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(C) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.Here, the length vector is $\vec{L} = L \hat{i}$.G

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$B_0(\hat{i} + \hat{j} + \hat{k})$

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(C) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.Here, the length vector is $\vec{L} = L \hat{i}$.Given $\vec{F} = I B_0 L(\hat{k} - \hat{j})$.Substituting these into the formula: $I B_0 L(\hat{k} - \hat{j}) = I(L \hat{i} 	imes \vec{B})$.Dividing by $I L$, we get: $B_0(\hat{k} - \hat{j}) = \hat{i} 	imes \vec{B}$.Let $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$.Then $\hat{i} 	imes (B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) = B_y \hat{k} - B_z \hat{j}$.Comparing this with $B_0(\hat{k} - \hat{j})$, we get $B_y = B_0$ and $B_z = B_0$.Since the cross product with $\hat{i}$ eliminates any component along $\hat{i}$, $B_x$ can be any value, but looking at the options, $B_x = B_0$ satisfies the condition.Thus, $\vec{B} = B_0(\hat{i} + \hat{j} + \hat{k})$.

$A$ wire of length $L$ carries a current $I$ along the $X$-axis. The magnetic force acting on the wire is given by $\vec{F} = I B_0 L(\hat{k} - \hat{j})$. The existing magnetic field $\vec{B}$ is

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