Finding the direction of magnetic fields analytically

Question

The integral form of Ampere's law: $$ I=\oint \vec{H}\cdot d\vec{\ell} $$ States that the current enclosed by an imaginary closed path is equal to the field's line integral over the length. By choosing a closed path such that the magnetic field is constant everywhere on the loop, the $H$ can be taken out of the integral: $$ I = ||H||\oint d\vec{\ell} = H\ell $$ Where $\ell$ is the length of the closed path. Computing the length and solving for $H$ gives us the magnitude of the field.

How do you find the direction of the field? There is the differential form of Ampere's law that states: $$ \nabla \times \vec{H} = \vec{J} $$ Where $J$ is the current density (and magnetisation is zero).

---

As an example, I tried computing the magnetic field around an infinite wire with a constant and uniform current. I found that $||\vec{H}||=\frac{I}{2\pi r}$ where $r$ is the distance from the wire, but I am struggling to find the direction. I know of the right hand rule, but I would like to try arriving at the result analytically.

Unfortunately, the curl of a vector field is not one-to-one, so there exists no inverse of the curl. Another identity that popped into mind was: $$ \nabla \times (\nabla \times \vec{H}) = \nabla(\nabla\cdot \vec{H})-\nabla^2\vec{H} $$ The curl of a constant is the zero vector, so if $\vec{H}$ is source free, we could obtain two equations: $$ -\nabla^2 \vec{H} = 0 $$ I am not, however, sure whether this is the case. Is it a rule that magnetic fields have to be source free? If so, then is there an explanation of this?

Is there any way to analytically determine the direction of a magnetic field, or is it only determined empirically?

Apparently, the argument that $\vec{H}$ is constant along the chosen path already implies symmetry, which implies that the field is a spin field. Why is a spin field symmetric though? What is meant by symmetric? It doesn't occur to me why outward fields are not considered symmetric in this context.

Answer 1

This question had bothered me for a long time. When taking an Electromagnetics course out of Zangwill's "Modern Electrodynamics", there was a section that was immensely helpful in Section 10.3 ("Ampere's Law").

The infinite wire is a good example to demonstrate the method. Take $ \vec{I}=I\hat{z} $, with $\vec{I}$ lying on the z-axis. To determine the components of $\vec{B}$, the idea is to show that it's impossible to have z- or r- components of $\vec{B}$ (in cylindrical coordinates).

Assume $\vec{B}=(B_z)\hat{z}$, and look specifically at the location $(x,y,z)=(R,0,0)$. Now reflect the system over the y-z plane (I picture just putting a mirror on the y-z plane, and seeing the +x axis now extending through the mirror in the opposite direction). Now, the location of the transformed vector $\vec{B}'$ is at $(-R,0,0)$.

Now, because $\vec{B}$ is a pseudovector, after the reflection, $\vec{B}'=(B_z)(-\hat{z})$.

Now, transform $\vec{B}$ a different way: rotate $\vec{B}$ about the z-axis 180 degrees. The location of this transformed vector $\vec{B}''$ is also $(-R,0,0)$.

Unlike inversion though, rotations act the same on vectors and pseudovectors. After rotation, $\vec{B}''=(B_z)(+\hat{z})$.

Both of these transformations do not alter $\vec{I}$, so the magnetic field it produces everywhere should also be the same in each transformed system. Then it should be that $\vec{B}'=\vec{B}''$, and equating z-components, $B_z = -B_z$. The only way this equality can be true is if $B_z=0$.

---

The same analysis can be done to show that the r-component must be zero, and that the $\phi$ component works $\big($for the $\phi$ component, you can use $\vec{B}=(B_y)\hat{y}$ at the point $(R,0,0)$$\big)$.

---

Helpful (or not so helpful) footnotes:

To do this with math, you can apply rotation and inversion matrices to $\vec{B}$. Note that the inversion matrix is different for pseudovectors: you have to multiply the inversion matrix by a constant -1 (the determinant of a general inversion matrix = -1).

Here is more info if when doing the above math, you wonder (like I did) how to rotate both the coordinate position of a vector, along with rotating its components. For example, in rotating $\vec{B}=(B_z)\hat{z}$, one needs to transform the coordinate position $(R,0,0)$ as well as rotate the vector itself.

Here's a nice video on pseudovectors.

Let me know if the explanation isn't clear -- I can include pictures at some point. I hope this is helpful!

Answer 2

You cannot use $\oint \vec H\cdot d\vec \ell$ to deduce $\vert \vec H\vert$ unless you can argue that the scalar product $\vec H\cdot d\vec \ell$ is constant on the circuit so that $\oint \vec H\cdot d\vec \ell= \vert\vec H\vert \ell$ where $\ell$ is the length of your circuit. This implies you already know the direction of $\vec H$ (usually using a symmetry argument), and the unknown is $\vert \vec H\vert$.