DC motor working principle and torque equation

Working principle

A DC motor converts electrical energy into mechanical energy. It works on the principle that a current-carrying conductor placed in a magnetic field experiences a mechanical force:

$$F = BIl\ \text{newton}$$

The direction of the force is given by Fleming's left-hand rule (forefinger = field, middle finger = current, thumb = force).

       N                       S
   +--------+             +--------+
   |   ->   |   (x) F up  |        |
   |  field |  (.) F down |        |
   +--------+   armature  +--------+
              conductors

1. The field winding produces flux $\Phi$ in the air gap.
2. Current from the supply enters the armature conductors through the brushes and commutator.
3. Conductors under the N pole carry current in one direction and those under the S pole in the opposite direction, so forces on both sides act in the same sense and produce a torque that rotates the armature.
4. The commutator reverses the current in a conductor as it passes from one pole to the next, so the torque stays unidirectional.
5. The rotating conductors cut flux and induce a back emf $E_b$ opposing the supply, so $V = E_b + I_aR_a$.

Torque equation

Let $\Phi$ = flux per pole, $Z$ = armature conductors, $P$ = poles, $A$ = parallel paths, $I_a$ = armature current, $r$ = armature radius, $l$ = effective length.

Current per conductor $= I_a/A$. Force on one conductor $F = B\,l\,\dfrac{I_a}{A}$. Torque of one conductor $= F\times r$.

Total torque of $Z$ conductors, with average flux density $B = \dfrac{P\Phi}{2\pi r l}$:

$$T_a = Z\cdot B\,l\,\frac{I_a}{A}\cdot r = Z\cdot\frac{P\Phi}{2\pi r l}\cdot l\cdot\frac{I_a}{A}\cdot r$$ $$T_a = \frac{ZP}{2\pi A}\,\Phi I_a = 0.159\,\frac{ZP}{A}\,\Phi I_a\ \text{N·m}$$

Writing $K_a = \dfrac{ZP}{2\pi A}$:

$$T_a = K_a\Phi I_a$$

Equivalent form from power: $T_a\omega = E_bI_a$ with $E_b = \dfrac{\Phi ZNP}{60A}$ gives the same result. For a shunt motor ($\Phi$ constant) $T \propto I_a$; for an unsaturated series motor ($\Phi\propto I_a$) $T \propto I_a^2$.