An alpha particle moves through a uniform magnetic field of 1.5 T directed along the positive z-axis. What is the magnitude of the magnetic force on the alpha particle when its velocity is \( \vec{v}=(3\hat{\imath}+4\hat{\jmath}+5\hat{k})\times10^{4}\,\mathrm{m\,s^{-1}} \)? (Charge of an alpha particle \( =3.2\times10^{-19}\,\mathrm{C} \).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.4\times10^{-14}\,\mathrm{N} \)
The magnetic force on a moving charge is \( \vec{F}=q\,(\vec{v}\times\vec{B}) \). Only the part of the velocity perpendicular to the field contributes, so the \( \hat{k} \) component of \( \vec{v} \), along \( \vec{B} \), gives no force.
With \( \hat{\imath}\times\hat{k}=-\hat{\jmath} \), \( \hat{\jmath}\times\hat{k}=\hat{\imath} \) and \( \hat{k}\times\hat{k}=0 \), the cross product is:
\( \begin{aligned} \vec{v}\times\vec{B} &= (3\hat{\imath}+4\hat{\jmath}+5\hat{k})\times10^{4}\times1.5\,\hat{k} \\ &= 1.5\times10^{4}\,(3\,\hat{\imath}\times\hat{k}+4\,\hat{\jmath}\times\hat{k}) \\ &= 1.5\times10^{4}\,(4\hat{\imath}-3\hat{\jmath})\,\mathrm{T\,m\,s^{-1}} \end{aligned} \)
Its magnitude is:
\( \begin{aligned} |\vec{v}\times\vec{B}| &= 1.5\times10^{4}\times\sqrt{4^{2}+3^{2}}\,\mathrm{T\,m\,s^{-1}} \\ &= 7.5\times10^{4}\,\mathrm{T\,m\,s^{-1}} \end{aligned} \)
The magnitude of the force is:
\( \begin{aligned} F &= q\,|\vec{v}\times\vec{B}| \\ &= 3.2\times10^{-19}\,\mathrm{C}\times7.5\times10^{4}\,\mathrm{T\,m\,s^{-1}} \\ &= 2.4\times10^{-14}\,\mathrm{N} \end{aligned} \)