Q.
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
NTA AbhyasNTA Abhyas 2022
Solution:
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, $F=ILB=\frac{\mu _{0} I_{1} I L}{2 \pi r}$
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.
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, $F=ILB=\frac{\mu _{0} I_{1} I L}{2 \pi r}$
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.