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A square loop $A B C D$, carrying a current $I_2$, is placed near and coplanar with a long straight conductor $X Y$ carrying a current $I_1$, as shown in figure. The net force on the loop will be
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Verified Answer
The correct answer is:
$\frac{2 \mu_0 I_1 I_2}{3 \pi}$
Force on arm $A B$ due to current in conductor $X Y$ is
$F_1=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2 L}{(L / 2)}=\frac{\mu_0 I_1 I_2}{\pi}$ acting towards the wire in the plane of loop.
Force on arm $C D$ due to current in conductor $X Y$ is
$F_2=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2 L}{(3 L / 2)}=\frac{\mu_0 I_1 I_2}{3 \pi}$ away from the wire the plane of loop.
$\therefore \quad$ Net force on the loop $=F_1-F_2$
$=\frac{\mu_0 I_1 I_2}{\pi}\left[1-\frac{1}{3}\right]=\frac{2}{3} \frac{\mu_0 I_1 I_2}{\pi}$
$F_1=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2 L}{(L / 2)}=\frac{\mu_0 I_1 I_2}{\pi}$ acting towards the wire in the plane of loop.
Force on arm $C D$ due to current in conductor $X Y$ is
$F_2=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2 L}{(3 L / 2)}=\frac{\mu_0 I_1 I_2}{3 \pi}$ away from the wire the plane of loop.
$\therefore \quad$ Net force on the loop $=F_1-F_2$
$=\frac{\mu_0 I_1 I_2}{\pi}\left[1-\frac{1}{3}\right]=\frac{2}{3} \frac{\mu_0 I_1 I_2}{\pi}$
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