## 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.