LH vs Gaussian units in Biot-Savart Law

In summary, the discussion is about the relationship between the Biot-Savart formula in Gaussian and Lorentz-Heaviside units. In Gaussian units, there is a ##\frac{1}{c}## outside the integral, while in LH units there is a ##\frac{1}{4\pi}\frac{1}{c}##. However, the transformation between the two units for B is ##B_{LH}=\frac{B_G}{\sqrt{4\pi}}##. The extra ##\sqrt{4\pi}## in the denominator comes from the transformation of the current, ##I=qv##, where ##q_{LH}=\sqrt{4\pi}q
  • #1
avikarto
56
9
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.

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  • #2
avikarto said:
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
it comes from I.
 
  • #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?
 
  • #4
avikarto said:
This seems to brings an extra 4π−−√4π\sqrt{4\pi} into the numerator, not the denominator, and would therefore remove the factor all together.
Check your math. Write the equation in one set of units, and then substitute. The factor is correct.
 

What is the difference between LH and Gaussian units in Biot-Savart Law?

The main difference between LH and Gaussian units is the system of measurement used for electric and magnetic quantities. LH units are based on the cgs (centimeter-gram-second) system, while Gaussian units are based on the mks (meter-kilogram-second) system. In terms of the Biot-Savart Law, LH units use the magnetic field strength in units of gauss (G), while Gaussian units use the magnetic flux density in units of tesla (T).

Which units are more commonly used in the Biot-Savart Law?

Gaussian units are more commonly used in the Biot-Savart Law, particularly in the field of electromagnetism and in engineering applications. This is because the mks system is the preferred system of measurement in most scientific disciplines, and it is easier to work with in calculations compared to the cgs system.

How do you convert between LH and Gaussian units in the Biot-Savart Law?

To convert between LH and Gaussian units, you can use the following conversions: 1 G = 0.1 mT and 1 mT = 10 G. This means that if you have a value in G, you can multiply it by 0.1 to get the equivalent value in mT, and vice versa.

Are there any advantages or disadvantages of using LH or Gaussian units in the Biot-Savart Law?

There are no significant advantages or disadvantages of using either LH or Gaussian units in the Biot-Savart Law. Both systems are equally valid, and the choice of which units to use often depends on personal preference or the requirements of a particular application.

Which system of units should I use in the Biot-Savart Law?

As mentioned earlier, Gaussian units are more commonly used in the Biot-Savart Law and in most scientific disciplines. However, if you are working in a field where the cgs system is the preferred system of measurement, then using LH units may be more appropriate. Ultimately, it is important to be consistent with the units you use and to make sure that all calculations are done in the same system.

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