Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 6.25

Imagine two charged magnetic dipoles (charge $q$ , dipole moment $m$ ), constrained to move on the $z -$ axis (same as Problem 6.42a, 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: $4.72 \times 10^{- 13}$ m).

Part (a)

The electric force is

F_{electric} = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{z^{2}} \hat{z}

While the magnetic force is

\begin{aligned} (Eq. 6.3) & F & = \nabla (m \cdot B) \\ (See Eq. 5.88) & = - \frac{3}{2} \frac{μ_{0} m^{2}}{π} \frac{1}{z^{4}} \hat{z} \end{aligned}

Equilibrium happens when these forces add up to zero:

\frac{1}{4 π ϵ_{0}} \frac{q^{2}}{z^{2}} - \frac{3}{2} \frac{μ_{0} m^{2}}{π} \frac{1}{z^{4}} = 0

Solve for $z$ , and find:

z_{eq.} = \frac{m \sqrt{6 ϵ_{0} μ_{0}}}{q}

Part (b)

To evaluate this, it's useful to use the relation between $c$ (the speed of light) and $ϵ_{0}$ and $μ_{0}$ :

\frac{1}{c} = \sqrt{ϵ_{0} μ_{0}}

A quantity that's more familiar to us. We recall the values:

\begin{aligned} q & = 1.6 \times 10^{- 19} C \\ c & = 3 \times 10^{8} m \\ m & = 9.22 \times 10^{- 24} A \cdot m^{2} \end{aligned}

We get:

z_{eq.} = 4.70506 \times 10^{- 13} m

Part (c)

No. Although we have an equilibrium, it is not a stable one.

Where, besides the values used in part (b), it is relevant to recall:

\begin{aligned} ϵ_{0} & = 8.854187 \times 10^{- 12} \frac{F}{m} \\ μ_{0} & = 1.256637 \times 10^{- 6} \frac{N}{A^{2}} \end{aligned}