In a region of space having uniform magnetic field \( \vec{B} = (3\hat{\imath} + 6\hat{\jmath} - 3\hat{k})\ \mathrm{T} \), a charge of \( 3\ \mu\mathrm{C} \) experiences a force \( \vec{F} = (x\hat{\imath} - 2\hat{\jmath} + 3\hat{k})\ \mathrm{N} \). The value of \( x \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 7
The magnetic force on a moving charge is \( \vec{F} = q(\vec{v}\times\vec{B}) \). A cross product is perpendicular to each of its factors, so \( \vec{F} \) must be perpendicular to \( \vec{B} \):
\( \vec{F}\cdot\vec{B} = 0 \)
\( (x)(3) + (-2)(6) + (3)(-3) = 0 \)
\( 3x - 12 - 9 = 0 \)
\( x = 7 \)
The charge (3 μC) and the unknown velocity are not needed. The perpendicularity condition alone fixes \( x \).