 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/37b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Again let

α = qE / (m c)

. For

B = (0, 0, E / c)

\overset{u}{˙}^{0} = α u^{x}, \overset{u}{˙}^{x} = α (u^{0} + u^{y}), \overset{u}{˙}^{y} = - α u^{x} .

u^{0} + u^{y}

is constant and initially equals

c

. Therefore

\overset{u}{˙}^{x} = α c, u^{x} = α c τ, u^{0} = c (1 + \frac{α ^{2} τ ^{2}}{2}), u^{y} = - \frac{c α ^{2} τ ^{2}}{2} .

A further integration gives

c t = c τ + \frac{c α ^{2} τ ^{3}}{6}, x = \frac{c α τ ^{2}}{2}, y = - \frac{c α ^{2} τ ^{3}}{6}, z = 0.

Eliminating

τ

between

x

and

y

yields

9 c y^{2} = 2 α x^{3},

a semicubical parabola, with the physical branch selected by

y \leq 0

for the signs shown.

 New to topics? Read the docs here!