Imagine two charged magnetic dipoles (charge q, dipole moment m), constrained to move on the z axis (same as Problem 6.23(a), but without gravity). Electrically they repel, but magnetically (if both m's point in the z direction) they attract.

(a) Find the equilibrium separation distance.

(b) What is the equilibrium separation for two electrons in this orientation. [Answer: 4.72x10^-13m.]

(c) Does there exist, then, a stable bound state of two electrons?

The charge is q.

The dipole moment is m.

Two charges move on the z axis.

(a)

The magnetic field due one charge is given by,

$B_1=\frac{{\mu }_{0}m}{4{\mathrm{\pi z}}^3}\hat{z}$

Here z is the distance.

The magnetic field due another charge is given by,

$B_2=\frac{{\mu }_{0}m}{4{\mathrm{\pi z}}^3}\hat{z}$

The total magnetic field due to both the charge is given by,

$B=B_1+B_2$

Substitute $\frac{{\mu }_{0}m}{4{\mathrm{\pi z}}^3}\hat{z}$ for B₁ and B₂ into above equation.

$$\begin{gathered} B=\frac{{\mu }_{0}m}{4{\mathrm{\pi z}}^3}\hat{z}+\frac{{\mu }_{0}m}{4{\mathrm{\pi z}}^3}\hat{z} \\ B=2\frac{{\mu }_{0}m}{4{\mathrm{\pi z}}^3}\hat{z} \\ B=\frac{{\mu }_{0}m}{4{\mathrm{\pi z}}^3}\hat{z} \end{gathered}$$

The force on the dipole is given by,

$F=\nabla (m·B)$

Substitute $\frac{{\mu }_{0}m}{2{\mathrm{\pi z}}^3}\hat{z}$ for B into above equation.

$$\begin{gathered} F=\nabla \left(m·\frac{{\mu }_{0}m}{2{\mathrm{\pi z}}^3}\hat{z}\right) \\ F=\frac{\partial }{\partial z}\left(m·\frac{{\mu }_{0}m}{2{\mathrm{\pi z}}^3}\hat{z}\right) \\ F=\frac{3m^2{\mu }_0}{2{\mathrm{\pi z}}^4}\hat{z} \end{gathered}$$

The expression for the Coulomb’s for between the 2 electrons is given by,

role="math" localid="1657716216750" $F_e=\frac{q^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{z}}^2}\hat{z}$

The net force on the charge is given by,

F + F_e= 0

substitute $\frac{q^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{z}}^2}\hat{z}$ for F_e and role="math" localid="1657716398268" $-\frac{3m^2{\mu }_0}{2{\mathrm{\pi z}}^4}\hat{z}$ for F inti above equation.

$\begin{array}{rcl}-\frac{3m^2{\mu }_0}{2{\mathrm{\pi z}}^4}\hat{z}+\frac{q^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{z}}^2}\hat{z}& =& 0\\ \frac{q^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{z}}^2}\hat{z}& =& -\frac{3m^2{\mu }_0}{2{\mathrm{\pi z}}^4}\hat{z}\\ \frac{q^2}{2{\epsilon }_0}& =& -\frac{3m^2{\mu }_0}{z^2}\\ z^2& =& -\frac{3m^2{\epsilon }_0{\mu }_0}{q^2}\end{array}$

Solve further as,

$$\begin{gathered} z^2=\frac{6m^2}{q^2\left({\displaystyle \frac{1}{{\epsilon }_0{\mu }_0}}\right)} \\ z=\sqrt{\frac{6m^2}{q^2\left(\frac{1}{{\epsilon }_0{\mu }_0}\right)}} \\ z=\frac{\sqrt{6}m}{q\left(\frac{1}{\sqrt{{\epsilon }_0{\mu }_0}}\right)} \end{gathered}$$

Substitute c for $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$into above equation.

role="math" localid="1657716909912" $z=\frac{\sqrt{6}m}{qc}$

Hence the equilibrium separation distance is $\frac{\sqrt{6}m}{qc}$.

(b)

The charge on electron,$q=1.6\times {10}^{-19}C$

The speed of the light,$c=3\times {10}^{8}m/s$

The dipole moment,$m=9.28\times {10}^{-24}J/T$

The separation equilibrium distance,

$z=\frac{\sqrt{6}m}{qc}$

Substitute $1.6\times {10}^{-19}C$ for q, $3\times {10}^{8}m/s$ for c and $9.28\times {10}^{-24}J/T$for m into above equation.

role="math" localid="1657717399234" $$\begin{gathered} z=\frac{\sqrt{6}\times 9.28\times {10}^{-24}}{1.6\times {10}^{-19}\times 3\times {10}^8} \\ z=\frac{22.731\times {10}^{-24}}{4.8\times {10}^{-11}} \\ z=4.73\times {10}^{-13}m \end{gathered}$$

Hence the equilibrium separation distance is $4.73\times {10}^{-13}m$.

(c)

Since the equilibrium point is unstable, therefore, the existence of the stable bound for two electrons does not exist.

Hence there is no existing stable bound state for two electrons.