A charge q moves with a velocity of \( 2\,\mathrm{m\,s^{-1}} \) along x-axis in a uniform magnetic field \( \vec{B} = \hat{\imath} + 2\hat{\jmath} + 3\hat{k} \), then charge will experience a force.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → In z-y plane
The magnetic force on a moving charge is \( \vec{F} = q\,(\vec{v}\times\vec{B}) \), which is perpendicular to both \( \vec{v} \) and \( \vec{B} \).
\( \vec{v} = 2\hat{\imath} \)
\( \displaystyle \vec{v}\times\vec{B} = 2\hat{\imath}\times(\hat{\imath} + 2\hat{\jmath} + 3\hat{k}) = 0 + 4\hat{k} - 6\hat{\jmath} \)
\( \displaystyle \vec{F} = q\,(-6\hat{\jmath} + 4\hat{k}) \)
The force has no x-component (it must be perpendicular to \( \vec{v} \)), but it has both a y- and a z-component. So it lies in the y-z plane, whatever the sign of \( q \).
NCERT (2026-27):
Magnetic Field, Lorentz Force – “… (ii) The magnetic force q [ v × B ] includes a vector product of velocity and magnetic field. … The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field.”