The magnetic field for given potential

$$ \vec{B} = \nabla \times A $$ Note several useful constants:

$$ \hbar c = 197.326 MeV fm $$ $$ e = \sqrt{4\pi \frac{1}{137}} $$ $$ \beta = \sqrt{1-1/\gamma^2} $$

To determine the field at point given by $\vec{x}$ $$ x = \vec{x}_0 / \hbar c $$ $$ y = \vec{x}_1 / \hbar c $$ $$ z = \vec{x}_2 / \hbar c $$

Then let: $$ R = \sqrt{ x^2 + y^2 + (z\gamma)^2} $$

Now the Magnetic field is : $$ \vec{B} = \beta \gamma ( x, y, z ) \times \vec{B_0} $$ $$ \vec{B} = (y \beta \gamma B_0, -x \beta \gamma B_0, 0) $$

where $B_0$ is: $$ B_0 = \frac{e^2}{4 \pi} \frac{Z_{Au} }{R^3} \frac{\int_0^{R/\hbar c} r^2 \rho(r) dr}{ \rho^\star } \hat{z} $$ $$ \rho^\star = \int_0^{\inf} r^2 \rho(r) dr $$

Compute $B(r)$ with $\vec{x} = (r, 0, 0)$ : $$ |B(r)| = r \beta \gamma \frac{e^2}{4 \pi} \frac{Z_{Au} }{R^3} \frac{\int_0^{R/\hbar c} r^2 \rho(r) dr}{ \rho^\star } $$

Using a spherically symmetric Woods-Saxson distribution: $$ \rho(r) = \frac{\rho_0 ( 1 + w^2 r^2 / R_{Au}^2 )}{ 1 + e^{ (r-R_{Au}) / a } } $$ with $$ \rho_0 = \frac{3 A} { 4 \pi R_{Au}^3 } $$

where: $a$ is the skin depth, $w$ is the ???, and $R_{Au}$ is the radius of the nucleus.