Calculating Magnetic Field Lines

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To calculate magnetic field lines for an arbitrary current distribution, the equations ∇×H = J and ∇·H = 0 are essential. A common approach involves first determining the magnetic field H and then deriving the field lines from it using differential equations. In a source-free region, Laplace's equation can be applied, allowing for the definition of a magnetic scalar potential, which simplifies the problem to a two-dimensional analysis when J has only a z component. However, if calculations need to be performed within the source region, the complexity increases, and working with the vector potential becomes necessary. Ultimately, understanding the geometry of the field lines requires careful consideration of the magnetic field's behavior in relation to the current distribution.
thegreenlaser
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I guess this is maybe more of a vector calculus question, but here it goes. Say I have an arbitrary current distribution \vec{J} with the corresponding magnetic field given by
\vec{\nabla}\times\vec{H} = \vec{J}
\vec{\nabla}\cdot\vec{H} = 0
What is the best way to calculate the field/flow lines of \vec{H} (contours to which \vec{H} is always tangent)? Do I need to calculate the magnetic field and then find the field lines from that, or is there an easier way to directly extract information about the geometry of the field lines from the equations above if I don't care about the magnitude of the magnetic field?

Edit: If it makes things easier, I'd still be interested to see the answer to this question with the added assumption that everything is uniform in the z-direction and \vec{J} only has a z component (i.e. the magnetic field is 2D: it has no z-dependence and no z-component)
 
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If you want H in a source-free region (no currents or magnetization), then Laplace's equation holds and you can define an effective magnetic scalar potential that is analogous to the usual electric potential. Equipotential lines and streamlines then play the same role as in electrostatics problems: you get orthogonal potential lines and field lines. This is by far the easiest approach.

Since J has only a z component, the problem is two-dimensional problem, further simplifying it.
 
marcusl said:
If you want H in a source-free region (no currents or magnetization), then Laplace's equation holds and you can define an effective magnetic scalar potential that is analogous to the usual electric potential. Equipotential lines and streamlines then play the same role as in electrostatics problems: you get orthogonal potential lines and field lines. This is by far the easiest approach.

Since J has only a z component, the problem is two-dimensional problem, further simplifying it.

That does make sense, but unfortunately I do care about being able to do it inside the source region as well. Is there any simple technique there?
 
No, nothing simple. You typically work with the vector potential. Complexity will depend on the source geometry and boundary conditions.
 
thegreenlaser said:
I guess this is maybe more of a vector calculus question, but here it goes. Say I have an arbitrary current distribution \vec{J} with the corresponding magnetic field given by
\vec{\nabla}\times\vec{H} = \vec{J}
\vec{\nabla}\cdot\vec{H} = 0
What is the best way to calculate the field/flow lines of \vec{H} (contours to which \vec{H} is always tangent)? Do I need to calculate the magnetic field and then find the field lines from that, or is there an easier way to directly extract information about the geometry of the field lines from the equations above if I don't care about the magnitude of the magnetic field?

The way I know how to do this requires you to calculate the field first. Once you know \mathbf{H}(\mathbf{r})then you just construct the equations for the field lines. Let d\mathbf{r} be a differential vector parallel to the magnetic field; then we know d\mathbf{r} \times \mathbf{H} = 0. This yields differntial equations for the field lines. For example, in Cartesian coordinates this yields
<br /> \frac{dx}{H_x} = \frac{dy}{H_y} =\frac{dz}{H_z}<br />
and in spherical coordinates this is
<br /> \frac{dr}{H_r} = \frac{rd\theta}{H_\theta} =\frac{r \sin\theta d\phi}{H_\phi}<br />
jason
 
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