 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/36a/c/solution

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

Because

ε_{ij}

and

μ_{ij}

are symmetric and time-independent, the linear anisotropic dielectric relations give

E_{i} \partial_{t} D_{i} = ε_{ij} E_{i} \partial_{t} E_{j} = \frac{1}{2} \partial_{t} (ε_{ij} E_{i} E_{j}),

and similarly

H_{i} \partial_{t} B_{i} = \frac{1}{2} \partial_{t} (μ_{ij} H_{i} H_{j}) .

Integrate the identity from part (b) over

V

and apply the divergence theorem to obtain

\frac{\partial}{\partial t} \int_{V} \frac{1}{2} (ε_{ij} E_{i} E_{j} + μ_{ij} H_{i} H_{j}) d V + \int_{S} (E \times H) \cdot d S = - \int_{V} E \cdot J d V .

When

J = 0

, this is conservation of electromagnetic energy. The Poynting theorem in a linear anisotropic medium identifies

u = \frac{1}{2} (ε_{ij} E_{i} E_{j} + μ_{ij} H_{i} H_{j})

N = E \times H

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