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(C) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{\ell} 	imes \vec{B})$.Since the wire is along the $x$-axis,its length

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$\sqrt{13} I \ell B$

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(C) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{\ell} 	imes \vec{B})$.Since the wire is along the $x$-axis,its length vector is $\vec{\ell} = \ell \hat{i}$.Given $\vec{B} = B(\hat{i} + 2\hat{j} - 3\hat{k})$.Substituting these into the formula: $\vec{F} = I(\ell \hat{i}) 	imes B(\hat{i} + 2\hat{j} - 3\hat{k})$.$\vec{F} = I \ell B [(\hat{i} 	imes \hat{i}) + 2(\hat{i} 	imes \hat{j}) - 3(\hat{i} 	imes \hat{k})]$.Using cross product rules: $\hat{i} 	imes \hat{i} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,and $\hat{i} 	imes \hat{k} = -\hat{j}$.$\vec{F} = I \ell B [0 + 2\hat{k} - 3(-\hat{j})] = I \ell B (3\hat{j} + 2\hat{k})$.The magnitude of the force is $F = |\vec{F}| = I \ell B \sqrt{3^2 + 2^2}$.$F = I \ell B \sqrt{9 + 4} = \sqrt{13} I \ell B$.

$A$ wire carrying current $I$ along the $x$-axis has length $\ell$ and is kept in a magnetic field $\vec{B} = (\hat{i} + 2\hat{j} - 3\hat{k}) B \text{ Wb/m}^2$. The magnitude of the magnetic force acting on the wire is:

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