Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 10C / ii

Table of contents
- Solution ii

Solution

Write

v = \dot{r}

. The original PDF contains the cross product

\dot{r} \times r

; the TeX aid drops its velocity dot. Since

v \times v = 0

, differentiation and the equation of motion give

\frac{d}{d t} [m (v \times r) \cdot B] = m (\ddot{r} \times r) \cdot B = - e [(v \times B) \times r] \cdot B .

The radial restoring force contributes no torque. The vector triple-product identity yields

[(v \times B) \times r] \cdot B = B^{2} v \cdot r - (v \cdot B) (r \cdot B) .

Meanwhile

\frac{d}{d t} [\frac{e}{2} ∣ r \times B ∣^{2}] = e (r \times B) \cdot (v \times B) = e [B^{2} r \cdot v - (r \cdot B) (v \cdot B)] .

The derivatives cancel, proving

m (\dot{r} \times r) \cdot B + \frac{e}{2} ∣ r \times B ∣^{2} = constant .

This is the magnetic axial angular-momentum invariant. Its first term is the negative of the ordinary angular momentum projected on

B

; retaining that sign is essential.

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