{-1}$ along $x$-axis enters a region where magnetic field of induction
$\vec{B}$ is along the $y$-axis and an electric field of magnitude $10^{4}V{m}^{-1}$ is along the negative z-axis.
If the charged particle continues moving along $x$-axis, the magnitude of $\vec{B}$ is :

• A. $16\times 10^{3}Wb{m}^{-2}$
• B. $2\times 10^{3}Wb{m}^{-2}$
• C. $1\times 10^{3}Wb{m}^{-2}$
• D. $4\times 10^{3}Wb{m}^{-2}$

Answer 1

Answer: (C)

Since particle is moving undeflected
So, $q_E=qvB$
$\Rightarrow B=\frac{E}{V}=\frac{10^4}{10}=10^{3}wb/m^2$