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

• A. a net repulsive force away from the conductor.
• B. a net torque acting upward perpendicular to the horizontal plane.
• C. a net torque acting downward normal to the horizontal plane.
• D. a net attractive force towards the conductor.

Answer 1

Answer: (D)

Magnetic field due to a long current carrying conductor $B=\frac{{\mu }_0{I}_1}{2\pi r}$ .
Force on a current carrying conductor placed in a magnetic field, $F=ILB=\frac{{\mu }_0{I}_{1}IL}{2\pi r}$
We can find the direction of magnetic field in the loop by using right-hand thumb rule.
Field will be perpendicular to the loop in the downward direction.
$F_1>F_2$ as $F\propto \frac{1}{r}$ and $F_3$ and $F_4$ are equal and opposite. Hence, the net attraction force will be towards the conductor.
The forces acting are in the plane of the loop. So, no net torque is produced.