 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ib/paper-3/17a/c/solution

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

The Poynting vector in the annular region is

S = \frac{1}{μ _{0}} E \times B = \frac{V I}{2 π r ^{2} ln ( b / a )} e_{z} .

It points from the battery towards the resistor with the preceding current convention. Integrating through an annular cross-section gives

P = \int_{a}^{b} S_{z} 2 π r d r = \frac{V I}{l n ( b / a )} \int_{a}^{b} \frac{d r}{r} = V I = V^{2} / R .

The energy is transported by the electromagnetic field in the space between the ideal conductors; the integration recovers the power dissipated at the terminating resistor.

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