Magnetic field for a cylindrical shell in which current flows

Question

If we have a cylindrical shell along the z-axis, with current density $\vec j= j_0 \vec e_z$ and small radius $R_i$ and big radius $R_a$. I tried to find the magnetic field inside by using Ampere's law and I got the following:

$$B=\frac{j_0}{2}\left(\rho -\frac{R_i}{\rho}\right)$$ where $\rho$ is the length of the position vector $\vec r$ projected in the xy-plane. This formula is corrent. I also can use the argument of the right hand rule to express my result as a vector : $$\vec B(\vec r)=\frac{j_0}{2}\left(\rho -\frac{R_i}{\rho}\right)\vec e_{\phi}$$.

But If I want to use only the Amper's Law without the right hand rule, how can I find the vector components of the magnetic field?

Answer 1

If you construct a circular loop around the cylindrical axis and argue by symmetry that the magnitude of the field is constant on that loop then Ampere's law can only tell you about the $\phi$ component of that field. You need further constraints and symmetry arguments to say what the other components might be. For instance, by symmetry and for an infintely long cylinder, we can say that the $\rho$ and $z$-components cannot vary with $z$ or $\phi$. We can also argue that the $\rho$ component cannot vary with $\rho$ otherwise that would require a non-zero divergence of the magnetic field. Taken together, this suggests that $B_\rho = 0$.

However, in general you could introduce any additional curl- and divergence-free magnetic field into the problem and it would not change the LHS of Ampere's law, since such a field always has a zero closed line integral around any loop.

Answer 2

The only way we can use amperes law to solve for this is by assuming the cylinder extends to infinity. We need this to use symetry arguments to pull the B field out of the integral

I have no idea how you solved this using amperes law without first knowing the direction of the magnetic field.

To solve this using amperes law you need to know that the B field only has a $\theta$ component to begin with, such that you can say B $\cdot$ dl for a circular gaussian ring is just |B| |dl| That is the only way you can solve it. So you may want to go over what you're actually doing when you pull the B field out of the integral

The way you determine that B forms concentric circles is using the definition of curl. Which ultimately IS the right hand rule.

Answer 3

The right hand rule (cross product) is part of Ampere's Law, so you'd still use it. But actually, the original "Maxwell's Equations" as J.C. Maxwell wrote them were broken into components, until Oliver Heaviside essentially invented vector calculus and put them in their current form. So you would solve for each component of the field $B_r, B_{ \phi}, B_z$ separately.