 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/15d/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

z

from the centre of the loop of radius

R

, toward the second loop. Between the loops the two axial fields have opposite directions. Their magnitudes are

B_{1} (z) = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + z ^{2} ) ^{3/2}}, B_{2} (z) = \frac{μ _{0} ( 2 I ) ( 2 R ) ^{2}}{2 {( 2 R ) ^{2} + ( D - z ) ^{2} } ^{3/2}} .

Equating them and cancelling common factors gives

{4 R^{2} + (D - z)^{2}}^{3/2} = 8 (R^{2} + z^{2})^{3/2} .

Taking the power

2/3

yields

(D - z)^{2} = 4 z^{2}

. Since

0 < z < D

, we have

D - z = 2 z

, and therefore

z = \frac{D}{3} .

Thus the magnetic-field cancellation between oppositely driven coaxial loops occurs one third of the separation from the smaller loop.

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