# LH vs Gaussian units in Biot-Savart Law

1. Jan 11, 2016

### avikarto

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. Jan 11, 2016

### Staff: Mentor

it comes from I.

3. Jan 11, 2016

### avikarto

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. Jan 11, 2016

### Staff: Mentor

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