\nabla \cdot \mathcal{B} = 0
\frac{3}{4 \pi} \sqrt{4 \cdot x^2 12}\\
\lim_{n \to \infty}
\sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}
e^{i \pi} + 1 = 0 \\
$ \begin{align*}
\int x^2 dx & = \frac{1}{3}x^3
\therefore\quad\int_0^1 x^2 dx &= \frac{1}{3}
\end{align*}
$