A square loop ABCD of side length L carrying | vedclass

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(B) The force on a current-carrying wire near a long straight conductor is given by $F = \frac{{\mu _0}IiL}{2\pi r}$.For the side of the loop nearest

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$\frac{{2{\mu _0}Ii}}{{3\pi }}$

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(B) The force on a current-carrying wire near a long straight conductor is given by $F = \frac{{\mu _0}IiL}{2\pi r}$.For the side of the loop nearest to the wire (distance $r_1 = L/2$),the current flows in the same direction as the wire $XY$. The force $F_1$ is attractive:$F_1 = \frac{{\mu _0}IiL}{2\pi (L/2)} = \frac{{\mu _0}Ii}{\pi}$.For the side of the loop farthest from the wire (distance $r_2 = L/2 + L = 3L/2$),the current flows in the opposite direction to the wire $XY$. The force $F_2$ is repulsive:$F_2 = \frac{{\mu _0}IiL}{2\pi (3L/2)} = \frac{{\mu _0}Ii}{3\pi}$.The forces on the top and bottom sides of the loop are equal and opposite,so they cancel out.The net force is $F_{net} = F_1 - F_2 = \frac{{\mu _0}Ii}{\pi} - \frac{{\mu _0}Ii}{3\pi} = \frac{{\mu _0}Ii}{\pi} (1 - 1/3) = \frac{2{\mu _0}Ii}{3\pi}$.

$A$ square loop $ABCD$ of side length $L$ carrying a current $i$ is placed near and coplanar with a long straight conductor $XY$ carrying a current $I$. The distance between the wire $XY$ and the nearest side of the loop is $L/2$. The net force on the loop will be:

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