 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-1/15d/c/solution

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

E

B / μ_{0}

, and use the stated vector identity. The result is the local balance law

\frac{\partial}{\partial t} [\frac{1}{2} (ϵ_{0} E^{2} + \frac{B ^{2}}{μ _{0}})] + \nabla \cdot S = - J \cdot E, S = \frac{1}{μ _{0}} E \times B .

Here

S

is the Poynting vector. Integration over

V

\frac{d U}{d t} = - \int_{\partial V} S \cdot n d S - \int_{V} J \cdot E d^{3} x .

The field energy decreases through outward electromagnetic energy flux and through work done on charges.

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