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"When a current passes through an electric motor, the magnetic force on the motor causes a torque on the loop of wire causing it to turn". However, when the loop rotates, there should also be a change in magnetic flux causing an induced emf (Faraday's law). However, my book makes no such reference.

my questions:

  1. Is there going to be an induced emf opposing the applied emf of the DC source in the electric motor?
  2. If yes, wouldn't the induced emf cancel the applied emf causing no magnetic force to be exerted and thus no longer converting electrical energy to mechanical energy?
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Yes, there is an induced EMF called "Back EMF" or "Counter EMF". This is proportional to the rate of rotation of the coil; the higher the velocity of the coil the higher the counter induced EMF.

The counter EMF can be calculated by subtracting the impressed voltage ($V$) in the coil from the supply EMF coming from the power source:

$e = E - IR$, and $V = IR$.

Where, the supply EMF is: $E$, the back EMF is: $e$, the resistance of the coil is: R and the current through the coil is: $I$.

-Since ($e$) is proportional to the angular speed ($\omega$ ) the greater $\omega$ the smaller $I$.

There must be slightly more impressed voltage than Back EMF, to allow enough current to flow, to overcome frictional losses. Once a load is added, all this goes out the window, because the motor slows down, reducing the Back EMF, and allowing greater current to flow, and greater power developed to drive the load.

If a motor has zero friction losses, the maximum velocity can be achieved when the counter EMF is equal to the supply EMF.

A simple test you could perform at your car: With the window closed, lift the switch of the electric window in your car that is running at idle, and hold it momentarily and notice the idle RPM drop. The electric motor in the door is stationary and therefore the inrush current will be very high. The alternator will try to provide for the large current which subsequently drags down the engine. As soon as the power window motor overcomes its inertia and starts spinning, back EMF will be produced, exerting less load on the alternator. Hence, the engine speed will return to the normal operation. (Wikipedia)

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  • $\begingroup$ this was extremely helpful. Just a few things: does the coil in a motor rotate with a constant $\omega$ , if not, does this mean that the back emf will vary? Also, are you referring to air resistance when talking about friction losses? Finally, max velocity is reached when the two EMFs equal each other, however, in that case, there will be no current, no magnetic torque and thus how would the coil rotate? $\endgroup$ – Eliza Feb 14 '14 at 11:03
  • $\begingroup$ @Eliza Yes, constant ω would cause the motor to act as a generator. Since the coil cross through the flux lines. If the coil's ω vary so will the back EMF. By resistance I meant all sorts of resistance, since this is not an ideal case. However, the resistance of the air is normally negligible compared to that from the bearings of the rotor (the moving part). $\endgroup$ – Ray Feb 14 '14 at 11:55
  • $\begingroup$ Ok. I get it now. also, the magnetic force on the current carrying wire, is it also a constant value? $\endgroup$ – Eliza Feb 14 '14 at 11:59
  • $\begingroup$ @Eliza As for the max velocity, even though it's a never possible case, since there will always be losses that could be counted as load, you are right there is no current. But if the motor slows down (as you might think for the no current), the back EMF (proportional with the velocity) reduces, thus the supply current will be excess again and make up for the loss in velocity, and with it increasing the back EMF and so on, to reach back the ideal condition. So with that said, it's not possible for the motor to slow down in this case. $\endgroup$ – Ray Feb 14 '14 at 12:02
  • $\begingroup$ @Eliza No, the magnetic field on the wire is not constant as long as the coil is rotating, since the direction of the flux lines always change with the angle of the coil as it rotates - If I understood your last question correctly. $\endgroup$ – Ray Feb 14 '14 at 12:59
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I believe you are correct to assume that and it is called back emf. Modern day motors have corrected for this (or minimized it). The back emf is always smaller than the applied, or should be if you want it to rotate and produce power.

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