Resistance and Magnetism  A stock motor has 27 turns of 22 AWG wire, with a length of 64 inches. 22 AWG copper wire has a resistance of .001345 Ohms per inch, so the 64 inches of wire has a total resistance of 0.086 Ohms (0.086=64*.001345). Using ohms law V=IR (V/R=I) we calculate that if you connect 7.2 Volts (6-cell pack) to this wire, you will have a current of 7.2 Volts / 0.086 Ohms = 84 Amperes. This means that although the maximum current this stock motor can pull is 84 amps this can only occur when it is stalled or restricted from rotating.
 The strength of the coils magnetic field is proportional to the amps flowing in the wire and the number of turns in the coil. More current and more turns generates a larger magnetic field. Maximum torque will be produced when the motor is stalled since this is when maximum current is pulled.
 A conductor moving in a magnetic field will have a current produced in it, this is basic electromagnetism. This is current is often called back-EMF. This back-EMF opposes the current being supplied by the battery. As the armature RPM increases the current induced in the coils from the permanent magnets will also increase. This increase directly opposes the current flowing from the batteries. This means that as RPM's go up, the positive current flow in the coils goes down as does the torque.
Confused? Don't be, lets just look at the numbers again. If the above motor is drawing 5 Amps at 7.2 Volts then the voltage induced in the armature coils needs to be 7.2 - (0.086. * 5) = 6.77 Volts. This leaves 0.43 volts from the battery to act through the 0.086 Ohms armature resistance -- 0.43 Volts = 5 Amps * 0.086 Ohms (V=IR) If you increase the load on the armature, the RPM will drop and the voltage induced in the armature (due to its rotation) will be less, allowing more current to flow from the battery and more torque to be produced. Conversely if you decrease the load on the armature, the RPM will go up allowing less current to be drawn and less torque to be produced. | 
