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How does B=μH really work?

  1. May 25, 2012 #1

    I am trying to figure this out. I figure it'd be best to do with an example.

    Say we have a bar magnet, N and S poles. Surrounded by vacuum with μ=μ0.

    CASE 1: Then. We put a high permeability material μr x just to the left of the magnet say. Now, from what I have been taught I understand that most of the flux will go through that since the flux will try to follow the path of least resistance.

    CASE 2: But now, I put another identical high permeability substance x to the right. With the same idea, the flux would try to go through this too. But since flux is limited (is it?), it will divide and half will go through the left and the other half through the right.

    So far so good?

    Ok. Now the part I don't understand is the equations don't add up the way I am doing it. If we do B=μH for CASE 1, say we get B=a. OK. But now if we do B=μH for CASE 2, we have 2 paths for the flux and ==> (B=μH=a) + (B=μH=a)=2a. Surely the flux density can't just double because I put another piece of metal around it.

    What I'm trying to get at is basically this, how does the equation B=μH "know" about the area outside the one it is measuring? How will it know if it is filled with very high permeability substance or just vacuum?

    A theory I have is that H will decrease accordingly as we increase the μ around said magnet. Don't know if it is right or wrong.

    Another doubt I have is. If H doesn't depend on μ, then is it always uniform and always present around a bar magnet irrelevant of its surroundings? (This doesn't sound right?). Is it like a theoretical field whereas B is the actual field in real life?

    What am I missing?
  2. jcsd
  3. May 25, 2012 #2
    Let's say you place an iron bar left to the magnet such that it touches it. So now nearly the entire magnetic flux will go through the iron. Flux is B * area. The total amount of flux can not change unless you demagnetize the magnet. If you add a second iron bar the flux will now be split between the 2 bars. That means the flux density is half of what it was before.
    Since B = µH, H also needs to be half of what it was. Remember that the iron actually produces an H field of it's own which cancels out the field of the magnet.
    Imagine the iron consisting of a large number of permanent magnets all of which point in random directions. They can freely rotate but are connected to a spring so they rotate back to their original position when no external field is present. Now that is a simplified model but it is good enough to explain your question.

    An important factor is the magnetization of the iron.
    It says there
    B = µ0(H+M)
    Since M = χmH you can also say B = µ0(H+χmH) = µ0*(1+χm)*H
    And then you say µ = µ0*(1+χm), so B = µH
    So what we call B is not a pure magnetic field. It's a combination of a magnetic field µ0*H and the magnetization of the material. Using the model I decribed, the magnetization represents the degree to which the magnets are aligned and thereby indirectly also the tension on the "springs". Because a change in magnetization induces an electric field just like a change in the H field does it made sense to combine the pure magnetic field with the magnetization into one field. That makes the B field very practical for calculations in electromagnetism

    The H field of the magnet goes from N to S outside the magnet, Inside the magnet it also goes from N to S. That means that inside the magnet H and B point in opposite directions.
    A magnetized bar of iron now behaves like a permanent magnet i.e. it produces it's own H field which goes from N to S, inside and outside. If you now superimpose the H fields of the magnet and the iron bar you see that they mostly cancel. The outside field of the magnet cancels the inside field of the iron and vice versa. That's why there is nearly no field outside the iron. It's all been canceled out. Of course that's not true for the B field since it consists mostly out of magnetization. So it is very strong and forms a circle through the magnet and the iron.
    When you attach the iron bar to the magnet the iron starts to become magnetized. The more it is magnetized the more the H field of the magnet gets weakened until it reaches a point where it is too weak to magnetize the iron any further. That remaining weak field is what you can measure outside the iron. That field is very weak but iron only requires a very weak internal H field to maintain a strong magnetization.
    So as I said - when you attach a second iron bar, the H field gets reduced even further so it is now at half it's previous strength and the B field will therefore also drop to half it's previous value. The total flux stays the same though.
    Last edited: May 26, 2012
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