A galvanometer coil has 500 turns and each turn has an average area of 3 × 10 –4 m² . If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 A is flowing through it, the strength of the field (in T) is ______.

Solution

Physics sir · 4 steps

Coil plane parallel to the field means its normal is perpendicular to it, so the torque is at its maximum.

1. Torque on a coil. A coil of N turns carrying current I has magnetic moment NIA. In a field B it feels a torque NIAB sin θ, where θ is the angle between the coil's normal and B.
2. Find the angle. The plane of the coil is parallel to the field, so its normal is perpendicular to the field. That gives θ = 90° and sin θ = 1.

   τ = NIAB ⇒ B = τ/NIA

3. Put in numbers. NIA = 500 × 0.5 × 3 × 10⁻⁴ = 0.075 A m². So B = 1.5/0.075.
4. So. B = 20 T, so the answer is 20.

The common slip: 10000. Leaving out the 500 turns gives B = 1.5/(0.5 × 3 × 10⁻⁴) = 10 000, not 20. Every turn adds to the torque, so N must be in the formula.

The idea to keep. In τ = NIAB sin θ, the angle θ is measured from the coil's normal, not its plane. A coil lying parallel to the field feels the maximum torque, NIAB.

Answer from the NTA final answer key. Solution by PrepWiser.