Electromagnetic induction > The motion of conductors in a constant magnetic field. Electric motors > Problem 11.1.27

Statement

$11.1.27.$

Why can an electric motor burn out if its rotor is stopped?

Solution

To solve this, let us first understand the normal operation (motor spinning)

· The motor has coils that rotate inside a magnetic field.
· According to Faraday's Law, as the coil rotates in the magnetic field, a voltage is induced in the opposite direction to the current feeding it. This voltage is called back electromotive force (back EMF, or CEMF).
· The back EMF acts as a "natural regulator": it opposes the source voltage and limits the current flowing through the coils.
· The resulting current is small (only what is needed to overcome the mechanical load and losses), and the motor runs stably without overheating.

Now let us analyze the case of a locked rotor (motor stalled)

If the rotor stops due to an overload or jam:

· There is no motion → no magnetic flux change → the back EMF disappears.
· With the opposition gone, the source voltage (V) is applied directly across the ohmic resistance (R) of the coils, which is very small.
· By Ohm's Law, the current surges to a very high value:
$ I_{\text{start}} = \frac{V}{R}$
· This current is 5 to 10 times higher than the rated current. All this electrical energy is converted into heat (Joule effect: $P = I^2 R)$.

As a result:
The extreme heat melts the insulating enamel on the wires, causes internal short circuits, and burns out the motor.

Answer

so the motor burn because the increment of current tranformed un heat by joule efect .