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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:
A
$\sqrt{3}\,IL$
B
$3\,IL$
C
$\sqrt{5}\,IL$
D
$5\,IL$
Explanation
$\vec{F}=I(\vec{L}\times\vec{B})$ has magnitude $5IL$.
Detailed Solution
$|\vec{F}|=|I(\vec{L}\times\vec{B})|$
$=|I[L\hat{i}\times(2\hat{i}+3\hat{j}-4\hat{k})]|$
$=5\,IL$