How do you find the magnetic field corresponding to an electric field?

Question

If we are given the electric field $\vec E$ how can I find the corresponding magnetic field? I think I can use Maxwell's equations? In particular, $\nabla\times \vec E= -{\partial \vec B\over \partial t}$? But is it completely determined? Since we only have a partial derivative?

Answer 1

You can not, you also need to know the current $\vec J$.

Any vector field is specified by ist transvere component and its parallel component (that is, its component with non zero curl and its component with non zero gradient). Therefore, knowing the curl and the gradient of a vector field one can compute the field itself. From the II and IV maxwell equation we have $$ \vec\nabla\cdot\vec B = 0 \qquad \vec\nabla \times \vec B=\epsilon\mu\frac{\partial \vec E}{\partial t}+\mu \vec J $$

Should $\vec J=0$ and the charge density $\rho=0$(e.g. you are talking about electromagnetic waves), this yelds $\vec E=\vec B\times \vec v$. See wikipedia for details.

Answer 2

The magnetic field is not completely specified. A boundary condition is required. A simple decomposition, using Green's functions and various integral theorems, shows this:

$$\begin{align*}B(r) &= \int_V \frac{r-r'}{4\pi|r-r'|^3} \times \mu_0 \epsilon_0 \frac{\partial E(r')}{\partial t} \; dV' \\ &+ \oint_{\partial V} \frac{r-r'}{4\pi |r-r'|^3} B(r') \cdot dS' \\ &+ \oint_{\partial V} \frac{r-r'}{4\pi |r-r'|^3} \times [B(r') \times dS']\end{align*}$$

where $r,r'$ are vectors. The last two integrals correspond to a vector field that obeys $\nabla^2 = 0$--it is harmonic, or rather, it is a homogeneous solution to this differential equation, while the first term is the particular solution.

If you can choose a surface on which the magnetic field is zero (for example, at infinity), then the field is completely specified by the first integral term. However, while this boundary condition is almost always implicit in EM theory, using it here makes the first integral very difficult to calculate unless one uses some cleverness.

If the charge and current densities are zero everywhere, then it's well known that the resulting solutions are EM waves, for which the E and B fields are entirely orthogonal, of equal magnitude (within factors of constants), and mutually orthogonal with the direction of propagation. This, I believe, is what Ferdinando was referring to.