induction, magnetism and conductivity‏

MarkoniF

You have no idea and yet you make statements as if you are certain about it, that's strange. There would be induced current in a straight wire. Try it out.


http://www.ndt-ed.org/EducationResou...inductance.htm
- Faraday's Law for an uncoiled conductor states that the amount of induced voltage is proportional to the rate of change of flux lines cutting the conductor. Faraday's Law for a straight wire is shown below.




http://www.sweethaven.com/sweethaven....asp?iNum=0201
- Faraday's Law for a Straight Wire: The amount of induced voltage is proportional to the rate of change of flux lines cutting the conductor.



Help yourself and learn some basics. My question stands, and as I explained it's not addressed by Faraday's law.

Andrew Mason

Ok. I can see where the problem arises. There is a lot of misinformation and large misunderstandings out there about Faraday's law, as Prof. Lewin mentions. It is not your fault. Your attitude, however, could use a improvement if you are serious about learning physics. I don't mind, but others on PF will take you to task for it!

A lot of 'theory' written for high school students and engineers is wrong or at best confusing or misleading and it is difficult to sort out what is right and what is wrong.

I am not sure where this idea originates that the voltage induced in a wire is determined by the number of lines of flux per second 'cut' by the wire. Faraday's law states that the voltage induced around a closed path is equal to the time rate of change of the magnetic flux through a surface enclosed by the path:

[tex]\oint \vec{E}\cdot \vec{dl} = \frac{d}{dt}\int \vec{B}\cdot d\vec{A}[/tex]

Engineers may have developed a short-hand conceptual way of thinking about this using flux 'cutting' a conductor but you should stick to Maxwell's version because it is always correct and clear. The engineering approach appears to have made its way into high school and engineering texts but, in my view, it is confusing if not wrong.

For starters, it suggests that a current can be induced in a conductor that does not form part of a circuit. The authors certainly know that cannot happen but that is a conclusion that might be drawn from what they have written.

But here is the real problem with that explanation. Flux is magnetic field x area. So the rate at which flux 'cuts' a conductor involves some concept of a magnetic field and area. It is not clear what the area part is.

If you take a square loop of conductor wire and accelerate it through a uniform magnetic field each side of the loop will 'cut' lines of flux at an increasing rate. But the voltage induced in the wire is always 0 according to Faraday's law because the rate of change of [itex]\vec{B}\cdot \vec{A}[/itex] through the area enclosed by the loop is necessarily 0. Now, of course, using the "rate at which the conductor cuts lines of flux" approach you would say that is because the directions of the voltage in each of the opposite sides is the same so they cancel each other out. That works. But I am not sure that it is correct.

I will start a thread and get some discussions going on this point, which is an interesting one.

AM

cabraham

Actually, a straight wire such as a dipole antenna when subjected to a time-varying magnetic field of high enough frequency will have induced current. A closed path or loop is not needed for current to exist. I'm old enough to remember 1970's cars equipped with windshield antennae. They were 2 L shaped wires not touching, neither wire forming a closed loop. Current was induced into the wires. Look it up, current does not need a closed path in the ac domain. It's called displacement current.

Claude

MarkoniF

It's not just a theory, it's reality. You can induce electric current in a straight wire by moving permanent magnet around it. Try it out.



What are you talking about? You seem to have problem visualizing how a straight wire can be a part of a circuit.


No, that's not correct. Direction of the current would not be opposite at opposite sides, it would be either clockwise or anticlockwise along the whole loop.

Andrew Mason

You are quite correct that a current does not need a closed path in the ac domain - but an induced current does. An electric dipole antenna is a good example.

My understanding is that the electric dipole antenna responds to the electric vector in the electromagnetic wave. In other words, the current flows in response to the time-dependent electric field of the electromagnetic wave, and is not induced by the time-dependent magnetic field. The antenna must also be coupled to a capacitor, otherwise there would be no flow at all.

My understanding is that the same applies to the displacement current. The displacement current derives from Ampere's law not Faraday.

AM

Andrew Mason

You can certainly induce a current in a straight wire by moving a magnet around it, provided the straight wire is connected to a circuit that encloses the magnetic flux of the magnet (ie. encloses an area through which there is some component of the magnetic field at right angles to the area). I think I have made that part clear.

The physical direction would be the same (eg. left to right) but since they form part of a circuit this would be in opposite directions around the circuit. In the example I gave, there is no current because there is no change in the magnetic flux enclosed by the loop.

AM

MarkoniF

No. Try it, learn it.


No. That's terribly wrong. There is no any left and right or up and down, only clockwise and anticlockwise.

zoobyshoe

Here's an example of a current being induced in a straight wire in an experiment by Faraday. The straight wire is part of a circuit but the only effectively moving part is the straight wire. The sides and bottom don't "cut" the field lines as the wire is flipped over. The magnetic field is, in this case, supplied by the earth:



http://www.gutenberg.org/files/14986...86-h.htm#toc_8

Andrew Mason

So you are telling us that Faraday's law is wrong???

There is no current. You have to read carefully what I wrote. I gave the example of a rectangle of wire sitting in a uniform magnetic field with the plane of the wire is perpendicular to the direction of the magnetic field. The rectangle is accelerated in a direction parallel to the plane of the wire. There is no induced current because d∅/dt is 0 ie. BA = constant. So Faraday's law says that there is 0 induced emf around the circuit and no current.

Now the "rate at which the conductor cuts flux" explanation would say that the leading edge has an induced voltage (eg. say clockwise - it depends on the direction of B and the acceleration) and an equal induced voltage in the trailing edge (which is necessarily counter-clockwise) in order to achieve that 0 current. Same result but a bit different explanation. The question is whether this is a correct model of the physics.

