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LH vs Gaussian units in Biot-Savart Law

  1. Jan 11, 2016 #1
    I have a question regarding the relationship between the Biot-Savart formula in Gaussian and Lorentz-Heaviside units. In Gaussian, we have a ##\frac{1}{c}## outside the integral, but in LH units we have a ##\frac{1}{4\pi}\frac{1}{c}##. This does not make sense, considering the transformation between Gaussian and LH for B is that ##B_{LH}=\frac{B_G}{\sqrt{4\pi}}##. Where does the extra ##\sqrt{4\pi}## come from? Thanks.

    Source
     
  2. jcsd
  3. Jan 11, 2016 #2

    Dale

    Staff: Mentor

    it comes from I.
     
  4. Jan 11, 2016 #3
    If we take ##I=q v##, then using that ##q_{LH}=\sqrt{4\pi} q_G##, we would have that ##I_{LH}=\sqrt{4\pi} I_G##. This seems to brings an extra ##\sqrt{4\pi}## into the numerator, not the denominator, and would therefore remove the factor all together. I don't understand how this resolves the issue. Besides, the transformation on B should be the transformation on B, without also having to transform its subcomponents additionally. Could you please elaborate?
     
  5. Jan 11, 2016 #4

    Dale

    Staff: Mentor

    Check your math. Write the equation in one set of units, and then substitute. The factor is correct.
     
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