Week 6 · Chapter 2 (C): worked force problems, the torque on a rectangular coil (τ = BIAN sin θ, the magnetic dipole moment) and the moving-coil galvanometer (construction, radial field, sensitivity) · 3 lessons
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The moving-coil galvanometer
Physics Sec 3 Term 1, Chapter 2: pp. 41–45
GoalExplain how a moving-coil galvanometer works and calculate its sensitivity in degrees per µA.
New words
radial field · a field whose lines point straight towards the axis; with curved poles and a soft-iron core the sides of the coil stay at right angles to it at every anglerestoring torque · the torque of the spiral springs, which grows with the angle turned and balances the magnetic torquesensitivity · the scale deflection per unit current, in degrees per µA
Explanation
A galvanometer detects very small currents and shows their direction. A coil of many turns on a light aluminium frame is pivoted on jewel bearings between the curved poles of a permanent magnet, around a soft-iron core. Because the poles are curved, the field is radial: at every angle the sides of the coil are at right angles to it, so the magnetic torque is BIAN whatever the position, and it is proportional to I. Two spiral springs oppose the turning (they also carry the current in and out). Their torque is proportional to the angle φ turned: τ(spring) = cφ, with c in N·m per degree. The coil stops where the torques are equal, BIAN = cφ, so φ = (BAN ÷ c) × I. The deflection is proportional to the current, the scale is evenly spaced, and the pointer moves the other way if the current is reversed. The sensitivity φ ÷ I = BAN ÷ c grows with more turns, a stronger magnet or a larger area, and with weaker springs.
Part
What it does
coil on an aluminium frame
carries the current and turns
curved poles and soft-iron core
make the field radial and strong
two spiral springs
give the restoring torque and carry the current
pointer and scale
show the deflection
Worked examples
A galvanometer deflects 45° for a current of 90 µA. Sensitivity = 45° ÷ 90 µA = 0.5 °/µA.
Its sensitivity is 0.5 °/µA and the deflection is proportional to the current. For 60 µA: φ = 0.5 × 60 = 30°. To deflect it 15°: I = 15 ÷ 0.5 = 30 µA. If a current of 40 µA flows the other way, φ = −20°, which means 20° on the other side of zero.
B = 0.08 T, N = 250 turns, A = 1.5 × 10⁻⁴ m² and c = 3.0 × 10⁻⁸ N·m per degree. Sensitivity = BAN ÷ c = 0.08 × 1.5 × 10⁻⁴ × 250 ÷ (3.0 × 10⁻⁸) = 3.0 × 10⁻³ ÷ 3.0 × 10⁻⁸ = 1 × 10⁵ °/A = 0.1 °/µA. A current of 100 µA gives 10°. With twice the turns and everything else equal, the sensitivity doubles.
Class activity
Paper practice. Draw a scale with zero in the middle, like a galvanometer face. For a sensitivity of 0.5 °/µA, mark where the pointer rests for 0, +40 µA and −40 µA. What would the pointer do if the springs were twice as stiff?
Practice
1
A galvanometer deflects 36° for 120 µA. Find its sensitivity.
36° ÷ 120 µA = 0.3 °/µA.
2
Its sensitivity is 0.25 °/µA. What is the deflection for 64 µA?
φ = 0.25 × 64 = 16°.
3
With the same 0.25 °/µA, the pointer rests at 22.5°. Find the current.
I = 22.5 ÷ 0.25 = 90 µA.
4
B = 0.12 T, N = 150 turns, A = 2.0 × 10⁻⁴ m² and c = 2.4 × 10⁻⁸ N·m per degree. Find the sensitivity in °/µA.