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
2
The torque on a current-carrying coil
Physics Sec 3 Term 1, Chapter 2: pp. 41–45
GoalCalculate the torque on a coil from τ = BIAN sin θ and use the magnetic dipole moment m = IAN.
New words
torque (τ) · the turning effect of the couple: one force times the perpendicular distance between the two forces; unit N·mnormal to the coil · the line at right angles to the plane of the coil; it points the way a right-hand screw advances when turned like the currentmagnetic dipole moment (m) · m = IAN, a vector along the normal; its unit is A·m²
Explanation
With the plane parallel to the field, each long side feels BINℓ and the perpendicular distance between the two forces is the width w of the coil. So τ = F × w = BINℓw = BIAN, because A = ℓw. In any position, τ = BIAN sin θ, where θ is the angle that the normal to the coil makes with the field. With the plane parallel to the field the normal is at 90°, sin 90° = 1 and the torque is largest. As the coil turns the torque falls, and when the plane is perpendicular to the field the normal lies along the field, θ = 0 and the torque is zero. The product m = IAN is the magnetic dipole moment, so τ = mB sin θ. Careful: θ is measured from the normal, not from the plane.
Plane of the coil
Angle θ of the normal
sin θ
Torque
parallel to the field
90°
1
BIAN (largest)
30° to the field
60°
0.866
0.866 BIAN
60° to the field
30°
0.5
0.5 BIAN
perpendicular to the field
0°
0
zero
Worked examples
N = 50 turns, each 0.20 m × 0.20 m, I = 2 A, B = 0.3 T, plane parallel to the field. A = 0.20 × 0.20 = 0.04 m². τ = BIAN = 0.3 × 2 × 0.04 × 50 = 1.2 N·m.
The same coil is turned so that its plane makes 60° with the field. The normal then makes 30° with the field (90° − 60°). τ = 1.2 × sin 30° = 0.6 N·m.
A coil of N = 100 turns measures 6 cm × 5 cm and carries I = 0.5 A. A = 0.06 × 0.05 = 3 × 10⁻³ m². Its dipole moment is m = IAN = 0.5 × 3 × 10⁻³ × 100 = 0.15 A·m². In a field of 0.2 T with the normal at right angles to the field, τ = mB = 0.15 × 0.2 = 0.03 N·m; with the normal at 30°, τ = 0.015 N·m.
Class activity
Paper practice. Draw the coil edge-on in four positions, with its plane at 0°, 30°, 60° and 90° to the field. For each, mark the angle θ of the normal (90°, 60°, 30°, 0°) and use sin θ to fill in the torque as a fraction of the largest value (1, 0.87, 0.5, 0). Which position gives no torque?
Practice
1
A coil of 80 turns measures 0.15 m × 0.10 m and carries 1.2 A in a field of 0.5 T, with its plane parallel to the field. Find the torque.
The same coil is turned until its normal makes 30° with the field. Find the torque.
τ = 0.72 × sin 30° = 0.36 N·m.
3
A coil has N = 200 turns, area 2.5 × 10⁻³ m² and carries 0.04 A. Find its magnetic dipole moment.
m = IAN = 0.04 × 2.5 × 10⁻³ × 200 = 0.02 A·m².
4
A coil with m = 0.15 A·m² is in a field of 0.4 T. Its largest torque is 0.06 N·m. At what angle θ of the normal (between 0° and 90°) is the torque 0.03 N·m?
