Solving for Magnetic Fields from time varying Electric fields

by GarageDweller
Tags: electric, fields, magnetic, solving, time, varying
 P: 834 That the source of the magnetic field the time-varying electric field and not an actual current makes no difference to Maxwell's equations. Because $\nabla \cdot B = 0$ still, you can remove the derivative through its inverse, the free-space Green's function $r/4\pi|r|^3$. What you get is $$B(r) \propto \int_V \frac{\hat z \times (r - r')}{4 \pi |r - r'|^3} \; d^3r'$$ This result (when the proper proportionality constants are in) is the Biot-Savart law. The problem, however, is that if this electric field is time-varying everywhere, then I doubt this integral converges. You'll likely have to confine yourself to a 2D plane and invoke symmetry, or even simplify the source electric field to being nonzero only at a specific point, unless you're explicitly interested in the case of E being nonzero throughout a particular region.
 P: 834 Solving for Magnetic Fields from time varying Electric fields You're not familiar with the use of Green's functions to solve differential equations? I'm not terribly surprised, I guess; the topic is sometimes treated like it's esoteric. Here's the basic idea: Many differential equations are of the form $dA/dx = J$. It's first order, and it's simple. The Green's function for $d/dx$ is defined such that $dG/dx = \delta(x)$--the delta function. This isn't immediately useful until you use it with the fundamental theorem of calculus, which can be written as $$\left. A(x) \right|_a^b = \int_a^b \frac{dA}{dx} \; dx = \int_a^b J(x) \; dx$$ Now, using the fundamental theorem above (with $dA/dx = J$), just for fun, consider $$\left. A(x') G(x-x') \right|_{x'=a}^b = \int_a^b \frac{dA(x')}{dx'} G(x-x') + A(x') \frac{dG(x-x')}{dx'} \; dx' = \int_a^b J(x') G(x-x') - A(x') \delta(x-x') \; dx'$$ Now, if we let $a \to -\infty$ and $b \to \infty$, we cover the whole real line. We generally require the solution for $G(x)$ to be well-behaved there, so the term on the left-hand side goes to zero. Remember also that when you integrate over a delta function, it forces the argument to zero. This leads to $0 = \int_{-\infty}^\infty J(x') G(x-x') \; dx' - A(x) \implies A(x) = \int_{-\infty}^\infty J(x') G(x-x') \; dx'$ So, if you know the form of the Green's function for a particular differential operator and the source that generates the field, you can solve for the field through this integration. Extending this to 3D isn't too hard, but it requires some mathematical formalism that may obscure the point. The Green's function for $\nabla$ is well known--you use it every time you talk about the field from a point charge, because we model point charges as delta function sources anyway.