Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polarity of Induced EMF in a Conducting Ring

  1. Mar 13, 2012 #1
    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.
  2. jcsd
  3. Mar 13, 2012 #2
    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.
  4. Mar 13, 2012 #3
    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.
  5. Mar 14, 2012 #4

    Philip Wood

    User Avatar
    Gold Member

    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.
  6. Mar 14, 2012 #5
    Thanks Philip. The issue is with polarity. Typically a current flow from high to low potential.

    Is it because the electric field here is non-conservative (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.
  7. Mar 14, 2012 #6

    Philip Wood

    User Avatar
    Gold Member

    As you say, the electric field is non-conservative. I would say that the concept of p.d. is inapplicable, though I stand to be corrected. We still have: emf = IR.
  8. Mar 15, 2012 #7
    I agree with you and thanks for the clarification. Much appreciated.
  9. Mar 15, 2012 #8


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    An emf, by definition, has no current. It's based on Faradays Law, which is one of the fundamental Maxwell equations of electromagnetism. Using Heaviside-Lorentz 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 right-hand rule: Directing the thumb of your right hand in direction of the surface-normal 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.
  10. Mar 15, 2012 #9

    Philip Wood

    User Avatar
    Gold Member

    vanhees71, as always, is spot-on. 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!]
  11. Mar 15, 2012 #10
    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?
  12. Mar 15, 2012 #11

    Philip Wood

    User Avatar
    Gold Member

    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.
  13. Mar 15, 2012 #12
    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.
  14. Mar 15, 2012 #13
    The emf does not "cause" a current. In order to create the time-changing emf, a current takes place. The emf cannot exist w/o displacement current. Even if the loop is open, there is still a current. In the ac domain, an ac voltage cannot exist w/o an ac current. The 2 are mutual & inclusive. Neither exists alone. Neither is the cause nor the effect of the other. Does this help? BR.

  15. Mar 15, 2012 #14

    Philip Wood

    User Avatar
    Gold Member

    The only way I can make sense of this, and the rest of your post, is by assuming that you are using 'current' to mean not just conduction current but so-called displacement current as well. In my posts I have been using 'current' to mean ordinary conduction current. I like to think this is the normal usage.
  16. Mar 15, 2012 #15
    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
  17. Mar 15, 2012 #16
    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.
  18. Mar 15, 2012 #17
    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.

  19. Mar 16, 2012 #18

    Philip Wood

    User Avatar
    Gold Member

    Whatever our views on displacement current, I think it is misleading to use the unqualified word 'current' to include both displacement current and ordinary conduction current. That's all.
    Last edited: Mar 16, 2012
  20. Mar 16, 2012 #19
    I can't believe that there is no Lorentz force. For an emf to be induced, charges must be moved. Without Lorentz force, charges won't move in the 1st place. If charges can't move, there can be no induced emf. When moving a magnet, the magnetic field is accompanied by an electric field.

    Not to be offensive, but how extensively have you studied e/m fields? BS, MS, PhD, phy or EE? Just wondering what you base your statements on. You seem to be stating the following. Moving a magnet towards a loop results in NO Lorentz force, yet an emf is induced which results in a current if the path is closed. If that is your belief, then it does not agree with known laws. I have trouble understanding how electrons move w/o Lorentz force.

    Remember that under time changing conditions, E fields & H fields (or B if you prefer) cannot exist independently. A stationary magnet has only H/B, no D/E. But once in motion, E & H are both present. A stationary electron feels a force due to Lorentz force via the E field, F = qE. Although vXB = 0 since v=0, the E field still accounts for a Lorentz force.

    https://www.physicsforums.com/showthr...46#post2163546 [Broken]

    Please refer to the above link. In circuit terms I computed the relation between flux (internal & external), frequency, current, & voltage. If you disagree, then please provide the correction you feel is needed. BR.

    Last edited by a moderator: May 5, 2017
  21. Mar 16, 2012 #20
    Also, I forgot to mention that although the electrons are not initially moving, the magnet is moving wrt the electrons. There is relative motion between electron and B field, & I recall learning that relative motion is all that is needed for Lorentz force. An electron moving past a magnet, or a magnet moving past an electron, will both result in Lorentz force.

    Am i missing something?

    Last edited: Mar 16, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook