A charged particle is projected with velocity \( \vec{v} = (2\times10^{6}\ \frac{\mathrm{m}}{\mathrm{s}})\hat{\imath} \) into a region in which magnetic field \( \vec{B} = [(0.2\ \mathrm{T})\hat{\imath} + (0.5\ \mathrm{T})\hat{\jmath}] \) exists. If the charge to mass ratio for the charged particle is \( 4.8\times10^{7}\ \mathrm{C\,kg^{-1}} \), its acceleration (in \( \mathrm{m\,s^{-2}} \)) will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4.8\times10^{13}\hat{k} \)
The magnetic Lorentz force is \( \vec{F} = q(\vec{v}\times\vec{B}) \), so the acceleration is \( \vec{a} = (q/m)(\vec{v}\times\vec{B}) \). The component of \( \vec{B} \) parallel to \( \vec{v} \) gives no force.
\( \vec{v}\times\vec{B} = (2\times10^{6}\,\hat{\imath})\times(0.2\,\hat{\imath} + 0.5\,\hat{\jmath}) \)
\( \vec{v}\times\vec{B} = 0 + (2\times10^{6})(0.5)(\hat{\imath}\times\hat{\jmath}) = 10^{6}\,\hat{k} \)
\( \vec{a} = (4.8\times10^{7}\ \mathrm{C\,kg^{-1}})(10^{6}\,\hat{k}) = 4.8\times10^{13}\,\hat{k}\ \mathrm{m\,s^{-2}} \)