A circular loop carrying a steady current lies flat on a smooth horizontal table. A uniform magnetic field is then set up in the region. Which of the following is correct?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → No uniform field can make the loop turn about a vertical axis.
The torque on a current loop is \( \vec{\tau}=I\vec{A}\times\vec{B} \). The area vector \( \vec{A} \) is normal to the loop, so for a loop lying flat on the table it is vertical. A cross product is always perpendicular to \( \vec{A} \).
Option (1) A vertical field makes the loop turn about a vertical axis. — Incorrect. A vertical field is parallel to \( \vec{A} \). Then \( \vec{A}\times\vec{B}=0 \), so there is no torque and the loop does not turn at all.
Option (2) A horizontal field makes the loop turn about a vertical axis. — Incorrect. A horizontal field gives a torque along a horizontal line. Such a torque tends to tilt the loop off the table about a horizontal axis.
Option (3) The loop slides across the table because a net force acts on it. — Incorrect. In a uniform field the forces on opposite parts of the loop are equal and opposite. The net force on the loop is zero, so it does not slide.
Option (4) No uniform field can make the loop turn about a vertical axis. — Correct. For any uniform field, \( I\vec{A}\times\vec{B} \) is perpendicular to the vertical vector \( \vec{A} \), so the torque always lies in the horizontal plane of the loop. Turning about a vertical axis would need a vertical torque.
NCERT (2026-27):
Example 4.11 – “(a) No, because that would require τ to be in the vertical direction. But τ = I A × B, and since A of the horizontal loop is in the vertical direction, τ would be in the plane of the loop for any B.”