ABC Magnetic Field derivation

[Nakul Aggarwal]

Can somebody please post the derivation of the ABC magnetic field components whose magnetic field components in the three cartesian directions(B1, B2, B3) are given by the following:
B1=A*sin(z)+C*cos(y)
B2=B*sin(x)+A*cos(z)
B3=C*sin(y)+B*cos(x)

 

[Charles Link]

An interesting magnetic field pattern. I'm going to presume this is not a homework problem because it would then belong in the homework section. $\\$ The first thing I observed is that $\nabla \cdot \vec{B}=0$. $\\$ To analyze what it is, you can set $B=0$ and $C=0$, and see what $A$ does by itself. The $A$ term causes a magnetic field with constant amplitude, (amplitude =$A$), in the x-y plane that spirals around as $z$ is increased. The cycle of the spiral for $z$ is $\Delta z=\frac{1}{2 \pi }$. The $B$ and $C$ terms behave similarly, and the results are superimposed, (with $B$ acting in the y-z plane, and $C$ in the x-z plane). Perhaps someone else can see something else of interest. One item is it seems to have infinite extent, and does not appear to represent a real physical system.

 

[Nakul Aggarwal]

True, it is a force-free helical steady-state solution of the Euler's equation in fluid dynamics. It does represent a physical situation as if you set A=1,B=1 and C=1, it represents a kinetic dynamo and this field is present in the solar corona as well. It has a regular and chaotic trajectory after plotting its Poincare map and calculating its Liapunov Exponents. I just need how it is derived from $\nabla\cross\vec{B}=constant*\vec{B}$ and using Euler's Equation

 

[Charles Link]

Given $\nabla \times \vec{B}=k \, \vec{B}$, this differential equation does have an integral solution. Sometimes there are additional homogeneous solutions that need to be added, but it is basically of the Biot-Savart form: Maxwell's equation $\nabla \times \vec{B} =\mu_o \vec{J}$ has the Biot-Savart solution: $\vec{B}(\vec{x})=\frac{\mu_o}{4 \pi} \int \frac{\vec{J}(\vec{x'}) \times (\vec{x}-\vec{x'})}{|\vec{x}-\vec{x'}|^3} \, d^3x'$. $\\$ If your last equation in the previous post is correct, it should be possible to at least show that the solution is consistent, i.e. if you replace $\mu_o \vec{J}$ by $k \vec{B}$, with the form as provided, you should get consistency on both sides. $\\$ Edit: The better way would be to simply put in the solution of $\nabla \times \vec{B}=k \, \vec{B}$, and solve for $k$. The $A$ term satisfies it for $k=1$ , and I believe the $B$ and $C$ do also. Starting with $\nabla \times \vec{B}=\vec{B}$, it really is only necessary to show that your solution satisfies the differential equation. It is unnecessary to derive anything.