Torque on a coil and the moving-coil galvanometer
Answers are for parents and teachers.
1
A wire 0.28 m long carries 7.5 A in a field of 0.12 T. Find the force when the wire is at right angles to the field and when it makes 30° with it.
At 90°: F = 0.12 × 7.5 × 0.28 = 0.252 N. At 30°: F = 0.252 × 0.5 = 0.126 N.
2
A coil of 120 turns measures 8.0 cm × 5.0 cm and carries 0.60 A in a field of 0.35 T. Find the torque when the plane is parallel to the field, and when the normal makes 30° with the field. Give 3 significant figures.
A = 0.08 × 0.05 = 4.0 × 10⁻³ m². Plane parallel: τ = 0.35 × 0.60 × 4.0 × 10⁻³ × 120 = 0.101 N·m. Normal at 30°: τ = 0.101 × 0.5 = 0.0504 N·m.
3
A coil has a magnetic dipole moment of 0.25 A·m². The largest torque on it in a certain uniform field is 0.05 N·m. Find the field. Then find the angle θ of the normal (between 0° and 90°) at which the torque is 0.025 N·m.
τmax = mB, so B = 0.05 ÷ 0.25 = 0.2 T. Then sin θ = 0.025 ÷ 0.05 = 0.5, so θ = 30°.
4
A galvanometer deflects 52.5° for 175 µA. Find its sensitivity. What current gives a deflection of 12°? What happens to the pointer if that current is reversed?
Sensitivity = 52.5 ÷ 175 = 0.3 °/µA. For 12°: I = 12 ÷ 0.3 = 40 µA. Reversed, the pointer deflects 12° on the other side of zero.
5
A galvanometer has B = 0.06 T, N = 500 turns, A = 1.0 × 10⁻⁴ m² and springs with c = 1.5 × 10⁻⁸ N·m per degree. Find its sensitivity in °/µA. What is it if the springs are replaced by springs twice as stiff?
BAN ÷ c = 0.06 × 1.0 × 10⁻⁴ × 500 ÷ (1.5 × 10⁻⁸) = 3.0 × 10⁻³ ÷ 1.5 × 10⁻⁸ = 2 × 10⁵ °/A = 0.2 °/µA. With c doubled, the sensitivity is halved: 0.1 °/µA.