A particle of charge \( -16\times10^{-18}\,\mathrm{C} \) moving with a velocity of \( 10^{6}\,\mathrm{m\,s^{-1}} \) along the x-axis enters a region where a magnetic field B is along the y-axis and an electric field of magnitude \( 10^{4}\,\mathrm{V\,m^{-1}} \) is along the negative z-axis. If the particle continues to move along the x-axis, the magnitude of B is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 10^{-2}\,\mathrm{T} \)
The particle keeps moving along the x-axis only if the total Lorentz force \( q\,(\vec{E}+\vec{v}\times\vec{B}) \) on it is zero. So the electric and magnetic forces must be equal and opposite.
The charge is negative and \( \vec{E} \) is along −z, so the electric force \( q\vec{E} \) is along +z.
With \( \vec{v} \) along +x and \( \vec{B} \) along +y, \( \hat{\imath}\times\hat{\jmath}=\hat{k} \). For the negative charge, the magnetic force \( q\,\vec{v}\times\vec{B} \) is along −z, opposite to the electric force.
Equating the magnitudes of the two forces:
\( |q|E=|q|vB \)
\( \begin{aligned} B &= \frac{E}{v} \\ &= \frac{10^{4}\,\mathrm{V\,m^{-1}}}{10^{6}\,\mathrm{m\,s^{-1}}} \\ &= 10^{-2}\,\mathrm{T} \end{aligned} \)