Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying a steady current I₁ at a distance d from the conductor as shown in figure. The loop will experience

a net repulsive force away from the conductor

a net torque acting upward perpendicular to the horizontal plane

a net torque acting downward normal to the horizontal plane

a net attractive force towards the conductor

Step-by-Step Solution

Torque Analysis: The magnetic field $\vec{B}$ produced by the long straight wire is perpendicular to the horizontal plane containing the loop. The magnetic moment $\vec{m}$ of the current loop is also perpendicular to the area of the loop (i.e., perpendicular to the horizontal plane). Since $\vec{m}$ and $\vec{B}$ are parallel (or anti-parallel), the angle $\theta$ between them is $0^{\circ}$ or $180^{\circ}$ . The torque $\tau = mB \sin\theta$ is therefore zero . This eliminates options related to torque.
Force Analysis: The net force is determined by the interaction between the straight wire and the sides of the square loop parallel to it. According to Ampere's law, parallel currents attract and anti-parallel currents repel .
Magnitude Dependence: The magnetic field strength decreases with distance ( $B \propto 1/r$ ) . The side of the loop closer to the wire experiences a stronger force than the side farther away.
Net Force: Assuming the standard configuration for this problem (where the current in the nearest side matches the direction of $I_1$ to align with the 'Probable Answer'), the attractive force on the near side ( $F_{near}$ ) dominates the repulsive force on the far side ( $F_{far}$ ). Thus, $F_{net} = F_{near} - F_{far}$ is directed towards the conductor .

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