Wave Equation from Maxwell's Equations

Maxwell's equations for a region with no charge or current are, in differential form:

$$\nabla\cdot{{\bf B}} = 0$$ (1)

$$\nabla\times{{\bf B}} = -{1\over c^2}{\partial{\bf E}\over\partial t}$$ (2)

$$\nabla\cdot{{\bf E}} = 0$$ (3)

$$\nabla\times{{\bf E}} = {\partial{\bf B}\over\partial t}$$ (4)

Here I have assumed that the the charge density $\rho$ and current density ${\bf J}$ are zero, and that the electric displacement vector can be expressed as ${\bf D} = \epsilon{\bf E}$ and the magnetic flux can be expressed as ${\bf H} = \mu{\bf B}$, which are common assumptions. Thus, the speed $c$ in the equations above refer to the speed of light in the particular medium $c = 1/\sqrt{\epsilon\mu}$, and in the case of a vacuum, $c$ is the standard speed of light in vacuum.

In this problem I will focus exclusively on the magnetic field components. Since the above equations are coupled, it suffices to solve either for the electric field ${\bf E}$ or magnetic field ${\bf B}$, and then the other vector can be inferred. This leads to a set of vector wave equation:

$$\nabla^2{\bf B}= {1\over c^2}{\partial^2{\bf B}\over\partial t^2}$$ (5)

which holds for each component of the vector ${\bf B}$. The above equation can be verified by using the vector differential identity,

$$\nabla^2{\bf F} = \nabla(\nabla\cdot{\bf F}) - \nabla\times{(\nabla\times{\bf F})}$$ (6)

When taking $\nabla^2{\bf B}$, one can use $\nabla\cdot{{\bf B}} = 0$ to eliminate the first term on the right hand side, leaving only the second curl-of-curl term. The interior curl can be found by again using Maxwell's equations, leaving

$$\nabla^2{{\bf B}} = - \nabla\times{\left(-{1\over c^2}{\partial{\bf E}\over\partial t}\right)}.$$ (7)

Upon exchanging the order of differentiation on the right hand side, and using Maxwell's equations one last time for $\nabla\times{{\bf E}}$, I verify the the wave equation (5) is correct.