A square loop ABCD of side length L,carrying | vedclass

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

Vedclass · Accepted Answer

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

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(A) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2\pi r}$.The force on a current-carrying wire segment is $F = i(L 	imes B)$.For arm $AB$ (at distance $r_1 = L/2$): The current flows upwards. The magnetic field is directed into the plane. The force $F_1$ is attractive towards the wire.$F_1 = i L B_1 = i L \left( \frac{\mu_0 I}{2\pi (L/2)} ight) = \frac{\mu_0 Ii}{\pi}$.For arm $CD$ (at distance $r_2 = L/2 + L = 3L/2$): The current flows downwards. The magnetic field is directed into the plane. The force $F_2$ is repulsive away from the wire.$F_2 = i L B_2 = i L \left( \frac{\mu_0 I}{2\pi (3L/2)} ight) = \frac{\mu_0 Ii}{3\pi}$.The forces on arms $BC$ and $AD$ are equal and opposite,so they cancel out.The net force $F_{net} = F_1 - F_2 = \frac{\mu_0 Ii}{\pi} - \frac{\mu_0 Ii}{3\pi} = \frac{\mu_0 Ii}{\pi} (1 - 1/3) = \frac{2\mu_0 Ii}{3\pi}$.

$A$ 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 conductor and the side $AB$ is $L/2$. The net force on the loop will be:

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