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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?

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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.

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Thanks, Ferdinando. $\vec J, \rho$ are indeed $0$. May I ask what is $\vec v$ and why is that relation true? – Gerry Oct 28 '12 at 14:01
@Gerry: Hello Gerry, Ferdinando actually talks about Magnetic Lorentz force $F=q(\vec{v}\times \vec{B})=Bqvsin\theta$ which $\implies E=\frac{F}{q}=Bvsin\theta$ – Waffle's Crazy Peanut Oct 28 '12 at 14:05
Of course. $\vec v$ is the speed of the wave. For electromagnetic wave it has the direction of the wave vector and modulus $v=1/\sqrt{\epsilon\mu}$. I do not remember a demonstration of it, but you can find it here on wikipedia @CrazyBuddy: the equation you wrote is true, but mine is as well (for electromagnetic waves). – Ferdinando Randisi Oct 28 '12 at 14:11
@FerdinandoRandisi: Hello Ferdinando, I think it's already available in Maxwell equations & several other related links... – Waffle's Crazy Peanut Oct 28 '12 at 14:14
@CrazyBuddy: I do not see the equation I wrote in the Maxwell Equations page. But I see it in the page on electromagnetic waves (i linked the wrong page in my previous comment): it is the last single-line equation. – Ferdinando Randisi Oct 28 '12 at 14:23

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.

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