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(C) The magnetic field is $\vec{B} = 0.2(1 + 2x) \hat{k} 	ext{ T}$. The loop is a square with side length $L = 0.5 	ext{ m}$. The loop is placed fro

Vedclass · Accepted Answer

$50$

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(C) The magnetic field is $\vec{B} = 0.2(1 + 2x) \hat{k} 	ext{ T}$. The loop is a square with side length $L = 0.5 	ext{ m}$. The loop is placed from $x = 2 	ext{ m}$ to $x = 2.5 	ext{ m}$ and $y = 2 	ext{ m}$ to $y = 2.5 	ext{ m}$.For the segments parallel to the $x$-axis,the forces are equal and opposite,so they cancel out.For the segments parallel to the $y$-axis,the force is given by $\vec{F} = I \int (d\vec{l} 	imes \vec{B})$.At $x = 2 	ext{ m}$,the current flows in the negative $y$-direction: $\vec{F}_1 = I L \hat{j} 	imes \vec{B}(x=2) = 0.5 	imes 0.5 	imes [0.2(1 + 2(2))] \hat{j} 	imes \hat{k} = 0.25 	imes 1 \hat{i} = 0.25 	ext{ N}$ (in $+x$ direction).At $x = 2.5 	ext{ m}$,the current flows in the positive $y$-direction: $\vec{F}_2 = I L (-\hat{j}) 	imes \vec{B}(x=2.5) = 0.5 	imes 0.5 	imes [0.2(1 + 2(2.5))] (-\hat{j}) 	imes \hat{k} = 0.25 	imes 1.2 (-\hat{i}) = -0.30 	ext{ N}$ (in $-x$ direction).The net force is $F_{	ext{net}} = |F_1 - F_2| = |0.25 - 0.30| = 0.05 	ext{ N} = 50 	ext{ mN}$.

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