# Why is a DC motor's back EMF constant for a given RPM

I was trying to prove this mathematically from the definition of EMF being the change in magnetic flux over time.

Starting with

$$\Phi=BAcos(\theta)$$

Then differentiate w.r.t time (assuming B field doesn't change with time)

$$\frac{d\Phi}{dt}=-BA\omega sin(\theta)$$

Where $$\\omega\$$ is angular velocity. Thus,

$$emf=BA\omega sin(\theta)$$

Now if we consider the motion of a single armature loop $$\\theta\$$ will be the angle between the normal vector to the plane of the armature then $$\\theta\$$ is going to constantly change as the loop rotates (hence why $$\\theta\$$ was treated as a function of time when differentiating to give $$\\omega\$$).

So the EMF follows a sine wave relationship as the loop rotates, therefore it is always changing.

Where have I gone wrong?

• Isn't back EMF usually defined as an RMS value? When we've used back EMF as a surrogate for motor speed (so that a separate tachometer is not needed), that's what was processed. Commented Jan 29, 2022 at 12:54
• "Now if we consider the motion of a single armature loop ..." Have you forgotten the commutator and brushes? Commented Jan 29, 2022 at 13:16
• $\theta=\omega t$
– Chu
Commented Jan 29, 2022 at 13:16
• @SteveSh if the brushes are correctly aligned, it's a peak value (and approximately DC)
– user16324
Commented Jan 29, 2022 at 13:18