# Eddy current induce EMF

Q1 Suppose there is a wire having AC current so as ac current is alternating it creates back EMF whether that back emf produces current?

If yes then that current is same as eddy current or not?

And that current will be alternating too so can it will creates another back emf and till this process continues or not?

Q2 genratore produces Ac current so that AC current be able to produce a motor action or not

Q3 motor have is continously moving in a magnetic flux be able to produce back EMF? If yes then that back emf will again produce current and according to law current carrying conductor have force acting on it in a magnetic field the force will again act on it. Whether that force be opposite to previous force or not

• Back EMF is not a current but a voltage. How much current that voltage creates depends on the resistance of the loop. Eddy currents are not in the wire loop but in conducting metal parts that are in the magnetic field it creates. The eddy currents do produce fields that are causing EMF in the wire and all of these do act on the dynamics of electrical machines. What that does, however, would be better discussed in electrical engineering as it is dependent on the design of the machines. There are electrical machines that use these eddy currents as their main functional principle. – CuriousOne Feb 1 '16 at 15:26

This is an answer to question 3 relating to simple motors and dynamos.

A simple motor and a simple generator are one and the same thing When a current flows through the coil of a motor the coil rotates in a magnetic field and so the coil acts as a generator with the induced electromotive force (emf from Faraday’s Law) in the opposite direction to the supply voltage $\mathcal{E}_{\text{supply}}$ (Lenz’s Law).

This is called the back emf, $\mathcal{E}_{\text{back}}$, and it is proportional to the speed of rotation of the coil $\omega$ - $\mathcal{E}_{\text{back}} = \text{constant} \times \omega$.

So we have $\mathcal{E}_{\text{supply}} I = I^2 R_{\text{coil}} + \mathcal{E}_{\text{back}} I$
which is power input from the power source making the motor rotate equal to the power lost due to the resistance of the coil $I^2 R_{\text{coil}}$ plus the external work done by the motor $\mathcal{E}_{\text{back}} I$.

$I$ is the current in the circuit and $R_{\text{coil}}$ is the resistance of the rotating coil.

Another useful way of writing this relationship is $I = \dfrac{\mathcal{E}_{\text{supply}} – \mathcal{E}_{\text{back}} }{R_{\text{coil}}}$

Just to show this is reasonable suppose the coil had no resistance and there were no losses due to friction at the bearings etc and the motor was doing no work.

You turn on the power supply the motor draws current and speeds up.

As it speeds up the back emf increases until it reaches a value equal to the supply voltage.
$I = \dfrac{\mathcal{E}_{\text{supply}} – \mathcal{E}_{\text{back}} }{R_{\text{coil}}}$ and $\mathcal{E}_{\text{supply}} – \mathcal{E}_{\text{back}} = 0$
No more current flows through the coil and the coil rotates at constant speed.
No power is needed to make the coil continuing to rotate because it has reached a constant rotational speed and there are no losses.

Now suppose that the coil has resistance and there is friction at the bearings of the coil.
The coil is at rest and the power supply is switched on.
The coil will speed up and finally reach a constant speed of rotation.

$\mathcal{E}_{\text{supply}} I = I^2 R_{coil} + \mathcal{E}_{\text{back}} I$

$\mathcal{E}_{\text{supply}} I$ is the power delivered by the power supply;
$I^2 R_{\text{coil}}$ is the power loss as heat in the coil resistance;
$\mathcal{E}_{\text{back}} I$ is the power required to do work against the friction at the bearings.

Now let's make the motor do some work by getting it to lift a load.
How does this work in terms of our equation?

$\mathcal{E}_{\text{supply}} I = I^2 R_{\text{coil}} + \mathcal{E}_{\text{back}} I$

Easy.
The rotational speed of the coil decreases.
So $\mathcal{E}_{\text{back}}$ decreases because for a generator the emf generated is proportional to the speed of rotation of the coil.
But $I = \dfrac{\mathcal{E}_{\text{supply}} – \mathcal{E}_{\text{back}}}{R_{\text{coil}}}$ so reducing $\mathcal{E}_{\text{back}}$ which increases the current $I$ in the circuit.
The power delivered by the power supply $\mathcal{E}_{\text{supply}}I$ increases.
The heat loss in the coil $I^2R_{\text{coil}}$ increases.
$\mathcal{E}_{\text{back}} I$ increases because $I$ increases proportionately more than $\mathcal{E}_{\text{back}}$ decreases.
This power does work against the bearing friction and also lifts the load.

Have you not noticed when using an electric drill the reduction in the sound frequency when the drill is asked to do some work like making a hole in a wall?

How do you make this motor a generator and make power go into the power supply.?

Easy.
$I = \dfrac{\mathcal{E}_{\text{supply}} – \mathcal{E}_{\text{back}}}{R_{\text{coil}}}$
If you can make $\mathcal{E}_{\text{back}} > \mathcal{E}_{\text{supply}}$ then current will flow in the opposite direction and flow into the positive terminal of the power supply – you have a generator.
You can do this by making the coil rotate faster by applying an external torque until the back emf exceeds the power supply voltage.
Electric vehicles use this idea when going downhill to recharge the batteries.
This also produces a breaking effect because the kinetic energy of the vehicle is being used to generate the electricity which is charging the batteries.

And finally, when is a simple motor most likely to be destroyed?
Answer - when it is first connected to the power supply.
Why?
Because when the coil is not rotating no back emf is produced and the current in the circuit is very high.

$I = \dfrac{\mathcal{E}_{\text{supply}} – 0}{R_{\text{coil}}}$

So in turn the heating effect due to the resistance of the coil is very large and the coil might burn out.
This is the reason you should not stall an electric motor.