electromagnetism - DC motors, back emf, mechanical power output of a DC motor

In the case of DC motors why is the mechanical power output equal to back emf multiplied by armature current.

Answer

You probably have no trouble with the concept of a simple rotating coil generator.
A coil is rotated in a magnetic field and an emf $\mathcal E$ is generated.

If there is a resistance $R$ across the terminals of the generator then a current $I$ flows in the circuit.

To simplify matters assume that the frictional forces are very small and the coil resistance is $r$.
Then $\mathcal E = I(R+r) \Rightarrow \mathcal EI = I^2R + I^2r$ where $I$ is the current flowing in the circuit.
The $I^2R$ term represents the useful power out and the $I^2r$ represents the wasted power.

So what is $\mathcal E I$?
It represents the mechanical power input which keeps the coil rotating at a constant speed.

Now suppose that the resistor $R$ is replaced by a battery of emf $E$ with $E < \mathcal E$
In this case the situation is not much different from that when the resistor was in place with $\mathcal EI - EI = I^2r \Rightarrow \mathcal EI = EI + I^2r$ where $EI$ now represents the power supplied to the battery (to recharge it?) whereas with a reistor in place that power is dissipated as heat.

If you increase the emf of the battery such that $E> \mathcal E$ you now have a reversal of the current thorough the coil and $EI - \mathcal EI = I^2r \Rightarrow EI = \mathcal EI + I^2r$ where $EI$ is the power supplied by the battery and as before $I^2r$ represent the power losses in the coil.

And what of $\mathcal E I$?
Well you call $\mathcal E$ the back emf because it is is trying to drive current in the opposite direction to that produced by the battery and $\mathcal EI$ represents the mechanical power output from the motor.