Solution
= Solution
In vacuum, curl <Faraday's law> and use the <Ampère-Maxwell equation>:
$$
\nabla\times(\nabla\times\mathbf B)
=\mu_0\varepsilon_0\frac{\partial}{\partial t}
(\nabla\times\mathbf E)
=-\mu_0\varepsilon_0\mathbf B_{tt}.
$$
Since $\nabla\cdot\mathbf B=0$, the left side is $-\nabla^2\mathbf B$. Hence
$$
\boxed{\mathbf B_{tt}=c^2\nabla^2\mathbf B,
\qquad c=(\mu_0\varepsilon_0)^{-1/2}}.
$$