🌳 Concept tree

├── Observation

├── Magnitude of the force

1. ├── Force between two parallel current-carrying wires
2. └── Torque on a rectangular coil in a magnetic field

• Take a rectangular coil abcd (sides ab = cd = L, the other sides of width w) of N turns, carrying current I, free to rotate in a uniform field B.

• The sides parallel to the field (ad and bc) feel no force, because θ = 0 for them.

• The sides ab and cd, perpendicular to the field, feel equal and opposite parallel forces of magnitude F = BIL (per turn), forming a couple: the forces are equal, opposite and not along the same line, so they turn the coil about its axis without moving it as a whole.

• When the plane of the coil is parallel to the field, the torque = force × perpendicular distance = BIL × w = BI A per turn (A = L w). For N turns:

  τ = B I A N

• The coil has a magnetic dipole moment m = N I A, a vector perpendicular to the plane of the coil in the direction a right-hand screw advances when turned with the current (the direction from the south face to the north face of the coil). Then τ = B m.

• If the normal to the coil (the direction of m) makes an angle θ with the field:

  τ = B I A N sin θ

  The torque is maximum (θ = 90°, plane parallel to the field) and zero when the plane is perpendicular to the field (θ = 0°, the normal along B). In that position the coil is in equilibrium, with the dipole moment lined up with the field.

• This torque is the principle of the moving-coil galvanometer and the electric motor.
• Example 6. A coil of 200 turns, 5 cm × 8 cm, carries 0.3 A and is placed with its plane parallel to a 0.5 T field. A = 4×10⁻³ m². τ = 0.5 × 0.3 × 4×10⁻³ × 200 = 0.12 N·m.
• Example 6a. A coil of 100 turns measures 6 cm × 4 cm (A = 2.4×10⁻³ m²) and carries 0.5 A in a field of 0.3 T with its plane parallel to the field: τ = 0.3 × 0.5 × 2.4×10⁻³ × 100 = 0.036 N·m. When the normal to the coil makes 30° with the field, τ = 0.036 × sin 30° = 0.018 N·m.
• Example 6b — finding the current. A coil of 250 turns and area 3×10⁻³ m² in a field of 0.2 T (plane parallel to the field) must experience a torque of 0.09 N·m. I = τ / (B A N) = 0.09 / (0.2 × 3×10⁻³ × 250) = 0.09 / 0.15 = 0.6 A.