Q103 / 180 · AIIMS 2017 · Physics

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 Figure

  • A
    $\frac{\mu_0 I_1 I_2}{2 \pi}$
  • B
    $\frac{\mu_0 I_1 I_2 L}{2 \pi}$
  • C
    $\frac{2 \mu_0 I_1 I_2 L}{3 \pi}$
  • Correct answer
    D
    $\frac{2 \mu_0 I_1 I_2}{3 \pi}$

Solution

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 $=F_1-F_2$ $=\frac{\mu_0 I_1 I_2}{\pi}\left[1-\frac{1}{3}\right]=\frac{2}{3} \frac{\mu_0 I_1 I_2}{\pi}$

Hence, the correct answer is option: 'D'.

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