I couldn't find the explanation for this query of mine. Why back emf in dc motor is nearly equal to the applied voltage on no-load? I tried to search on the internet but all in vain.
You can think of the motor as behaving like a resistor connected to a generator (the latter corresponding to the back EMF). It produces a torque that is proportional to the current through the resistor.
The higher the difference between the back EMF and the input voltage, the higher the current. The higher the current, the higher the torque.
If the motor is stalled, the entire input voltage appears across the resistor and a lot of torque is produced. Since it only takes a bit of torque to overcome losses such as bearing friction and windage and magnetic losses when there is no load, the motor spins up until the back EMF is almost equal to the input voltage and the torque has dropped until the motor is no longer accelerating significantly. The current resulting from the remaining difference (multiplied by the input voltage) gives you the power consumption with no load. Efficiency is zero, of course, with no load since it is producing zero output power at the shaft.
The losses in the copper windings are the current squared times the resistance (which is the winding resistance) aka \$I^2R\$ losses.
If the windings had zero resistance the back EMF would always be equal to the input voltage, (even under load) since any difference would allow infinite current to flow, generating infinite torque.
Because electric motors are pretty efficient.
When a motor spins, it acts like a generator. When it spins due to a applied voltage, the spinning generates a voltage internally that is opposite of what you applied. What keeps the motor spinning is the difference of the applied voltage minus this internal EMF.
This is why the no-load motor speed self-regulates as a function of voltage. At first the motor is not spinning, so all the applied voltage goes to making it spin. As the motor speeds up, the back EMF becomes stronger, so there is less and less voltage left to spin the motor. Eventually, the motor goes so fast that the back EMF cancels enough of the applied voltage to only keep it spinning at that speed.
The more efficient the motor, the closer that back EMF is to the applied voltage. Since the motor is theoretically doing no work, it should take no voltage to keep going. Real motors have real friction and other losses, so it will always take some finite fraction of the applied voltage to keep it spinning even with no load.
All the replies in this post are correct however no one seems to want to realise or admit that when a DC electric motor runs at high speed it wastes most of the applied voltage to the curse of back EMF. It is only the magnetising current flowing thru the coils that is doing the work of producing mechanical output power, the very same current is also producing heat in the coils equal to (I2R), the power that is actually doing work is only the voltage drop in the coils multiplied by the current, the back EMF destroys all the other input power. We normally say the copper losses cause waste heat production, but I say it is all the "loss" that is simultaneously doing the work.
Anyone can prove i am correct by doing a locked rotor test on a motor, put a calibrated test current thru the motor and test the voltage drop of the stator and field coils and do you own calculations. Standard electric motor are horrendously inefficient! Electric vehicle manufacturers would do much better to focus on the problems with Electric motors. Every motor should be able to exceed 100%, the laws of thermodynamics are seriously flawed.