 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018/s7/1/b/solution

q A \cdot \dot{r}

. Thus

L = \frac{m}{2} (\overset{x}{˙}^{2} + \overset{y}{˙}^{2} + \overset{z}{˙}^{2}) - \frac{q B _{0}}{2 L} x^{2} \overset{z}{˙} .

The velocity-dependent magnetic term changes canonical momentum but does not contribute to the mechanical energy.

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