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(D) The magnetic force on a current-carrying wire is given by $\overrightarrow{F} = I(\vec{L} 	imes \overrightarrow{B})$.Since the wire is along the

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$5$

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(D) The magnetic force on a current-carrying wire is given by $\overrightarrow{F} = I(\vec{L} 	imes \overrightarrow{B})$.Since the wire is along the positive $x$-axis,its length vector is $\vec{L} = L\hat{i}$.Given $\overrightarrow{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) 	ext{ T}$.Substituting these into the formula:$\overrightarrow{F} = I [ (L\hat{i}) 	imes (2\hat{i} + 3\hat{j} - 4\hat{k}) ]$$\overrightarrow{F} = IL [ (\hat{i} 	imes 2\hat{i}) + (\hat{i} 	imes 3\hat{j}) + (\hat{i} 	imes -4\hat{k}) ]$Using cross product rules ($\hat{i} 	imes \hat{i} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,$\hat{i} 	imes \hat{k} = -\hat{j}$):$\overrightarrow{F} = IL [ 0 + 3\hat{k} - 4(-\hat{j}) ]$$\overrightarrow{F} = IL (4\hat{j} + 3\hat{k})$.The magnitude of the force is $|\overrightarrow{F}| = IL \sqrt{4^2 + 3^2} = IL \sqrt{16 + 9} = IL \sqrt{25} = 5IL$.Thus,the magnitude is $5IL$.

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

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