# Why is there an induced EMF in this case?

• Salman72
In summary: I need to look at the entire loop?Yes, the physics still applies if you only look at each half of the path.
Salman72
Hey people

I was wondering why is an emf induced in a loop which moves perpendicular to a uniform magnetic field. I mean, say the loop is a square of area r^2, and moves in the xy plane, where there is a uniform magnetic field B in the z direction. Why will an emf be induced? The flux remains constant right? So if d(phi)/dt is 0, why is an EMF still induced??

Thanks

Does the loop remain completely inside the region where the B field exists? Or, to put it another way, does the B field always "fill" the loop? If this is true, then you are correct, there should be no induced EMF. Can you describe the example or situation that claims there is an EMF?

Thanks a lot!

Yes, the loop does remain completely within the field. It was just "coming" to my mind that an emf will be induced... however, I think it was because of a flawed intuition!... Anyway, will it be right to conclude that there will be no net EMF in the loop, but there will be equal and opposite EMFs (given by Blv) in the opposite sides of the loop?

And on a related note: If I move a straight piece of conductor in a uniform magnetic field, again an EMF is induced (proved using B*l*v I think). However, where is the flux change in that? I mean, there is no flux to begin with, since there is no loop- so how can we prove the EMF using Faradays Law?

Salman72 said:
Thanks a lot!

Yes, the loop does remain completely within the field. It was just "coming" to my mind that an emf will be induced... however, I think it was because of a flawed intuition!... Anyway, will it be right to conclude that there will be no net EMF in the loop, but there will be equal and opposite EMFs (given by Blv) in the opposite sides of the loop?

And on a related note: If I move a straight piece of conductor in a uniform magnetic field, again an EMF is induced (proved using B*l*v I think). However, where is the flux change in that? I mean, there is no flux to begin with, since there is no loop- so how can we prove the EMF using Faradays Law?
If a straight conducting rod is moving in a uniform magnetic field, in a direction perpendicular to the field, the induced emf is not due to a change in flux. It is because charges collect at the edges because of the Lorentz force. This is called motional emf.

Chandra Prayaga said:
If a straight conducting rod is moving in a uniform magnetic field, in a direction perpendicular to the field, the induced emf is not due to a change in flux. It is because charges collect at the edges because of the Lorentz force. This is called motional emf.

But it is due to a change in flux. The conducting rod forms only one part of the circuit that forms a complete loop.

cabraham
stedwards said:
But it is due to a change in flux. The conducting rod forms only one part of the circuit that forms a complete loop.

No, it is due to charges moving in a magnetic field. Change in flux is a rule of thumb. Notice that if the charges all move in one direction (i.e. down, not clockwise) then the emfs of each half of the loop will cancel. You'll just have some charges collecting at the top/bottom of the loop.

No, it is due to charges moving in a magnetic field. Change in flux is a rule of thumb. Notice that if the charges all move in one direction (i.e. down, not clockwise) then the emfs of each half of the loop will cancel. You'll just have some charges collecting at the top/bottom of the loop.

I have no idea what any of this means. Apparently Faradays Law $\oint E \cdot s +d\Phi_{B}/dt$ is a "rule of thumb".

Faraday's Law in integral form requires a closed path. It can't be applied to an open path.

jtbell said:
Faraday's Law in integral form requires a closed path. It can't be applied to an open path.

Yes, that's why I used the $\oint$. The open path wasn't made clear to me in the post. Thanks for clarifying.

stedwards said:
I have no idea what any of this means. Apparently Faradays Law $\oint E \cdot s +d\Phi_{B}/dt$ is a "rule of thumb".

What this means is that the physics is simply about moving charges in magnetic fields.

Yes, apparently Faraday's Law is a rule of thumb, since it only describes conveniently closed paths. Do you think the physics breaks down if I only look at each half of the path independantly?

What this means is that the physics is simply about moving charges in magnetic fields.

Yes, apparently Faraday's Law is a rule of thumb, since it only describes conveniently closed paths. Do you think the physics breaks down if I only look at each half of the path independantly?

Again, say what you mean. I'm not a mind reader. Do you have anything to add beyond what jtbell said?

stedwards said:
Again, say what you mean. I'm not a mind reader. Do you have anything to add beyond what jtbell said?

I suppose I am not being eloquent enough for you. My apologies.

I am trying to make you get it, namely that Faraday's Law is a rule of thumb for predicting the net result of the Lorentz forces on the charged particles in a closed loop of conducting material. This is not much beyond what jtbell and Chandra said, but it seemed like you were still having difficulty in accepting it.

To re-illustrate the example I used: Say you have a uniform magnetic field pointing out of the page everywhere. If you drag a loop of conducting wire that is normally oriented to it towards the right, there will be a downward force on the positive charges, and an upward force on the negative ones, as per the Lorentz force law.

The charges will migrate up and down respectively, establishing an electric field that will balance the effect of the magnetic field and movement, i.e. charges will migrate until the Lorentz force F = q [ E + v x B ] is equal to zero.

So you will have a motional emf as Chandra said, a voltage difference between the top and bottom of the loop. You will not have an emf around the circuit though, since the endpoints of any closed path will have the same voltage. This is the meaning of Faraday's Law, as jtbell said: it only applies to the whole closed loop -- when flux is constant as in this case there is no net emf around the loop.

I hope that helps!

stedwards said:
But it is due to a change in flux. The conducting rod forms only one part of the circuit that forms a complete loop.
If it is a complete loop, and the loop is entirely within the uniform magnetic field, moving the loop in a direction perpendicular to the magnetic field does not change the flux, and there is no EMF. EMF is there only if the flux changes, as stated in Faraday's law.

## 1. Why is there an induced EMF in this case?

There are several possible reasons for an induced EMF (electromotive force) in a given case. One common cause is the movement of a conductive material through a magnetic field, creating a change in magnetic flux. Another possible cause is a changing magnetic field itself, which can induce an electric field and thus an EMF.

## 2. How does induction create an EMF?

Induction is the process by which a changing magnetic field can create an electric field in a conductor. This electric field can then cause a flow of electrons, creating an induced EMF. This phenomenon is described by Faraday's law of induction.

## 3. What factors affect the magnitude of the induced EMF?

The magnitude of the induced EMF depends on several factors, including the strength of the magnetic field, the speed and direction of the movement of the conductor, and the angle between the magnetic field and the conductor. The number of turns in a coil of wire can also affect the magnitude of the induced EMF.

## 4. Why is induced EMF important in electrical engineering?

Induced EMF is important in electrical engineering because it is the basis for many devices and technologies, such as generators, transformers, and motors. It also plays a crucial role in the transmission of electricity through power lines, where changes in magnetic fields are used to induce currents.

## 5. How is induced EMF different from a regular EMF?

Induced EMF is different from a regular EMF in that it is created by a changing magnetic field, rather than by a chemical reaction or a battery. Induced EMF is also temporary, lasting only as long as the changing magnetic field is present, whereas a regular EMF can be maintained as long as the power source is active.

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