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I Field Energy of Permanent Magnets vs. Magnetic Monopoles

  1. Feb 19, 2016 #1
    When two permanent magnets attract, their magnetic fields reinforce each other, whether they are stacked on their ends, in which case they are pointing the same direction, or whether they are stuck on their sides, in which case the two magnets are pointing in opposite directions. Everyone here knows that a compass will line up in the same direction as the applied magnetic field.

    Since the total magnetic field energy is based on the square of the magnetic field strength, it would appear that the mean square of this field would increase as these magnets approached each other (or at least rotated to have mutual magnetic alignment), maximizing the average alignment between the magnetic fields over the whole spatial domain. Since this process releases kinetic energy, I would have to assume that magnetic potential energy should have gone down in this process. But from what I see according to the definition of the magnetic energy density based on (1/2)B^2/mu_0, which is not preceded at all by a negative sign, the energy density in the magnetic field, according to this, would have increased, not decreased. If both the magnetic field energy and the kinetic energy are increasing as two permanent magnets approach each other, it would beg the question of where the compensating energy change lies.

    It is quite clear though if I have two electromagnets that in order to generate a magnetic field I have to expend electrical energy to develop the field in the first place, and then when I allow the solenoids to attract each other, the increase in the magnetic field energy energy due to (1/2)B^2/mu_0 is not without the introduction of back-emf which is clearly necessary to preserve the energy conservation. In this case, it is very clear that (1/2)B^2/m_0 (as well as kinetic energy) increased at the expense of my input electrical energy.

    With the permanent magnets, we do not have this simple explanation as to why both these occur. Potential energy is apparently expended in order to bring the magnets to acceleration, but how could both the magnetic potential energy and kinetic energy increase in this process without something else in the picture? Something that is somehow expended as the permanent magnets are brought together?

    I still think it makes sense that magnetic potential energy decreases as the permanent magnets come close together under attraction, just as we might find with approaching solenoids whose back-emf reduces the magnetic field of each solenoid, reducing the power consumption, but not the total energy already invested into creating the magnetic field. The problem is that there does not appear to be any mechanism by which the attracting and approaching permanent magnets could generate emfs into each other that could in any way act against the intrinsic electron magnetic moments. So I am left with a suspicion that the magnetic potential energy change would be negative.

    Now, if I look at the potentials formulation of electromagnetism, it is quite evident to see where we might get a negative change in potential energy, say, if we have a moving charge q with velocity v in a solenoid acting against the magnetic field of a permanent magnet, which possesses a vector potential A. Reason being, the sign of the potential energy depends on the sign of q and A, as well as the cosine of the angle between v and A. However, using (1/2)B^2/mu_0 for the magnetic energy density appears to only give us a positive change in the energy density as the two magnets approach each other under attraction. The same problem arises if we use the expression for the magnetic energy density for linear, non-dispersive materials, (1/2)(BH), where B lines up with H.

    To make matters worse, with permanent magnets, instead of a charge q with velocity v, we instead just have many pairs of electrons where one electron with its intrinsic magnetic moment orients in line with the field of the other (and vice versa), in a process which should release kinetic energy. But from what? And how does that fit into the potentials definition of the magnetic field energy?

    Lastly, let us consider the case of magnetic monopoles, which are said to be analogous to electric charge, in that opposite poles attract and like poles repel. But how do we know that the opposite is not true? Remember that in the case of attracting permanent magnets, when they come closer together (or simply come into alignment), the mean square magnetic field rises, and therefore the magnetic field energy also rises. I would think this would be a binding energy, even if one were to replace the two permanent magnets with two solenoids with current. So the attractive forces which bring the magnets together will over time lead to decrease of the potential energy, if by potential energy we actually mean the potential energy that is converted into kinetic energy. If we applied the same logic to magnetic monopoles, then like magnetic monopoles should attract, as their merger would increase with the mean square magnetic field, analogous to the attracting permanent magnets. If so, we would, in converse, expect that opposite magnetic monopoles should repel. This makes magnetic monopoles less like an analog to electrical charges exerting Coulomb forces, and more like an analogy to gravitating masses. I have never found this to be described in the literature.

    I know these are a lot of questions and concerns, but if you don't mind to answer at least some of them, that would be greatly appreciated. Thanks to all in advance.
  2. jcsd
  3. Feb 20, 2016 #2


    Staff: Mentor

    Actually, when they atract their magnetic fields mostly cancel each other.

    Which is precisely the reason that the original statement must be untrue.
  4. Feb 20, 2016 #3


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    The magnetic energy density is not based on B2, but on B2/μ ( not μ0 )

    ½*B2/μ = ½ * B * B/μ = ½ * B * H, so the magnetic energy density is also based on μ.

    If you approach a piece of unmagnetized iron to a magnet, the iron will fill a volume wherein the magnetic energy density is smaller than outside said volume, because the relative permeability as for iron is, say 1000. So outside said volume, the magnetic energy density will be

    Edens = ½ * B * B/(μ0)

    whereas inside said volume it will be

    Edens = ½ * B * B/(μ0 * 1000)

    So approaching the piece of iron, the magnetic energy density will be weakened in said volume, hence the total energy.

