I'm doing a high school physics project where we need to investigate the relationship between the speed and magnetic flux of a DC motor.

We built a simple DC motor which basically looks like this, with brushes and a split-ring commutator. So our observation is that the further apart the magnets, i.e. the smaller the flux density, the lower the rpm.

We are given the torque equation t=nIABsinΘ so it seems like torque has decreased in this case. But everywhere on the internet is saying that a DC motor's speed is inversely proportional to torque and that higher flux gives lower speed, e.g. on wikipedia, which all contradict our result.

There must be some flaws in our understanding so can someone please explain to me what it is? Any help would be appreciated.

  • 3
    \$\begingroup\$ excellent work so far! I'd like to welcome the up-and-coming generation of electrical engineers to this board. \$\endgroup\$ Commented Apr 4, 2020 at 12:08
  • \$\begingroup\$ You need to overcome basic friction problems before you can adequately see the relationship. \$\endgroup\$
    – Andy aka
    Commented Apr 4, 2020 at 12:29
  • \$\begingroup\$ Here's a quite new case where the motor as electrical circuit model is discussed electronics.stackexchange.com/questions/490895/… Of course it's not an answer for your apparent problem but it can be useful. \$\endgroup\$
    – user136077
    Commented Apr 4, 2020 at 13:45

2 Answers 2


Your understanding (higher flux = higher torque = lower speed) is not wrong, but it's based around a motor whose construction is reasonably close to the ideal, which is difficult to achieve with simple hand construction as seen in that video.

Now consider a motor with much less ideal construction, where bearings and brushes have significant friction, and flux is adjustable from low to very very low.

At "low" flux, the motor can turn at a few hundred RPM against the accumulated friction, but as you reduce the flux, yes the torque reduces, but instead of increasing speed, the friction simply stalls the motor.

So what can you do?

I would suggest using a much more "ideal" construction as a starting point for some more experiments. Toy motors can be found at several for a dollar, or even free in the right sort of old junk. (But at this stage, avoid anything that isn't a brushed motor) Bend a couple of metal tabs with pliers or screwdriver and you can pull them apart. If you lose a couple at that price, who cares?

You get accurately shaped magnets, curved to concentrate flux in the accurately made rotors, and good bearings. Mount those magnets on your stands as close as you can to that rotor.

Keep one set of magnets in the original can to see how the ring of iron around them both traps flux where you want it. Does adding a thicker soft iron ring outside the original preserve more flux?

Pull the wire off a couple of rotors and rewind with more turns, or fewer turns of fatter wire. (requires careful soldering to keep the commutator intact) What do more turns or fewer do to the rotor?

Measure voltage and current. If you can, measure RPM too.

  • \$\begingroup\$ Thanks for the answer! You said that under a lower flux the motor is stalled, so I assume you mean it stops after running for a while. But by my observation after the flux is lowered, rather than stopping it still runs at a near uniform speed but slower. It seems to me that in this case the generated torque still balances out the load torque plus the friction so it's not stalled. Then why is decreasing speed happening here? \$\endgroup\$
    – djaniuiin
    Commented Apr 6, 2020 at 4:04
  • \$\begingroup\$ I meant it was nearer the stall condition. In that state, if it can start it'll keep on going, but its speed will be sensitive to torque load. \$\endgroup\$
    – user16324
    Commented Apr 6, 2020 at 9:41

How does magnetic flux affects the speed of a DC motor

It's all about how does current get to flow in the rotor...

Scenario 1

You have high flux and the rotor is spinning constantly at a moderate speed on no load. The rate of change of flux (\$\frac{d\Phi}{dt}\$) seen by the rotor is such that \$\frac{d\Phi}{dt}\$ induces an equal and opposite voltage in the rotor coil to that which is coupled from the battery via the commutation. This means that virtually no current flows into the rotor because, no torque is needed as speed is constant and there is no mechanical load.

Scenario 2

If the magnets then move further away, in order to induce that same equal voltage, the rotor has to speed up. In other words \$\frac{d\Phi}{dt}\$ becomes maintained once the rotor has increased its speed. The rotor has no choice but to speed up because the induced voltage is not equal to the applied voltage and hence there is a significant current flow and that creates an acceleration torque.

So, the rotor speeds up because a higher current is drawn (via the commutation brushes) and that creates acceleration. Once in equilibrium (\$\frac{d\Phi}{dt}\$ is back to where it was) the induced voltage and applied voltage are equal but, of course, now, the rotor is spinning faster.


To overcome friction, there must be a constant torque generated by the rotor and, that means that the induced voltage is not quite what the applied voltage is. This causes current to flow into the rotor via the commutation.

Mechanical load

To overcome mechanical load (and friction) there must be a bigger torque generated by the rotor and that means that the induced voltage is a little more not quite what the applied voltage is.


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