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(D) The magnetic force on a current-carrying wire is given by the formula $\vec{F} = I(\vec{L} 	imes \vec{B})$.Here,the wire is along the $x$-axis,so

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$\sqrt{2} ILB_0$

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(D) The magnetic force on a current-carrying wire is given by the formula $\vec{F} = I(\vec{L} 	imes \vec{B})$.Here,the wire is along the $x$-axis,so the length vector is $\vec{L} = L\hat{i}$.The magnetic field is $\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k})$.Substituting these into the force equation:$\vec{F} = I(L\hat{i}) 	imes B_0(\hat{i} - \hat{j} - \hat{k})$$\vec{F} = ILB_0 [(\hat{i} 	imes \hat{i}) - (\hat{i} 	imes \hat{j}) - (\hat{i} 	imes \hat{k})]$Using the cross product rules ($\hat{i} 	imes \hat{i} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,$\hat{i} 	imes \hat{k} = -\hat{j}$):$\vec{F} = ILB_0 [0 - \hat{k} - (-\hat{j})]$$\vec{F} = ILB_0 (\hat{j} - \hat{k})$The magnitude of the force is $|\vec{F}| = ILB_0 \sqrt{(1)^2 + (-1)^2} = ILB_0 \sqrt{1 + 1} = \sqrt{2} ILB_0$.

$A$ wire of length $L$ carries current $I$ along the $x$-axis. $A$ magnetic field $\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k}) \text{ T}$ acts on the wire. The magnitude of the magnetic force acting on the wire is:

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