Verified Solution • 2025–26 Session moderate

QPage 229: Q6: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Page 229: Q6 Solution: A circular loop of area , carrying a current of 10 A, is placed …

⚠️ Common Exam Traps & Conceptual Pitfalls

When the field is said to be perpendicular to the plane of the loop, the area vector (normal to the plane) is actually parallel to the field. Thus the angle θ between B and the area vector is 0°, so sinθ=0 and τ=IAB sinθ or m×B vanishes. Avoid the mistake of treating the field as perpendicular to the area vector and assuming a maximum torque.

A circular loop of area 1 cm21 \text{ cm}^2, carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is

Multiple Choice Options

  • A zero
  • B 10−4 N m10^{-4} \text{ N m}
  • C 10−2 N m10^{-2} \text{ N m}
  • D 1 N m1 \text{ N m}

Step-by-Step Verified Solution

**Answer:** (a) zero. **Step-by-Step Explanation:** The magnetic field is perpendicular to the plane of the loop, so the area vector is parallel to B. Torque = m _ B = 0.

All Questions in this Chapter

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