# Can we express magnetic field strength similar to how we express electric field strength?

I'm currently studying magnetism and I'm looking to make some connections with electricity, namely how we define the two fields.

For the electric field, I know that the electric field strength is: $$\vec E = \frac{\vec F_e}{q_o}$$ where $$F_e$$ is the electric (Coulomb's) force and $$q_0$$ is a test charge.

For the magnetic field, I'm confused about what to call the magnetic field strength and if there's a way to express it. In some literature, it's $$\vec B$$ and in other, it's $$\vec H$$. Are these the same thing?

Also, about relating it back to electric field strength - I know we have to take into account that magnets are dipoles and that whereas electric charge monopoles exist, magnetic monopoles are yet to be discovered. Theoretically though, if they were to be discovered, could we express the magnetic field strength similar to the way we express the electric field strength?

• In the future, please only ask one question per post. Since you currently have different answers here answering different parts of your post, it is kind of hard to fix it for this post now, so just be aware for next time. Commented Dec 28, 2021 at 15:20
• Got it. Sorry about that. Commented Dec 28, 2021 at 22:57

Usually the magnetic field strength is called $$H$$. Its unit is A/m.

$$B$$ is magnetic induction in Tesla and is a consequence of a magnetic field. It is Given by $$H$$ mulitplied with the local permeability $$\mu$$.

Sometimes, because of a convenience or habit to use units of Tesla, the magnetic field is expressed as $$\mu_0H$$, especially when speaking about a magnetic field in air/vacuum or paramagnetic/diamagnetic materials, which have a relative permeability of about 1.

The equation you wrote is valid in the electrostatic case and defines the electric field using the static expression of Lorentz force; the complete expression for the force on a particle in a electromagnetic field would be: $$\boldsymbol{F}=q(\boldsymbol{E}+\boldsymbol{v}\times\boldsymbol{B})\tag{1}$$ This equation tells you the force when you know the electric and the magnetic field. You might wonder, how do we know the fields? The answer is given by Maxwell equations: $$\nabla\cdot\boldsymbol{E}=\frac{\rho}{\epsilon_0} \\ \nabla\cdot\boldsymbol{B}=0 \\ \nabla\times\boldsymbol{E}=-\partial_t\boldsymbol{B} \\ \nabla\times\boldsymbol{B}=\mu_0\boldsymbol{J}+\epsilon_0\mu_0\partial_t\boldsymbol{E} \tag{2}$$ That can be solved knowing the charge and current distribution and the boundary conditions. $$(1)$$ and $$(2)$$ contain all of classical electrodynamics: $$(2)$$ describes the evolution of the fields and $$(1)$$ describes the interaction between fields and charges. As you might have realized at this point, your definition for the electric field won't work in the electrodynamic case and what I wrote above is the general answer. If you wanted to fit your definition in this scheme, assuming only $$(1)$$ and $$(2)$$, you could derive the electric field of a point charge by Maxwell equations and then, in the static case ($$\boldsymbol{v}=\boldsymbol{0}$$), $$(1)$$ would become: $$\boldsymbol{F}=q\boldsymbol{E}$$

If monopoles discovered one would be able to write an equation for the magnetic field similar to the definition of the electric field:

$$\mathbf{H} =\frac{\mathbf{F}_m}{q_m} \tag{1}$$.

However, as magnetic monopoles have not been discovered yet, but very probably exist due to Dirac's argument, these monopoles are probably very far away from the earth in the universe related with some exotic phenomena. So writing down this equation is therefore of very little interest. One can find such an equation sometimes in textbooks if the symmetry of the Maxwell equations is discussed apart from this it is of no practical use. In this respect consider the Lorentz force as definition for the magnetic field as already done by @Feynman_00.

In order to clear up the question on the difference of $$\mathbf{B}$$ and $$\mathbf{H}$$, they are both related by the magnetisation of some magnetic material (for instance a bar magnet):

$$\mathbf{B} = \mu\mu_0 \mathbf{H} + \mathbf{M} \tag{2}$$

So if no magnetic material (with some non-zero magnetisation) is around $$\mathbf{B}$$ and $$\mathbf{H}$$ are the same apart from the scaling factor which is the product of $$\mu_0$$ the vacuum permeability and $$\mu$$ the permeability of the matter in which we observe the magnetic fields (for air $$\mu \approx 1$$ in very good approximation).

One has to be careful when the fields $$\mathbf{B}$$ and $$\mathbf{H}$$ are considered inside a magnetized material (as a bar magnet), in this case $$\mathbf{B}$$ and $$\mathbf{H}$$ take on very different values. In such a constellation as the bar magnet $$\mathbf{B}$$ and $$\mathbf{H}$$ show further differences. The $$\mathbf{H}$$ sees the bound (!) magnetic charges of at the ends of the bar magnet, but $$\mathbf{B}$$ stays completely divergence-free. Nevertheless as the relationship (2) is local, outside the magnetic material (e.g. the bar magnet) $$\mathbf{H}$$ and $$\mathbf{B}$$ are the same (apart from the permeability scaling factor).

Bound charges are created if a "normally" neutral material gets polarized. The material can then be seen as an assembly of many little dipoles which are orientated (if it is not a permanent one for instance in case of a bar magnet) along the field lines of the external electrical or magnetic field. The charges of the little dipoles inside the material are mutually compensated, but the outest layer of charge at the surface has no further layer of opposite charge. So these bound charges (necessarily opposite to the bound charge of the other end of the material which in total is completely neutral) have an effect on the field $$\mathbf{H}$$ which has to be taken into account.

So bound magnetic charges also exist and play an important role as source of the $$\mathbf{H}$$ field. (But $$\nabla \cdot \mathbf B =0$$ always).