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Charge Moving in Field of Magnetic Dipole

Physics Asked by TKT on August 23, 2021

I’m self-studying Kibble & Berkshire Classical Mechanics, and I’m completely stuck on Problem 11 in Chapter 12 on Hamiltonian mechanics.

It asks about a particle of mass m and charge q moving (in the equatorial plane) in the field of a magnetic dipole of moment $mu$ and described by a vector potential with the single non-zero component $A_phi = frac{mu_0 mu sin(theta)}{4pi r^2}$. Assuming the particle is approaching from a great distance with velocity $v$ and impact parameter $b$, show that the distance of closest approach to the dipole is $frac12(sqrt{b^2-a^2} + b)$ or $frac12(sqrt{b^2+a^2} – b)$ depending on whether $b > a$ or $b < a$, where $a^2 = frac{mu_0 q mu}{pi m v}$. It also asks to show that $P_phi$ and $v$ are constants of the motion.

Here’s my thinking up to the point I get stuck: For a particle in an electric field we have: $$L = frac12m dot{r}^2 + q dot{r} cdot A – q phi$$ where, working in spherical coordinates, $dot{r} = dot{r} hat{r} + rdot{theta}hat{theta} + r sin(theta) dot{phi} hat{phi}$, $A = frac{k sin(theta)}{r^2}$ (where $k = frac{mu_0 mu}{4pi}$) and $phi = 0$. Also, we know the motion is confined to the equatorial plane, $sin(theta) = 1$ and $dot{theta} = 0$. Subbing in, the Lagrangian becomes…

$$L = frac12 m dot{r}^2 + frac12 m r^2 dot{phi}^2 + frac{q k dot{phi}}{r} $$

Now, to find the Hamiltonian, we first calculate $P_r = m dot{r}$ and $P_phi = m r^2 dot{phi} + frac{q k}{r}$. Using $H = dot{r} P_r + dot{phi} P_phi – L$, we get…

$$H = frac{P_r^2}{2 m} + frac{P_phi^2}{2m r^2} – frac{qk}{m r^3}P_phi + frac{q^2 k^2}{2 m r^4}$$

But here is where I get stuck. I’m thinking that since H has no explicit time dependence, it should be conserved. And when the particle is approaching from a vast distance (and r is large) it’s clear that H = $frac{m v^2}{2}$. Also, when the particle is at it’s point of closest approach, we should have $P_r = 0$. So we should be able to solve…

$$frac{P_phi^2}{2m r^2} – frac{qk}{m r^3}P_phi + frac{q^2 k^2}{2 m r^4} = frac{m v^2}{2}$$

for r, where $P_phi$ is constant and should equal $mvb$. But this doesn’t look like it gives the given answer at all for r. Is there any chance someone could point me in the right direction or let me know if I’m missing something. Any help would be greatly appreciated!

EDIT:

In response to a comment below, here is what I get for the E-L equations:

$$m ddot{r} = m r dot{phi}^2 – frac{q k dot{phi}}{r^2} quad quad m r^2 dot{phi} + frac{q k}{r} = constant$$

I’ve played around with this a bunch and still can’t see how to get the answer the book is looking for for the distance of closest approach, which is: $frac12(sqrt{b^2-a^2} + b)$ or $frac12(sqrt{b^2+a^2} – b)$ depending on whether $b > a$ or $b < a$, where $a^2 = frac{mu_0 q mu}{pi m v}$

References:

  1. T. Kibble & F. Berkshire, Classical Mechanics, 2004; problem 11 p. 303.

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