DC motor principle and torque equation

Working principle

A DC motor converts electrical energy into mechanical energy. Its action rests on this principle: a current-carrying conductor placed in a magnetic field experiences a force: F = B I l (newton), whose direction is found by Fleming's left-hand rule.

1. The field poles produce flux Φ in the air-gap.
2. The armature conductors, connected to the DC supply through brushes and commutator, carry current.
3. Each conductor experiences a force. The conductors under the N pole are pushed one way and those under the S pole the other way, which produces a torque on the armature and it rotates.
4. The commutator reverses the current in a conductor as it passes from one pole to the next, so the torque always acts in the same direction.
5. As the armature rotates, a back emf E_b is induced; the supply voltage must balance it: V = E_b + I_aR_a.

   N pole          S pole
  ( x )  force up   (.)  force down
   conductor current in   conductor current out
   -> torque turns armature

Torque equation

Let r = armature radius (m), l = active length of a conductor (m), Z = total conductors, A = parallel paths, I_a = armature current, P = poles, Φ = flux per pole.

• Current in each conductor: I_a / A
• Force on each conductor: F = B l (I_a/A); B = average flux density = PΦ / (2πrl)
• Torque by one conductor: F × r
• Total torque (Z conductors): T_a = Z × B l (I_a/A) × r = Z (PΦ/2πrl) l (I_a/A) r

T_a = (PZ / 2πA) Φ I_a newton-metre

= 0.159 (PZ/A) Φ I_a N·m

So T_a = KΦI_a, with K = PZ/(2πA). Torque is proportional to flux and armature current.

Check using power: E_b I_a = T_a ω, with E_b = PΦZN/(60A) and ω = 2πN/60, giving T_a = PΦZI_a/(2πA), which matches.