# How is there not an infinite regress in the EMF of a motor?

• OnAHyperbola
In summary: Force is related to stress in a similar way, in that they are both vectors. They are both measured in Newtons, and they both have a magnitude and a direction.
OnAHyperbola
Say we have a coil connected to a battery in a uniform magnetic field, perpendicular to the magnetic moment of the coil. This is a simple motor. There is a torque on the coil that varies with the angle θ between the field and the moment. Clearly, the angular velocity ω is not constant.

Here is my concern: since the flux through the coil is changing in time, there is a back emf in it. his will reduce the net emf, which will reduce the current. That will reduce the torque on the coil which in turn will change the angular velocity in time (I imagine it is the angular velocity that determines dΦ/dt in the first place) and that should change the back emf...one can see where this is going.

What is happening here? And how does one calculate dΦ/dt?

Thanks.

There are two back emf's:

1) The first one from self induction of the coil. This quickly drops to zero
2) The second one from the rotation of the coil in the external magnetic field (this is a mutual induction in case the external magnetic field is created by another coil). This second emf grows larger as angular velocity grows larger (due to the torque which is decreasing but still it is in the positive direction to increase angular velocity. In other words ##\frac{d\omega}{dt}## is decreasing but remains positive). At some point this emf is so large that it almost cancels the emf of the source, so the total current is very small and the torque is very small just to counterbalance any resistances/frictions.

cnh1995 and vanhees71
@Delta² Thank you for your answer! This is just the kind of explanation I was looking for! At the point where the torque is very small, is the angular velocity (more or less) at a maximum? Also, is there any way to calculate the back-emf as a function of time? I imagine that at this point, it too is at a maximum, and the current at its minimum, most stable value. Have I got it right?

Thanks so much!

Yes seems to me you got it right.

But to calculate exactly the back em as a function of time is not so easy. It depends mainly on the geometry of the coil (its exact shape, how many turns, area per turn) and the magnetic field of the stator.

The equation of the back emf is coupled to equation of the torque. if ##T(t)## is the torque then we ll have

##T(t)=Ia(t)=A\frac{d\omega}{dt}=A\frac{d^2\theta(t)}{dt^2}## (1)
where A is the moment of inertia of the coil and depends on the geometry of the coil as well.

Also we will have if E(t) is the back emf,

##E(t)=BScos(\theta(t))\frac{d\theta(t)}{dt}## (2)
where B is the magnetic field of stator and S is another constant that depends on the geometry of the coil(number of turns and area per turn) as well.

There are two more equations one involving how the Torque is generated from Laplace force due to the current in coil and it will be like

##T(t)=UBI(t)sin(\theta(t))## (3)
where U another constants that depends again on number of turns and radius of each turn in coil, and I(t) is the current in coil.

and the last equation is the differential equation for the current I(t) in the coil

##E-E(t)-L\frac{dI(t)}{dt}-I(t)R=0## (4)
where R is the ohmic resistance of coil

So all in all we have a system of four differential equations with four unknown functions, Torque T(t), back emf E(t), current I(t) and the angle ##\theta(t)##. As you see the equations are coupled in various ways, solving this system we can get all four of the unknown functions.

@Delta² I've got nothing to say, except that YOU ARE brilliant!
This had been bothering me for days, and I can finally have a good night's sleep.

Delta2
Delta² said:
B is the magnetic field of stator

Does B represent magnetic flux or magnetic flux density?

B is the magnetic field vector(not sure if we can call it magnetix flux density), the flux through a surface S will be BScos(theta), where theta the angle between the vector B and the normal n to the surface S.

I think I have a mistake in (2) and/or (3), both equations should have ##sin(\theta(t))## or ##cos(\theta(t)##, maybe there are other mistakes too, my post was a kind of "raw" attempt to show how even a simple motor involves solving a system of 4 differential equations.

Many thanks for your help. The physical quantity represented by B in textbooks seems to be sometimes flux (Wb), and other times flux density (T).

It may be analagous to force versus stress in mechanics. One is Newtons, and the other is Newtons per square metre.

David Lewis said:
The physical quantity represented by B in textbooks seems to be sometimes flux (Wb), and other times flux density (T).
In electrical machines, B is the flux density and 'Φ' is used to denote flux.

## 1. How does the EMF of a motor work?

The EMF (electromotive force) of a motor is a measure of the electric potential difference that is generated when a magnetic field and an electric conductor interact. In a motor, the magnetic field is created by the permanent magnets or electromagnets in the stator, and the electric conductor is the rotor, which carries current. The interaction between these two components results in the generation of an EMF, which can be used to power the motor.

## 2. Why is there not an infinite regress in the EMF of a motor?

The EMF of a motor is not an infinite regress because the magnetic field in the stator is constantly being replenished by the flow of electric current in the rotor. This creates a continuous cycle of energy conversion, where the magnetic field induces an electric current in the rotor, which in turn creates a new magnetic field, and so on. This process does not continue infinitely, as there are losses in the system that eventually lead to the motor's eventual stoppage.

## 3. How is the EMF of a motor related to the motor's speed?

The EMF of a motor is directly proportional to the motor's speed. This is because the faster the motor spins, the higher the rate at which the magnetic field in the stator is being replenished. This results in a higher induced electric current in the rotor, which leads to a higher EMF output. This relationship is known as Faraday's Law of Electromagnetic Induction.

## 4. Can the EMF of a motor be controlled or manipulated?

Yes, the EMF of a motor can be controlled and manipulated by changing the strength of the magnetic field in the stator or by changing the speed of the motor. For example, by increasing the number of coils in the stator or by increasing the strength of the permanent magnets, the EMF output of the motor can be increased. Similarly, by varying the speed of the motor, the EMF output can be adjusted to meet specific power requirements.

## 5. How is the EMF of a motor measured?

The EMF of a motor is measured using a voltmeter. The voltmeter is connected to the motor's terminals, and the voltage reading is taken. This voltage reading represents the EMF output of the motor. It is important to note that the EMF output of a motor may vary depending on the load on the motor, so multiple readings may be necessary to accurately measure the EMF output.

• Electrical Engineering
Replies
10
Views
1K
• Electrical Engineering
Replies
7
Views
2K
• Electrical Engineering
Replies
17
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
556
• Introductory Physics Homework Help
Replies
3
Views
1K
• Electromagnetism
Replies
27
Views
2K
• Electromagnetism
Replies
4
Views
1K
• Electromagnetism
Replies
10
Views
3K
• Electromagnetism
Replies
11
Views
2K