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[Proof] Magnetic field of a permanent magnet

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    What is the expression for the magnetic field of a magnet? Biot Soviet law cannot be used since there is no current. What should I do? I get stuck at the final step shown below.


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  3. May 11, 2013 #2


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    For magnets you don't have the simple relation ##\vec{M} = \chi \vec{H}##.

    One approach to this problem is to note that you can always write ##\vec{B}## in terms of a vector potential as ##\vec{B} = \vec{\nabla} \times \vec{A}##. Use this for ##\vec{B}## in the integral and invoke a vector identity to rewrite the integral.
  4. May 11, 2013 #3
    [itex]\int(\nabla \vec{A} \times \vec{H}) d\tau[/itex]=[itex]\int[(\nabla \times \vec{H}) \cdot \vec{A}] d\tau[/itex]+[itex]\int(\vec{A} \times \vec{H})da[/itex]

    I know the integral on the left is zero as
    [itex]\nabla \times \vec{H}=J_{f}[/itex]
    and there is no free current.

    I don't know if the same argument can be used on the second integral as A comes from current, no current means A=0.(Griffiths 3rd edition p235 )
    Last edited: May 11, 2013
  5. May 11, 2013 #4


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    Use the identity ##\vec{\nabla} \cdot (\vec{A} \times \vec{H}) = \vec{H} \cdot (\vec{\nabla} \times \vec{A}) - \vec{A} \cdot (\vec{\nabla} \times \vec{H})##

    As you noted, there are no free currents so ##\vec{\nabla} \times \vec{H} = 0##. However, no free currents does not mean that ##\vec{A} = 0##. Page 235 of Griffiths is assuming a situation where the B-field is produced only by free currents. See page 263 for a discussion of A due to magnetization of a material.
  6. May 11, 2013 #5
    However magnetization [itex]M[/itex] is not known in this question, in other words, finding [itex]A[/itex] doesn't work since p 263 talks about bounded current which requires [itex]M[/itex]?
  7. May 11, 2013 #6
    I used Gauss law on the second integral, now I need to prove either
    [itex]\vec{A}\times \vec{H}=0[/itex]
    [itex]\int \vec{A}\times \vec{H} da=0[/itex]

    I still cannot think of a exact method.

    One idea in my mind now (but not sure if it makes sense) is to use the given condition "the integral carried out over all space", then the surface integral can be a surface far away from the magnet in which there is no magnetization and thus [itex]\vec{A}=0[/itex]
  8. May 12, 2013 #7


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    In general, ##\vec{A}## is nonzero even at places where ##\vec{M}## is zero (outside the magnet). However, both ##\vec{A}## and ##\vec{H}## approach zero as you go infinitely far from the magnet. As you said, the surface integral is at infinity. You can argue that ##\vec{A}## and ##\vec{H}## approach zero "fast enough" as you go infinitely far from the magnet that the surface integral will be zero.
  9. May 12, 2013 #8
    I know [itex]\overrightarrow A [/itex] decreases in the order of [itex]\frac{1}{r}[/itex].
    However how does [itex]\overrightarrow H [/itex] decreases with ##r##?

    Base on the equation [itex]\overrightarrow H = \frac{1}{{{\mu _0}}}\overrightarrow B + \overrightarrow M [/itex], we conclude that [itex]\overrightarrow H [/itex] depends on [itex]\overrightarrow B [/itex].
    Also [itex]\overrightarrow B \propto \frac{1}{{{r^2}}}[/itex] implies [itex]\overrightarrow A \times \overrightarrow H \propto \frac{1}{r}\frac{1}{{{r^2}}}[/itex] which decreases faster than the surface integral thus the integral approaches zero [itex]r[/itex] approaches zero.

    Is this argument valid?
    Last edited: May 12, 2013
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