 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/34c/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

These potentials give the uniform magnetic field

B = B z

and zero electric field. With the signed frequency

ω_{c} = qB / m

\overset{x}{¨} = ω_{c} \overset{y}{˙}, \overset{y}{¨} = - ω_{c} \overset{x}{˙}, \overset{z}{¨} = 0.

Thus

\overset{x}{˙} + i \overset{y}{˙} = (\overset{x}{˙}_{0} + i \overset{y}{˙}_{0}) e^{- i ω_{c} t},

and explicitly

x (t) y (t) z (t) = x_{0} + \frac{x ˙ _{0} sin ( ω _{c} t ) + y ˙ _{0} [ 1 - cos ( ω _{c} t )]}{ω _{c}}, = y_{0} + \frac{y ˙ _{0} sin ( ω _{c} t ) - x ˙ _{0} [ 1 - cos ( ω _{c} t )]}{ω _{c}}, = z_{0} + \overset{z}{˙}_{0} t .

The transverse trajectory is a circle about the fixed guiding centre

(x + \overset{y}{˙} / ω_{c}, y - \overset{x}{˙} / ω_{c})

ω_{B} = \frac{∣ qB ∣}{m} .

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