Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 36D (Electrodynamics)

Table of contents
- a 36D
  - Solution a
- b 36D
  - Solution b
- c 36D
  - Solution c

a

Solution

Differentiating

g = ϵ_{0} E \times B

, substituting Maxwell equations, and using the stated vector identity gives the Maxwell stress tensor conservation law

\partial_{t} g_{j} = \partial_{i} [ϵ_{0} (E_{i} E_{j} - \frac{1}{2} E^{2} δ_{ij}) + \frac{1}{μ _{0}} (B_{i} B_{j} - \frac{1}{2} B^{2} δ_{ij})] - (ρ E + J \times B)_{j} .

Moving the divergence to the left identifies

a = - ϵ_{0}, b = \frac{ϵ _{0}}{2}, c = - \frac{1}{μ _{0}}, d = \frac{1}{2 μ _{0}} .

b

Solution

σ_{ij}

is the flux of the

j

th component of electromagnetic momentum across a surface normal to the

i

direction, with the sign convention used in the conservation law. Its diagonal components are normal momentum fluxes, interpreted as electromagnetic pressures or tensions. Its off-diagonal components are tangential momentum fluxes, interpreted as shear stresses.

c

Solution

Since

B = 0

,

E (r) = \frac{ϵ _{0} E ( r ) ^{2}}{2} = ⎩ ⎨ ⎧ \frac{q ^{2} r ^{2}}{32 π ^{2} ϵ _{0} R ^{6}}, \frac{q ^{2}}{32 π ^{2} ϵ _{0} r ^{4}}, r \leq R, r > R .

Radial integration gives

W_{in} = \frac{q ^{2}}{40 π ϵ _{0} R}, W_{out} = \frac{q ^{2}}{8 π ϵ _{0} R},

and therefore

W_{em} = \frac{3 q ^{2}}{20 π ϵ _{0} R} .

To first order in

v / c

,

E^{'} = E

and

B^{'} = - v \times E / c^{2}

. Thus

g_{z}^{'} = \frac{ϵ _{0} v}{c ^{2}} E^{2} sin^{2} θ .

Using the angular average

⟨ sin^{2} θ ⟩ = 2/3

and

\int ϵ_{0} E^{2} d V = 2 W_{em}

,

P_{z} = \frac{4}{3} \frac{W _{em}}{c ^{2}} v = \frac{q ^{2} v}{5 π ϵ _{0} R c ^{2}} .

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