Q 12-04-008NEETNEET 2023Top questionEasy
A wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k})$ T. The magnitude of the magnetic force acting on the wire is :
Answer: (C) $5\,IL$
$$\vec{F} = I\vec{L} \times \vec{B} = IL\,\hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k}) = IL(3\hat{k} + 4\hat{j})$$
(using $\hat{i}\times\hat{i} = 0$, $\hat{i}\times\hat{j} = \hat{k}$, $\hat{i}\times\hat{k} = -\hat{j}$).
$$|\vec{F}| = IL\sqrt{3^2 + 4^2} = 5\,IL$$
(The component of $\vec{B}$ along the wire, $2\hat{i}$, exerts no force.)
Solution by Sreeraj P, M.Sc Physics