    This lost magnetic energy is converted to kinetic energy in the iron.
  5. Feb 20, 2016 #4
    So if one were to insert a permanent magnet inside a coiled tube of wire and pass current through that wire, then what you are saying is that the permanent magnet would seek to cancel the cumulative magnet field generated by the wire, sort of like how an electric field generated inside capacitor's dielectric would cancel the electric field of the plates which polarized the dielectric? That's quite the opposite of how I was taught.
  6. Feb 20, 2016 #5
    When the iron is not magnetized, B should be zero inside of it. When it is magnetized, B should not be 0 inside of it. Since you are squaring B, then clearly the average square of B would rise when the iron is magnetized.

    In the case of my opening post, none of the scenarios involved situations where the differential permeability would necessarily differ significantly from unity. Example: Neodymium magnets (internal magnetization doesn't change much relative to the applied magnetic field [at room temperature]) vs. a magnetized steel rod (internal magnetization changes greatly to the applied magnetic field).
  7. Feb 20, 2016 #6


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    No, B is not physically squared, but mathematically:

    Edens = ½ * B * H , H = B / μ →

    Edens = ½ * B2 / μ

    The squared B is just a mathematical reformulation of the equation. Physically the B is not squared.
    Approaching a piece of iron to a cylindric magnet will not change the B-field significant in the surroundings of the magnet.

    Example using ohms law:

    P = V * I , I = V / R →

    P = V2 / R

    The voltage is not physically squared. There is no (square) transformer involved.
    Last edited: Feb 20, 2016
  8. Feb 20, 2016 #7
    So physically, is the B of the applied magnet inducing the H in the iron, or is the H of the applied magnet inducing the B on the iron?

    In the former case, I can see how the B of the applied magnet is a sense "reduced" by the soft magnetic material which has a relativity permeability above unity, so (1/2) B * H (using * for the dot product) should be less than (1/2) B * B / mu_0.

    In the latter case, the H of the applied magnet is in a sense "increased" by the presence of the soft magnetic material. This is clearly the case for transformer, where the magnetic field of the current (the applied field) is clearly less than the field generated into the permeable core, which stores energy analogous to how a water tank stores water.
  9. Feb 20, 2016 #8


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    H is inducing B, like voltage is inducing current.
    No, Ampere's law says:

    circulation H⋅ds = N*I

    Ampere says nothing about "soft magnetic material"/permeability. The H-field will be the same, high or low permeability.

    But B = μ * H will change due to changed permeability, when H is constant.
  10. Feb 20, 2016 #9
    So in other words, the energy density is:

    (1/2) mu_2 (H_1 * H_2), where "1" indicates the source of the applied field, and "2" indicates the permeable material affected (at the point where the energy density is measured).

    Equivalently, this is:

    (1/2) (H_1 * B_2)

    However, what you said before sounded like:

    (1/2) (B_1 * H_2)

    Since you were dividing by the permeability of the material (i.e. mu_2).

    Or did you mean:

    (1/2) (B_2 * H_2)
  11. Feb 20, 2016 #10


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    What do you mean by: B_1, H_2 , etc.?
  12. Feb 20, 2016 #11
    Say I have two permeable magnetic materials "1" and "2", so both have relative permeability greater than unity.

    As you say, the H fields are not in anyway dependent on the presence of permeable magnetic material.

    B_1 is different than mu_0 (H_1) inside "1", but it is the same outside of both "1" and "2".

    B_1 is a function of position. So B_1 is weaker outside of "1" than it is inside "1", for two reasons:

    1) The field strength decreases with distance

    2) The permeability is lower outside

    H_1 is a function of position. So H_1 is weaker outside of "1" than it is inside "1", for one reason:

    1) The field strength decreases with distance

    Permeability isn't a part of this.

    So we have the field H_1 of "1" at the position where we wish to calculate the energy density. Let's say that position is inside of "2".

    But say "2" has a relative permeability of 1000.

    So we have the applied field H_1 from "1" affecting a domain inside of "2". This domain, like the rest of object "2" has a high permeability.


    (1/2) mu_2 (H_1 * H_2)


    (1/2) (H_1 * B_2)

    is our energy density in this case, where H_1 is the local magnetizing field, and B_2 is the local flux density. As you can see here, this quantity would increase with greater permeability, just as you would expect with current wrapped around an iron nail.
  13. Feb 20, 2016 #12


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    I'm sorry, but I'm not able to reply on that long post, due to that english is not my mother tongue. ( See my profile ).

    But I have some comments:
    The B-field is to be compared to electric current. The currents through two different resistors in series are the same ( Kirchhoffs's 1. law ). So the B-fields will be the same just outside/inside "1" ( no significant difference in distance ), but the H-fields will not, due to different μ-values ). This surface, that separates inside from outside is a pole. A pole is a discontinuity in permeabilty. That's why there is no poles in a toroid, though it may contain a magnetic field.

    Say you have a liquid wherin a piece of iron is submerged. The relative permeability is the same in liquid/iron. Now you induce a strong inhomogenous magnetic field through the liquid/iron: The iron will never be attracted by this field because a movement of the iron will not make a difference in the energy density. There is no discontinuity as for the permeability.

