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Induction, magnetism and conductivity 
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#19
Dec1712, 10:24 PM

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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 shorthand 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 


#20
Dec1812, 01:00 AM

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Claude 


#21
Dec1812, 05:49 AM

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#22
Dec1812, 01:05 PM

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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 timedependent electric field of the electromagnetic wave, and is not induced by the timedependent 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 


#23
Dec1812, 01:23 PM

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AM 


#24
Dec1812, 04:28 PM

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#25
Dec1812, 05:06 PM

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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:



#26
Dec1812, 05:25 PM

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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 counterclockwise) 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 


#27
Dec1812, 05:58 PM

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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 & viceversa. 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 timevarying 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 timechanging 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 nonzero 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 


#28
Dec1812, 06:49 PM

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[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. AM 


#29
Dec1912, 10:58 AM

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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 


#30
Dec1912, 12:59 PM

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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.ndted.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/inducedemfi...docd258502729 You just need to "cut through flux lines". So if you move that wire updown there will be induced current, but not if you move it leftright. 


#31
Dec1912, 03:29 PM

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AM 


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