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Polarity of Induced EMF in a Conducting Ringby samirgaliz
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#1
Mar1312, 12:26 PM

P: 10

I have a question regarding a conducting loop of radius r in a changing magnetic field B.
I understand and can determine the direction of the induced current which implies the existing of an electric field that is tangential to the loop. Since this is a closed loop, I am having trouble determining the polarity of the induced emf(potential difference). How would it be possible to measure experimentally this voltage or even use it to power up a light bulb? Any help would be appreciated. Thanks 


#2
Mar1312, 01:08 PM

P: 617

In a purely theoretical sense, ungrounded AC circuits do not have a stationary polarity. The emf, and therefore the induced current, oscillates in direction back and forth, and therefore the polarity continually switches as the system oscillates. In practice, something special is done to one leg of the circuit (such as grounding it) that is not done to the other leg of the circuit, and we therefore externally impose an effective stationary polarity on the AC system.



#3
Mar1312, 02:03 PM

P: 10

Thanks for your speedy reply. I forgot to mention that the loop is closed and stationary and the magnetic filed is continuously increasing in one direction perpendicular to the loop and thus the induced current does not change direction.



#4
Mar1412, 05:49 AM

PF Gold
P: 964

Polarity of Induced EMF in a Conducting Ring
Unless I've missed something, the answer is very easy. The emf is in the same direction as the current.
The conventional current is in the same direction as the electric field at each point in the ring. So work is done on the conventional (positive) charge carriers by the field. So the emf is in the same sense as the current. 


#5
Mar1412, 04:01 PM

P: 10

Thanks Philip. The issue is with polarity. Typically a current flow from high to low potential.
Is it because the electric field here is nonconservative (line integral along a closed path in this case is not = 0) and is created due to the change in magnetic field and not by charge separation as in the case of a battery? Thanks again for your help. 


#6
Mar1412, 04:51 PM

PF Gold
P: 964

As you say, the electric field is nonconservative. I would say that the concept of p.d. is inapplicable, though I stand to be corrected. We still have: emf = IR.



#7
Mar1512, 03:02 AM

P: 10

I agree with you and thanks for the clarification. Much appreciated.



#8
Mar1512, 03:50 AM

Sci Advisor
Thanks
P: 2,552

An emf, by definition, has no current. It's based on Faradays Law, which is one of the fundamental Maxwell equations of electromagnetism. Using HeavisideLorentz units it reads
[tex]\vec{\nabla} \times \vec{E}=\frac{1}{c} \frac{\partial \vec{B}}{\partial t}.[/tex] This Law can be written in integral form by integrating over an arbitrary closed loop [itex]\partial A[/itex], encirceling an area [itex]A[/itex]. With help of Stokes's Law, one gets [tex]\int_{\partial A} \mathrm{d} \vec{x} \cdot \vec{E}=\frac{1}{c} \int_{A} \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}.[/tex] Here, by definition, the boundary [itex]\partial A[/itex] of the surface [itex]A[/itex] is orientied positive relative to the surface by the righthand rule: Directing the thumb of your right hand in direction of the surfacenormal vectors (whose direction can be chosen arbitrarily!), the fingers point into the positively oriented tangent vectors of the boundary curve, [itex]\partial A[/itex]. This uniquely defines the signs in Faraday's Law in integral form. 


#9
Mar1512, 04:10 AM

PF Gold
P: 964

vanhees71, as always, is spoton. And, of course, the magnitude and direction of the emf don't depend in any way on the presence of a current. The ring could be an insulator. Nonetheless, if you have been taught how to predict the sense of a current in a conducting ring, then you can use the same rule to give the sense of the emf in a ring whether conducting or not. This procedure is more convoluted than starting from Faraday's law, but it does deliver the goods. [Actually, the rule for current direction can be justified in terms of energy conservation, which is pretty fundamental!]



#10
Mar1512, 05:35 AM

P: 409

Actually I have a similar problem. When I use Faraday's law, I can't say ,from the result , that what is the direction of emf. I need to use "right hand lay" for that. But in a case of a moving conductor in static magnetic field, since the line integral has a direction , I can say the direction of the emf without using the right hand law.
I have read that, the Faraday's law had no "  " sign at the beginning. The negative sign was proposed by Lenz. What does the negative sign tell us about the direction of the emf? 


#11
Mar1512, 09:18 AM

PF Gold
P: 964

You're right about Lenz's contribution. Faraday himself didn't use algebra; his mathematics didn't go beyond arithmetic and the concept of proportionality (possibly some geometry, too, but I don't think we've much evidence on this). So he didn't write the law as an equation, let alone use a minus sign.



#12
Mar1512, 11:56 AM

P: 1,506

I agree with Philip Wood, Faraday did not use high level mathematics to convey his findings.
He invented magnetic flux lines of force to explain what he discovered. The facts are easy to state (Faraday's laws) Whenever a conductor experiences a changing magnetic flux linkage and emf is induced. The induced emf = rate of change of flux linkage The emf opposes the changing flux linkage (usually known as Lenz's law) The  sign in an equation indicates 'opposition' this occurs in other aspects of physics (SHM etc) If there is a complete circuit the induced emf causes a (induced) current to flow. 


#13
Mar1512, 02:34 PM

P: 1,045

Claude 


#14
Mar1512, 04:58 PM

PF Gold
P: 964




#15
Mar1512, 05:46 PM

P: 1,506

If there is not a complete circuit there is no conventional current.
I was of the understanding that, in physics, when we say current we mean conventional current 


#16
Mar1512, 05:54 PM

P: 1,506

I would also go back to the original post......How is it possible to measure the voltage and power a light bulb...... It is easy, break the loop and connect a voltmeter or a lamp......it will Be easier to demonstrate the truth with a coil of several turns and a moving magnet.
Standard, basic school demonstration of Faradays laws.......no maths needed at this level. 


#17
Mar1512, 09:08 PM

P: 1,045

Per Maxwell, current is current is current. Displacement current is as real as is conduction current. Maxwell's equations includes both  ref Ampere Law. My point was that the emf is not the "cause" of the current. The only way the loop can have no displacement current is when it is a perfect short. But there is no emf here, yet there is a conduction current. Current in this case exists w/o emf. I was just clarifying what is happening.
Current in the loop is not caused by the emf, but rather, both are dependent on Lorentz force. Claude 


#18
Mar1612, 04:08 AM

PF Gold
P: 964




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