Q16Engineering Physics

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Answer 1

In advanced classical electrodynamics, the fundamental transport of massive energy by an electromagnetic wave strictly through space is mathematically governed entirely by the Poynting Vector. It physically represents the exact directional power density (Watts per square meter) of the propagating wave.

Mathematical Deduction (Poynting's Theorem)

We strictly begin with the two fundamental time-varying Maxwell curl equations:

\nabla \times E = - \frac{\partial B}{\partial t} = - μ \frac{\partial H}{\partial t}

\nabla \times H = J + \frac{\partial D}{\partial t} = σ E + ϵ \frac{\partial E}{\partial t}

To systematically derive an equation exclusively concerning energy, we rigorously take the vector dot product of $H$ with the first equation, and $E$ with the second equation, and mathematically subtract them:

H \cdot (\nabla \times E) - E \cdot (\nabla \times H) = - H \cdot (μ \frac{\partial H}{\partial t}) - E \cdot (σ E + ϵ \frac{\partial E}{\partial t})

Applying the strict standard vector calculus identity: $\nabla \cdot (E \times H) = H \cdot (\nabla \times E) - E \cdot (\nabla \times H)$

\nabla \cdot (E \times H) = - \frac{1}{2} μ \frac{\partial}{\partial t} (H \cdot H) - σ ∣ E ∣^{2} - \frac{1}{2} ϵ \frac{\partial}{\partial t} (E \cdot E)

- \nabla \cdot (E \times H) = σ E^{2} + \frac{\partial}{\partial t} (\frac{1}{2} ϵ E^{2} + \frac{1}{2} μ H^{2})

This is Poynting's Theorem. The term $\frac{1}{2} ϵ E^{2} + \frac{1}{2} μ H^{2}$ is strictly the total electromagnetic energy density ( $u$ ). The term $σ E^{2}$ is the massive ohmic power dissipation (Joule heating). Therefore, by strict conservation of energy, the term $(E \times H)$ absolutely must represent the outward flux of electromagnetic energy strictly per unit area. This is defined exactly as the Poynting Vector, $S$ .

S = E \times H

Q16Engineering Physics

Question

Answer

Mathematical Deduction (Poynting's Theorem)