 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014/10/3/solution

In charge- and current-free vacuum,

\nabla \cdot E = \nabla \cdot B = 0

\nabla \times (\nabla \times E) = - μ_{0} ϵ_{0} \frac{\partial ^{2} E}{\partial t ^{2}} .

Since

\nabla \times (\nabla \times E) = \nabla (\nabla \cdot E) - \nabla^{2} E

\nabla^{2} E = μ_{0} ϵ_{0} \frac{\partial ^{2} E}{\partial t ^{2}}, \nabla^{2} B = μ_{0} ϵ_{0} \frac{\partial ^{2} B}{\partial t ^{2}} .

Substituting a plane wave with phase

k \cdot r - ω t

gives

k^{2} = μ_{0} ϵ_{0} ω^{2}

and therefore

c = \frac{ω}{k} = \frac{1}{μ _{0} ϵ _{0}} ≃ 3.00 \times 1 0^{8} m s^{- 1} .

The divergence equations make the waves transverse, and the curl equations fix

B = \hat{k} \times E / c

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