 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ib/paper-2/18b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Put

R = r - r^{'}

J (r^{'})

does not depend on the differentiated variable

r

\nabla_{r} \times \frac{J ( r ^{'} )}{R} = \nabla_{r} (R^{- 1}) \times J (r^{'}) = - \frac{R}{R ^{3}} \times J (r^{'}) = \frac{J ( r ^{'} ) \times R}{R ^{3}} .

Taking the curl under the source integral consequently gives the Biot-Savart law,

B (r) = \frac{μ _{0}}{4 π} \int \frac{J ( r ^{'} ) \times ( r - r ^{'} )}{∣ r - r ^{'} ∣ ^{3}} d^{3} r^{'} .

Outside the compact source this differentiation is ordinary differentiation of a smooth kernel. For regular volume currents, the locally integrable

R^{- 2}

derivative kernel gives the same formula at interior points.

For the loop, choose positive current circulating counterclockwise as viewed from the positive

z

-axis. Parametrize its source position and directed line element by

r^{'} = a (cos φ, sin φ, 0), d ℓ^{'} = a (- sin φ, cos φ, 0) d φ .

(0, 0, z)

the denominator is the constant

(a^{2} + z^{2})^{3/2}

, while

d ℓ^{'} \times R = (a z cos φ, a z sin φ, a^{2}) d φ .

The transverse terms integrate to zero around the loop. The Biot-Savart law therefore yields

B (0, 0, z) = \frac{μ _{0} I a ^{2}}{2 ( a ^{2} + z ^{2} ) ^{3/2}} e_{z} .

The magnetic field is axial, with the stated magnitude for

I > 0

; reversing the electric current reverses its direction.

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