 Solution

ID: past-exam-of-the-electromagnetism-course-of-the-university-of-oxford/2023/3/d/i/solution

K = 1/ (4 π ϵ_{0})

t_{0} = t - r / c

and

n = r / r

, and evaluate

p

p^{'}

p^{''}

t_{0}

. The chain rule gives

\nabla t_{0} = - \frac{n}{c}, \partial_{i} p_{j} = - \frac{n _{i}}{c} p_{j}^{'}, \partial_{i} n_{j} = \frac{δ _{ij} - n _{i} n _{j}}{r} .

In particular, for

u (t_{0})

m

\nabla (\frac{n \cdot u}{r ^{m}}) = \frac{u - ( m + 1 ) n ( n \cdot u )}{r ^{m + 1}} - \frac{n ( n \cdot u ^{'} )}{c r ^{m}} .

Apply this gradient identity to the electric potential

ϕ = K [(n \cdot p) / r^{2} + (n \cdot p^{'}) / (cr)]

\nabla ϕ = K [\frac{p - 3 n ( n \cdot p )}{r ^{3}} + \frac{p ^{'} - 3 n ( n \cdot p ^{'} )}{c r ^{2}} - \frac{n ( n \cdot p ^{''} )}{c ^{2} r}] .

A = K p^{'} / (c^{2} r)

\partial_{t} A = K p^{''} / (c^{2} r)

E = - \nabla ϕ - \partial_{t} A

E = \frac{1}{4 π ϵ _{0}} [\frac{3 n ( n \cdot p ) - p}{r ^{3}} + \frac{3 n ( n \cdot p ^{'} ) - p ^{'}}{c r ^{2}} + \frac{n ( n \cdot p ^{''} ) - p ^{''}}{c ^{2} r}] .

\nabla \times (\frac{p ^{'}}{r}) = \nabla (r^{- 1}) \times p^{'} + r^{- 1} \nabla \times p^{'} = - \frac{n \times p ^{'}}{r ^{2}} - \frac{n \times p ^{''}}{cr} .

B = \frac{μ _{0}}{4 π} (\frac{p ^{'}}{r ^{2}} + \frac{p ^{''}}{cr}) \times n .

These formulas hold for

r > 0

. The

r^{- 3}

r^{- 2}

terms are induction contributions, and the

r^{- 1}

terms describe electric dipole radiation. Retaining the dependence of

t_{0}

on position in every spatial derivative is what produces both the induction and radiation contributions.

 New to topics? Read the docs here!