NEET (UG) 2023 — Question 46

The magnetic force on a current-carrying wire is given by $\vec{F} = I \vec{L} \times \vec{B}$. The length vector is along the $x$-axis: $\vec{L} = L \hat{i}$. The magnetic field is $\vec{B} = 2\hat{i} + 3\hat{j} - 4\hat{k}$. Compute the cross product: $\vec{F} = IL \hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k}) = IL (0 \hat{i} + 3\hat{k} + 4\hat{j}) = IL (3\hat{k} + 4\hat{j})$. The magnitude is $|\vec{F}| = IL \sqrt{3^2 + 4^2} = IL \sqrt{25} = 5IL$. Thus the answer is $5IL$ (option B).