Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 1 / b / i

Table of contents
- Solution i

Solution

For the aligned magnetic dipole field,

B_{r} = \frac{2 μ c o s θ}{r ^{3}}, B_{θ} = \frac{μ s i n θ}{r ^{3}} .

Since

e_{z} = cos θ e_{r} - sin θ e_{θ}

,

Ω \cdot B_{0} = \frac{Ω μ}{r ^{3}} (3 cos^{2} θ - 1) .

The corotation velocity is

u = Ω r sin θ e_{ϕ}

. The current density is therefore the advected charge density,

j = n_{q} u = - \frac{Ω ^{2} μ}{2 π c r ^{2}} sin θ (3 cos^{2} θ - 1) e_{ϕ} .

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