A wire carrying a current $I$ along the positive x-axis has length $L$. It is kept in a magnetic field $\vec{B

Question

A wire carrying a current $I$ along the positive x-axis has length $L$. It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) 	ext{ T}$. The magnitude of the magnetic force acting on the wire is

Accepted Answer

The magnetic force on a current-carrying wire is $\vec{F} = I(\vec{L} 	imes \vec{B})$. Here $\vec{L} = L\hat{i}$. So $\vec{F} = I(L\hat{i} 	imes (2\hat{i} + 3\hat{j} - 4\hat{k})) = IL(2(\hat{i}	imes\hat{i}) + 3(\hat{i}	imes\hat{j}) - 4(\hat{i}	imes\hat{k})) = IL(0 + 3\hat{k} - 4(-\hat{j})) = IL(4\hat{j} + 3\hat{k})$. The magnitude is $|\vec{F}| = IL \sqrt{4^2 + 3^2} = IL \sqrt{16+9} = 5IL$.

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