    Say you place a piece of iron inside a homogenous magnetic field, the iron will not be attracted anywhere, because if the iron moves, it will just move the location of a lower/higher energy density, but the total energy as for the all over system will be the same. Hence no lost magnetic energy to be converted to mechanical energy.
    This is not right: If μr > 1 inside "1", the H-field will be weaker inside "1", because H = (constant) B / ( μ0 * μr ).

    I hope somebody, speaking better english, will give you a more detailed answer.
  14. Feb 20, 2016 #13


    Staff: Mentor

    The 1 and 2 subscripts don't matter. The energy density at each point is given by the fields and the material at each point regardless of whether the field is an "applied" field or not.
  15. Feb 20, 2016 #14


    Staff: Mentor

    Yes. The force on the magnet will be in the direction that minimizes the cumulative field energy.
  16. Feb 20, 2016 #15
    According to this then:

    If B is fixed, then an increase in permeability would cause H to fall. That's because B=mu*H. So for a given applied B, H would be inverse to the local permeability mu.

    If B were a function with time, then introducing a source of B closer to a separate material with permeability, would cause H to fall where the permeability is high, reducing the energy density according to (1/2) H * B, where H and B are those of the source of magnetism one is using to apply the field, assuming that B were the field being applied.

    However, consider that "hard" magnetic materials can have a magnetic permeability close to unity, and with those, there is no excess permeability (i.e. magnetic susceptibility) to reduce the field energy in the way you suggest here.

    In any case, how can B be independent from mu, when earlier you said:

    Where according to this, it is clear that H is being applied (i.e. it's the thing which is fixed).
  17. Feb 20, 2016 #16
    Do you agree that there is a potential energy (component) between two permanent magnets in some way based on the dot product between field lines, integrated over the volume of interest?


    |C|^2 = |A-B|^2 = |A|^2 + |B|^2 - 2|A||B|cos(theta_AB)

    Which is the simply the cosine law for a triangle.

    In a location that the field vectors A and B are unaligned, cos(theta_AB) would be negative.

    |A|^2 would be due to the source which generates A.

    |B|^2 would be due to the source which generates B.

    - 2|A||B|cos(theta_AB) would be the interaction energy, which would be positive in the case that A is opposed to B.

    |C|^2 would be due to taking the norm of the difference between A and B.

    - - - - - - - -

    But I was thinking that the energy in the field is really dependent on the squared-norm of the sum of A, B, etc., not the difference, since like you say, it doesn't matter which one is "1" and which one is "2".

    So instead consider:

    |C|^2 = |A+B|^2 = |A|^2 + |B|^2 + 2|A||B|cos(theta_AB)

    Which is the simply the cosine law for a triangle (with vector B swapped).

    In a location that the field vectors A and B are unaligned, cos(theta_AB) would be negative.

    |A|^2 would be due to the source which generates A.

    |B|^2 would be due to the source which generates B.

    + 2|A||B|cos(theta_AB) would be the interaction energy, which would be negative in the case that A is opposed to B.

    |C|^2 would be due to taking the norm of the sum between A and B.

    - - - - - - - -

    Based on what you are saying, is it the quantity (1/2)|C|^2/mu_(x,y,z) or the (1/2)mu_(x,y,z) |C|^2 that should be integrated over space, where "C" of the former is the "B field", and "C" of the latter is the "H field"?
    Last edited: Feb 20, 2016
  18. Feb 20, 2016 #17


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    H doesn't have to be fixed.

    Say you have two resistors in series ( different values, but equally sized ), the currents ( B-fields ) will be the same but the inside electric fields ( H-fields ) will be different.

    The previous mentioned Ampere's law says that the mean strength of the H-field = N * I / s, and can not be used to calculate the strength at a specific location.
  19. Feb 20, 2016 #18


    Staff: Mentor


    It wouldn't be C in either case, it would be the sum of A and B, not the difference. However, whether you use H or not depends if you are using the microscopic or macroscopic Maxwell's equations. As long as you are consistent you are fine.
  20. Feb 20, 2016 #19
    If I took two identical loops of wire with identical currents, and if I were to orient them in the same direction along their central axis (like two wheels on an imaginary axle), they would attract, would they not? If I measured their mutual inductance, would this not increase from 0 to some positive value in Henries as they approached each other from infinity? Wouldn't I have to apply more power to keep the current constant given the back-EMF we know will be induced? If I wrapped these loops around iron disks, would not their field be oriented in the same direction as each other? Would not these magnets reinforce each other's magnetic field the same way the loops did such that the total field was not zero but rather more or less the sum of the two fields which point the same way?
  21. Feb 20, 2016 #20


    Staff: Mentor


    No. They partially cancel each other's field. That is precisely why they attract.

    Near the south pole the radial component of the field points in and near the north pole it points out. These cancel each other out leaving a purely axial field which is much weaker and therefore has a much lower energy density.

    They don't point the same way everywhere. Remember, it is the cumulative field energy, integrated over the whole field.
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