A square loop,carrying a steady current I,is | vedclass

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(A) Let the side length of the square loop be $a$. The magnetic field produced by the long straight wire at a distance $r$ is $B = \frac{\mu_0 I_1}{2

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a net attractive force towards the conductor

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(A) Let the side length of the square loop be $a$. The magnetic field produced by the long straight wire at a distance $r$ is $B = \frac{\mu_0 I_1}{2 \pi r}$.The force on a current-carrying wire is given by $F = I \int (dl 	imes B)$.$1$. The top segment of the loop (at distance $d$) carries current $I$ in the direction opposite to $I_1$. By the right-hand rule,the force on this segment is attractive towards the wire: $F_{top} = \frac{\mu_0 I_1 I}{2 \pi d} 	imes a$ (upward).$2$. The bottom segment of the loop (at distance $d+a$) carries current $I$ in the same direction as $I_1$. The force on this segment is repulsive away from the wire: $F_{bottom} = \frac{\mu_0 I_1 I}{2 \pi (d+a)} 	imes a$ (downward).$3$. The two vertical segments of the loop experience equal and opposite forces,so they cancel out.Since $d Therefore,the net force is attractive towards the conductor.

$A$ square loop,carrying a steady current $I$,is placed in a horizontal plane near a long straight conductor carrying a steady current $I_1$ at a distance $d$ from the conductor as shown in the figure. The loop will experience:

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