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Calculate flux in a ferrite bead on a wire

  1. Aug 12, 2015 #1
    Suppose that a ferrite bead is put around a cable where a constant current [itex]I[/itex] flows, just like in this image.

    The coordinate system has the [itex]z[/itex] axis along the cable. Let's evaluate the current through the [itex](x,y)[/itex] plane: according to the Ampère's law, the only magnetic field component generated by that current is

    [itex]H_{\phi} = \displaystyle \frac{I}{2 \pi r}[/itex]

    where [itex]r[/itex] is the distance from the origin.
    In order to calculate the inductance of the wire with the ferrite bead, the magnetic flux through the ferrite should be calculated first.
    The ferrite bead has [itex]a[/itex] as internal radius and [itex]b[/itex] as external radius, so it is present in the [itex](x,y)[/itex] plane only for [itex]a \leq r \leq b[/itex]. It surrounds the conductor (which has of course a section diameter less than [itex]2a[/itex]) and the flux of the magnetic field through the ferrite bead should be:

    [itex]\Phi = \mu_0 \mu_r \displaystyle \frac{I}{2 \pi} \int_a^b \displaystyle \frac{1}{r} dr = \displaystyle \frac{\mu_0 \mu_r I}{2 \pi} \ln \left( \displaystyle \frac{b}{a} \right)[/itex]

    Then the inductance is [itex]L = \Phi / I[/itex] as usual.

    My question is: why is the flux calculated in such a way?? Should not be considered all the surface surrounded by the circuit or something similar? (I know, it is a linear cable and I didn't specify that area)
     
  2. jcsd
  3. Aug 12, 2015 #2

    marcusl

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    It makes sense to speak of the inductance of the wire/bead combination because the ferrite has a high permeability so its inductance dominates over that of the rest of the wire. Does that address your question?
     
  4. Aug 13, 2015 #3
    Thant means the thickness of the ferrite bead is unity
     
  5. Aug 13, 2015 #4
    Thanks to both of you. And for @marcusl yes, sure, your observations address my question.
     
  6. Aug 13, 2015 #5

    Philip Wood

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    I think there's a small slip. You need to integrate the flux through an area at right angles to the flux, so you need to multiply by the axial length of the bead. Alternatively you must interpret your Phi as flux per unit axial length of bead.
     
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