 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014/10/1/solution

\nabla \cdot D = ρ_{free}, \nabla \cdot B = 0, \nabla \times E = - \frac{\partial B}{\partial t}, \nabla \times H = J_{free} + \frac{\partial D}{\partial t} .

D = ϵ_{0} E

and

H = B / μ_{0}

, so the last equation is

\nabla \times B = μ_{0} J + μ_{0} ϵ_{0} \partial E / \partial t

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