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(A) The magnetic field produced by an infinite current-carrying sheet is uniform and given by $B = \frac{\mu_0 k}{2}$.Since the sheet is in the $y-z$

Vedclass · Accepted Answer

$\mu_0 k R i$

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(A) The magnetic field produced by an infinite current-carrying sheet is uniform and given by $B = \frac{\mu_0 k}{2}$.Since the sheet is in the $y-z$ plane and current is in the $+y$ direction,the magnetic field points in the $+x$ direction for $x > 0$,i.e.,$\vec{B} = \frac{\mu_0 k}{2} \hat{i}$.The force on a current-carrying wire is given by $\vec{F} = i(\vec{L}_{eff} 	imes \vec{B})$.For a semicircular loop of radius $R$ lying in the $x-y$ plane,the effective length vector $\vec{L}_{eff}$ is the straight line connecting the two ends of the semicircle.Since the loop lies in the $x-y$ plane and is centered at $(d, 0, 0)$,its diameter is parallel to the $y$-axis. Thus,$\vec{L}_{eff} = 2R \hat{j}$.Therefore,the net force is $\vec{F} = i(2R \hat{j} 	imes \frac{\mu_0 k}{2} \hat{i}) = i R \mu_0 k (\hat{j} 	imes \hat{i}) = -\mu_0 k R i \hat{k}$.The magnitude of the net force is $\mu_0 k R i$.

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