Question

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 of $d$ from the conductor as shown in figure. The loop will experience-

(A) a net attractive force towards the conductor
(B) a net repulsive force away from the conductor
(C) a net torque acting upwards perpendicular to the horizontal plane
(D) a net torque acting downward normal to the horizontal plane

Answer 1

When the square loop which carries the steady current of $I$, then the force in every side of the square loop conductor is perpendicular to the surface of each side of the square loop conductor. So, there is no torque to experience.

Complete answer:
From the given diagram of the square loop which carries current and it is laced in a horizontal plane near a long straight conductor carrying a steady current ${I_1}$ at a distance $d$ from the square loop conductor, then the force experienced in the square loop conductor diagram is given by,

From the above diagram, the force ${F_3}$ and the force ${F_4}$, will not have any effect because in the vertical plane there is no parallel current carrying conductor. If suppose any other conductor is placed in the vertical plane then the square loop will also have the attraction or repulsion force. Now, the force ${F_1}$ and the force ${F_2}$ are responsible for the movement of the square loop towards the long straight conductor. Because the force ${F_1}$ is very close to the long straight conductor, so the attraction force is very high.

Hence, the option (A) is the correct answer.

Note:
By comparing the force of ${F_1}$ and ${F_2}$, the force ${F_1}$ will have the attraction force towards the long straight conductor and the force ${F_2}$ will have the repulsion force against the long straight conductor. But the force ${F_1}$ is very close to the long straight conductor, then attraction is more.