A square loop ABCD carries a current i and is | vedclass

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(B) The magnetic field $B$ at a distance $r$ from a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$.The force on a current-carrying segmen

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$\frac{2\mu_0 Ii}{3\pi}$

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(B) The magnetic field $B$ at a distance $r$ from a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$.The force on a current-carrying segment of length $L$ in a magnetic field is $F = BiL$.For the square loop,the sides parallel to the wire $XY$ experience forces,while the sides perpendicular to the wire experience equal and opposite forces that cancel out.$1$. Force on the side closer to the wire (distance $r_1 = L/2$): $F_1 = \left( \frac{\mu_0 I}{2\pi (L/2)} \right) i L = \frac{\mu_0 Ii}{\pi}$. This force is attractive (towards the wire).$2$. Force on the side farther from the wire (distance $r_2 = L/2 + L = 3L/2$): $F_2 = \left( \frac{\mu_0 I}{2\pi (3L/2)} \right) i L = \frac{\mu_0 Ii}{3\pi}$. This force is repulsive (away from the wire).The net force $F_{net} = F_1 - F_2 = \frac{\mu_0 Ii}{\pi} - \frac{\mu_0 Ii}{3\pi} = \frac{2\mu_0 Ii}{3\pi}$.

$A$ square loop $ABCD$ carries a current $i$ and is placed near a long straight wire $XY$ carrying a current $I$ as shown in the figure. What is the net force on the loop?

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