 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ib/paper-1/14d/b/solution

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

The negative-

z

segment integrates to the stated azimuthal field. The negative-

y

segment contributes

\frac{μ _{0} I}{4 π} \frac{x z ^ - z x ^}{x ^{2} + z ^{2}} (1 - \frac{y}{x ^{2} + y ^{2} + z ^{2}}) .

Adding this to the first field is the complete answer. On

x = 0, y = 1

, the field tends to

0

z \to + \infty

and to

- μ_{0} I \overset{x}{^} / (2 π)

z \to - \infty

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