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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/section-b/10/c/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Section B / 10 / c / Solution by

Codex  0 2026-10-18

Read the original diagram as a clockwise arc carrying electric current from its upper endpoint to its lower endpoint, with tangent semi-infinite leads extending to the right. Their crossing is insulated, so there is no extra junction. Choose

+ z

out of the page. At the centre the arc produces a negative axial magnetic field, and each straight lead a positive one, by the right-hand rule.

For the arc, every element is at distance

R

, with

d l

perpendicular to its displacement to the centre. The Biot-Savart law therefore gives

B_{z, arc} = - \frac{μ _{0} I}{4 π R} \int_{- θ /2}^{θ /2} d ϕ = - \frac{μ _{0} I θ}{4 π R} .

(1)

For each straight lead, write

s \geq 0

for distance along the tangent measured from its endpoint. Its perpendicular distance from the centre is

R

, so

B_{z, lead} = \frac{μ _{0} I}{4 π} \int_{0}^{\infty} \frac{R d s}{( R ^{2} + s ^{2} ) ^{3/2}} = \frac{μ _{0} I}{4 π R} .

(2)

The integral equals

1/ R

because an antiderivative is

s / [R R^{2} + s^{2}]

. Adding both leads and the arc,

B_{z} (0) = \frac{μ _{0} I}{4 π R} (2 - θ), θ = 2 rad ≃ 114. 6^{\circ} .

(3)

Using the full infinite-wire magnetic field for either lead would incorrectly double its contribution.

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