A charge of \( 2\ \mu\mathrm{C} \) moves with a speed of \( 3\times10^{6}\ \mathrm{m\,s^{-1}} \) along the negative \( x \)-axis. A magnetic field \( \vec{B} \) of strength \( (0.2\hat{\jmath}+0.4\hat{k})\ \mathrm{T} \) exists in space. The magnetic force acting on the charge is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( (2.4\hat{\jmath}-1.2\hat{k}) \)
The magnetic force on a moving charge is the Lorentz force \( \vec{F} = q(\vec{v}\times\vec{B}) \); the cross product is worked out with \( \hat{\imath}\times\hat{\jmath} = \hat{k} \) and \( \hat{\imath}\times\hat{k} = -\hat{\jmath} \).
\( q = 2\times10^{-6}\,\mathrm{C},\ \vec{v} = -3\times10^{6}\,\hat{\imath}\ \mathrm{m\,s^{-1}},\ \vec{B} = (0.2\hat{\jmath}+0.4\hat{k})\,\mathrm{T} \)
\( \displaystyle \vec{v}\times\vec{B} = -3\times10^{6}\left[0.2(\hat{\imath}\times\hat{\jmath})+0.4(\hat{\imath}\times\hat{k})\right] \)
\( \displaystyle \vec{v}\times\vec{B} = -3\times10^{6}\left[0.2\,\hat{k}-0.4\,\hat{\jmath}\right] \)
\( \vec{v}\times\vec{B} = (1.2\times10^{6}\,\hat{\jmath}-0.6\times10^{6}\,\hat{k})\ \mathrm{T\,m\,s^{-1}} \)
\( \vec{F} = (2\times10^{-6})(1.2\times10^{6}\,\hat{\jmath}-0.6\times10^{6}\,\hat{k})\ \mathrm{N} \)
\( \vec{F} = (2.4\hat{\jmath}-1.2\hat{k})\ \mathrm{N} \)
Note: It seems that there is a mistake in the original NTA paper, which prints the options without a unit when they should read in newton, for example \( (2.4\hat{\jmath}-1.2\hat{k})\ \mathrm{N} \). All the data are in SI units, so the force comes out in newton and the question can still be solved.