<p>It entirely depends what system of units you are working in. <span class="math-container">$\mu_0 \simeq 4\pi\times 10^{-7}$</span> in S.I. units.</p>
<p>However, in <a href="https://en.wikipedia.org/wiki/Gaussian_units" rel="nofollow noreferrer">Gaussian units</a>, <span class="math-container">$\mu_0=1$</span> and so it isn't included.</p>
<p>In Gaussian units, the radial component of a magnetic dipole field is
<span class="math-container">$$B_r =\frac{2m}{r^3}\cos\theta $$</span>
measured in Gauss, where <span class="math-container">$m$</span> is a magnetic dipole moment.</p>
<p>In SI units, the same dipole has a radial field
<span class="math-container">$$B_r = \frac{2\mu_0 m}{4\pi r^3}\cos\theta$$</span>
in Tesla.</p>

CC BY-SA 4.0
2023-10-06T15:57:16.270Z
Thank you @ProfRob. However with m = (2\mu_0)}*R^3*g_1^0 is it not B_r = 2(R/3)^3* g_1^0 *cos(theta) which is the same as I wrote ?
Lunthang Peter
CC BY-SA 4.0
2023-10-06T21:11:30.723Z
I have no idea what your comment means. @LunthangPeter
ProfRob
CC BY-SA 4.0
2023-10-05T07:59:07.853Z
2023-10-05T13:23:40.247Z
2023-10-05T13:23:40.247Z
ProfRob
<p>To numerically analyse the Lorentz effect due to the magnetic field, say, radial component on a charge particles which of the expression would be correct, <span class="math-container">$\{ B_r = \frac{2R}{r^3} g_1^0 \cos(\theta) \}$</span> or <span class="math-container">$\{ B_r = \mu_0 \frac{2R}{r^3} g_1^0 \cos(\theta) \}$</span> ? Assuming there are no other particles and the particle is in the gravitational and magnetic field of the planet?</p>

CC BY-SA 4.0
2023-10-05T06:12:46.647Z
Show us your derivation, or a source which derives it and why you distrust one or the other solution.
planetmaker
<planet><astrophysics><magnetic-field><planetary-science><planetary-systems>
Is the magnetic permeability (mu_0) necessary in the expression for planetary magnetic field?
Lunthang Peter
# Is the magnetic permeability (mu_0) necessary in the expression for planetary magnetic field?

(Asked by: username_0 on 2023-10-05)

To numerically analyse the Lorentz effect due to the magnetic field, say, radial component on a charge particles which of the expression would be correct, $\{ B\_r = \frac{2R}{r^3} g\_1^0 \cos(\theta) \}$ or $\{ B\_r = \mu\_0 \frac{2R}{r^3} g\_1^0 \cos(\theta) \}$ ? Assuming there are no other particles and the particle is in the gravitational and magnetic field of the planet?

**Tags:** planet, astrophysics, magnetic-field, planetary-science, planetary-systems
> **Comments:**
> **username_1**: Show us your derivation, or a source which derives it and why you distrust one or the other solution.  

---

## 1 answers

### Answer by username_2 on 2023-10-05. Score: 3 | Accepted answer

It entirely depends what system of units you are working in. $\mu\_0 \simeq 4\pi\times 10^{-7}$ in S.I. units.

However, in [Gaussian units](https://en.wikipedia.org/wiki/Gaussian_units), $\mu\_0=1$ and so it isn't included.

In Gaussian units, the radial component of a magnetic dipole field is
$$B\_r =\frac{2m}{r^3}\cos\theta $$
measured in Gauss, where $m$ is a magnetic dipole moment.

In SI units, the same dipole has a radial field
$$B\_r = \frac{2\mu\_0 m}{4\pi r^3}\cos\theta$$
in Tesla.

> **Comments:**
> **username_0**: Thank you @ProfRob. However with m = (2\mu_0)}*R^3*g_1^0 is it not B_r = 2(R/3)^3* g_1^0 *cos(theta) which is the same as I wrote ?  
> **username_2**: I have no idea what your comment means. @LunthangPeter  

---

