Back emf and its role in DC motor

Back emf

When the armature of a DC motor rotates, its conductors cut the magnetic flux, so an emf is induced in them (generator action). By Lenz's law this emf opposes the supply voltage which is the cause of the current. It is called the back emf or counter emf:

$$E_b = \frac{\Phi ZNP}{60A}\ \text{V}$$

The voltage equation of the motor is

$$V = E_b + I_aR_a \quad\Rightarrow\quad I_a = \frac{V - E_b}{R_a}$$

How back emf adjusts the torque to the load

The torque is $T = K_a\Phi I_a$, so the motor must draw the armature current that its load requires. Back emf adjusts this current automatically:

 Load up -> speed N falls -> Eb falls -> (V - Eb) rises
        -> Ia rises -> torque rises -> speed stops falling
 Load down -> N rises -> Eb rises -> Ia falls -> torque falls

1. When the load on the shaft increases, the load torque exceeds the motor torque, so the speed falls.
2. $E_b\propto N$ falls, so $(V - E_b)$ increases and a larger armature current $I_a = (V - E_b)/R_a$ flows.
3. The torque $K_a\Phi I_a$ rises until it equals the new load torque, and the motor settles at a slightly lower speed.
4. When the load decreases, the reverse happens: speed rises, $E_b$ rises, $I_a$ falls and torque falls to match the lighter load.

So back emf acts as a self-regulating governor that makes the motor draw only the current needed by the load. At no load $E_b\approx V$ and the current is just enough to supply the losses. The power converted is $E_bI_a = T\omega$.

Back emf is also why the starting current is large: at start $E_b = 0$, so $I_a = V/R_a$ and a starter is required.