To my understanding, Back EMF will occur when a motor rotates.

This rotation inside a magnetic field will cause a change in flux, generating an EMF by Faraday's law. This EMF, the back EMF, will oppose the supply voltage by Lenz's law.

As such we can say: $$V_{net} = V_{supply} - \epsilon_{back}$$

I am confused as to where I will be able to measure such a $V_{net}$.

To understand my confusion, consider 2 cases of where I can measure back emf.

Let us say the supply voltage is 10V and the back emf is 3V. I have labelled this.

enter image description here

Thus I can see that the potential at the left purple point is going to be 10V and on the right is 0V. Thus the voltage measured here should be 10V. This makes sense.

However, if I probed the voltage INSIDE of the armature, it should definitely read less than 10V (for example 6V). This is because the net current will fall as the net voltage falls due to back emf increasing $V_{net} = V_{supply} - \epsilon_{back}$

enter image description here

So my ultimate question is, where does the back emf "start"? If we measure very far from the armature, we should read supply. If we measure inside the coil, we will measure $V_{supply} - \epsilon_{back}$

Does back emf start when the rectangular bit of the coil ends?

  • $\begingroup$ IMO, "Back EMF" is not real. (for some definition of "real.") Suppose as you say, the battery is forcing an actual voltage of 10V across the terminals of the motor, and suppose that the "back EMF" is 3V. If you could probe inside the motor, you would not find any place where the voltage suddenly jumps from 10V to 7V. But if you measure the current, then you should find it to be the same as if the motor was not turning (a.k.a., "stalled"), and the applied voltage was 7V. $\endgroup$ Commented Feb 22, 2023 at 18:52
  • $\begingroup$ ...OTOH, you could argue that it is real in the sense that, if you disconnected the battery from the motor, and if you forced the motor shaft to keep spinning at the same speed that created 3V of "back EMF," then you would measure an actual 3V at the motor's terminals. $\endgroup$ Commented Feb 22, 2023 at 19:12

1 Answer 1


You're confused because you are confusing net EMF in the whole circuit with voltage on the terminals.

Net EMF is due to both the battery and the induced EMF:

$$ \mathscr{E} = \mathscr{E}_{bat} + \mathscr{E}_{i}. $$ Kirchhoff's second circuital law, in this case, states that net EMF determines current in the motor circuit according to the equation

$$ \mathscr{E} = (R_{bat} + R_{winding}) I. $$

The induced EMF in motors is sometimes also called back-EMF, perhaps because it acts on current in the opposite way to the way the EMF of the power source does. But the induced EMF acts in all parts of the circuit where induced electric field is; it is not localized like the battery EMF.

Induced EMF and thus also net EMF in the motor circuit cannot be directly measured by a voltmeter or any other similar device, unless we wind the probing wires exactly like the motor circuit winding, with many turns. In effect, the voltmeter would not be measuring difference of potentials on the terminals (the usual intended use of voltmeter), but net EMF in the measurement circuit , which would be only due to $\mathscr{E}_i$ (this is not the usual intended use of voltmeter).

We know this because we believe in Kirchhoff's second circuital law, which in this case states

$$ \mathscr{E}_i = R_{measurement~circuit} I. $$

Such measurement is impractical but we can guess the result it would have, based on measuring current $I$ (which is easy) and using the above law.

In contrast measuring voltage on the terminals of the battery, or the motor, is easy - just connect the voltmeter probes to stationary terminals. If the two measuring wires are well placed (close to each other, to minimize induced EMF in the measurement circuit), the voltmeter will show reading close to difference of potentials between the terminals. Since the battery has internal resistance $R_{bat}$, this voltage on the terminals should be close to $$ \mathscr{E}_{bat} - R_{bat} I. $$


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