AM

cabraham

I can't believe what you're saying. Have you had formal e/m field theory classes? Induced current needs no closed path at all. An antenna current is induced. As far as electric & magnetic fields go, they are inseparable, as under ac conditions, neither one can exist w/o the other. The E & B fields both exert a force on the charges in the antenna loop. B exerts normal force, E exerts tangent force.

It is indeed induced. Is it induced by E or B is a moot question as they cannot exist alone. A good example is a static example where a pair of plates is charged forming a capacitor. A wire loop is inserted in between the plates. There is an E field between the plates due to charge polarization on the 2 plates. Said E field moves charges in wires so as to create polarization. The electrons move towards the positive plate & vice-versa. But current does not circulate in the loop.

Here we have charges moving under E w/o a B force, but continuous current is zero. The E field has no curl because no varying B is present. If the same loop is immersed in a time-varying B field (magnetic), the associated E field has curl, aka "rotation". Here charges circulate in the loop due to this type of E field. But a rotational E field does not exist w/o a time-changing B field. The charges flow in the loop due to both fields, E & B. E moves an electron tangentially, but in the case of the loop in between cap plates, E alone will not produce circulation. The B force is normal to charge particle velocity keeping it in a circulating motion.

To say that induction is due solely to E or to B is incorrect because you need them both. When we say that induction happens when the magnetic field is time varying, it is implicit & understood that there is an accompanying E field which exerts the tangential force needed for charge circulation. "Induction" does not imply that E is irrelevant.

As far as antennae needing to connect to a capacitor, that is incorrect as well. Any wire has capacitance between itself & the reference node, i.e. ground. Also, a dipole antenna in space has finite area in the ends of the wires, forming a capacitance.

A time varying B field is always accompanied by a non-zero rotational E field. Each field exerts a force on free charges in the loop moving them around the loop. They both are involved, E & B, neither being more or less important. The loop need not be closed, as some capacitance is always present. Induced current exists in an open or a closed path.

Claude

Andrew Mason

We were talking not about an antenna loop but a straight antenna wire. Perhaps you could explain how a stationary charge responds to anything but an electric field? Once a current starts flowing the charges will interact with the magnetic field through the Lorentz force. Is that what you are referring to?

But there is an electric force on the charges. Since the charges within the conductor distribute themselves to create equipotential surfaces in the wire and 0 field inside the conductor, I am not sure why anyone would expect a current. If the polarity of the plates kept changing (fast enough), would there not be tiny charge oscillations?

In fact the E field is related to the time dependent B field by Faraday's law:

[tex]\nabla \times \vec E =-\frac{\partial \vec B}{\partial t}[/tex]

I understand that. The antenna issue is a side issue to the point we were discussing and I will concede that I have not spent a lot of time studying antennas.

The point we have been discussing is whether one can measure the emf (or E field x length) induced in a straight conductor by passing a magnet near it by attaching a voltmeter of some kind (eg. galvanometer). My point was the induced voltage measured by a voltmeter depends on how you configure the leads and, in particular, an area enclosed by the leads and the conductor and the orientation of that area to the direction of the magnetic field. The voltage reading that you get depends on the way you connect the voltmeter. That was the point that seems to make MarkoniF apoplectic but it is just an obvious consequence of Faraday's law. If I am wrong on that, perhaps you could enlighten me.

Quite correct. I was trying to point out that the current flow requires capacitance which, in my understanding, is why capacitors are coupled to an antenna - to get a detectable current (other capacitors are needed to create a resonant circuit for tuning frequencies but that is a different function).

AM

Andrew Mason

This is an illustration of Faraday's law but it is also easily explained by the Lorentz force or Lorentz field (v x B) because you are moving the free charges in the top wire in a direction that is not parallel to the direction of B so v x B is non-zero. Since the charges are free to move around the circuit, the v x B E-field produces current in the wire which is detected by the galvanometer.

In the situation proposed by MarkoniF the wire is stationary and you draw a magnet past the wire. So there is no v x B electric field. Explanation requires Faraday's law. When applying Faraday's law, it becomes apparent that the length and orientation of the other wires will affect the reading on the galvanometer. So the galvanometer or voltmeter cannot be used to determine emf induced in the straight section of wire. The galvanometer reading reflects the induced emf in the entire loop.

AM

MarkoniF

Don't you have a voltmeter, piece of wire and a magnet? Try it out and you will see. How else do you expect anyone shall convince you? I already gave you two links that confirm current can indeed be induced in a straight wire. You think people around the world just hallucinated that and came up with the same equation? Here is some more:

http://www.patana.ac.th/secondary/sc...htm#inducedemf
- "The induced electromotive force across a conductor is equal to the rate at which magnetic flux is cut by the conductor."

http://www.ndt-ed.org/EducationResou...inductance.htm
- "Faraday's Law for an uncoiled conductor states that the amount of induced voltage is proportional to the rate of change of flux lines cutting the conductor. Faraday's Law for a straight wire is shown below."



http://www.kean.edu/~asetoode/home/tech1504/acdc/ii.htm
- "The amount of induced voltage is proportional to the rate of change of flux lines cutting the conductor."

http://www.schoolphysics.co.uk/age16...tor/index.html
- "When a straight conductor is moved through a magnetic field an e.m.f. is induced between its ends."

http://www.saburchill.com/physics/chapters/0059.html

http://www.xtremepapers.com/revision..._induction.php

http://ebookbrowse.com/induced-emf-i...doc-d258502729





You just need to "cut through flux lines". So if you move that wire up-down there will be induced current, but not if you move it left-right.

Andrew Mason

Faraday's law is a fundamental part of electromagnetic theory. If you think I am failing to appreciate that, you are mistaken. You do not seem to be able to read or to understand what I have written.

AM