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Magnetic field constant km?

  1. Nov 9, 2013 #1
    For a moving charge is km = μ0/2pi or is km = μ0/4pi?

    examples:
    Biot-Savart (magnetic field dB at a point P due to a length element ds that carrues a steady current I is:
    dB = (μ0/4pi) (I ds x r) / r^2

    While for an long straight wire, (≥10m?)
    B = (μ0 I)/(2pi r)

    What is the different use 4pi when length is less than 10meters?

    wikipedia says that km is
    "In another system, the "rationalized-metre-kilogram-second (rmks) system" (or alternatively the "metre-kilogram-second-ampere (mksa) system"), km is written as μ0/2π, where μ0 is a measurement-system constant called the "magnetic constant".[12] The value of μ0 was chosen such that the rmks unit of current is equal in size to the ampere in the emu system: μ0 is defined to be 4π × 10−7 N A−2.[5]"
    https://en.wikipedia.org/wiki/Magnetic_constant

    However in my notes from class I have Km = 1E-7 as well as Km = 2E-7.

    Which do i use for a charge particle. And why the difference in the stated formulas? is it the distance?

    Km is suppose to be a constant not a variable.
     
  2. jcsd
  3. Nov 9, 2013 #2

    Simon Bridge

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    Biot-Savart is telling you the relationship you need.

    It is the contribution to the magnetic field from an infinitesimal length of wire.
    The other one (>10m you say) is actualy the result of doing the calculation for an infinite length of wire.
    It is a good approximation where the distance from the wire is very small compared with the length of the wire.

    The ##\mu_0## I know is usually referred to as "the permiability of free space".
    Looking up "magnetic constant" and everyone says this is ##\mu_0## - including your reference.

    The ##k_m## is an attempt to make magnetic relations look like Coulombs Law.
    Like the wiki article says, there were many definitions, which basically sets the units for current, and the SI system settled on one. If $$F=k_m\frac{I^2}{r}$$ then $$k_m = \frac{\mu_0}{2\pi}$$ in SI units.

    This will be consistent with the Biot Savart Law - once you've done the calculus.
     
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