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

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$-i B_0 (2R) \hat{j}$

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(C) The magnetic force on a current-carrying wire is given by $\vec{F} = i (\vec{L}_{eff} 	imes \vec{B})$.Here,the effective length vector $\vec{L}_{eff}$ is the displacement vector from the starting point to the ending point of the wire within the magnetic field.From the figure,the current enters at the bottom left and exits at the bottom right of the semicircular loop. The displacement vector $\vec{L}_{eff}$ for the semicircular part is $2R \hat{i}$.The magnetic field is $\vec{B} = B_0 \hat{k}$.Therefore,$\vec{F} = i (2R \hat{i} 	imes B_0 \hat{k})$.Since $\hat{i} 	imes \hat{k} = -\hat{j}$,we get $\vec{F} = i (2R B_0) (-\hat{j}) = -2 i B_0 R \hat{j}$.

$A$ rigid wire consists of a semicircular portion of radius $R$ and two straight sections. The wire is partially immersed in a perpendicular magnetic field $B = B_0 \hat{k}$ as shown in the figure. The magnetic force on the wire if it carries a current $i$ is:

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