Solution

Since there are no currents,

\nabla \times B = 0

in each constant-permeability region, while

\nabla \cdot B = 0

gives

\nabla^{2} ψ = 0

. The displayed dipole forms are therefore harmonic. Regularity at the origin and the field at infinity give

b_{1} = 0, a_{3} = B_{0} .

For

ψ = (a r + b / r^{2}) cos θ

B_{r} = (a - 2 b / r^{3}) cos θ, B_{θ} = - (a + b / r^{3}) sin θ .

B_{r}

and

H_{θ}

, give the remaining four equations

a_{1} - \frac{2 b _{1}}{R _{1}^{3}} = a_{2} - \frac{2 b _{2}}{R _{1}^{3}}, \frac{a _{1} + b _{1} / R _{1}^{3}}{μ _{0}} = \frac{a _{2} + b _{2} / R _{1}^{3}}{μ},

a_{2} - \frac{2 b _{2}}{R _{2}^{3}} = a_{3} - \frac{2 b _{3}}{R _{2}^{3}}, \frac{a _{2} + b _{2} / R _{2}^{3}}{μ} = \frac{a _{3} + b _{3} / R _{2}^{3}}{μ _{0}} .

The two

R_{1}

equations yield

a_{2} = \frac{μ _{0} + 2 μ}{3 μ _{0}} a_{1}, b_{2} = \frac{μ - μ _{0}}{3 μ _{0}} R_{1}^{3} a_{1} .

Together with

b_{1} = 0

and

a_{3} = B_{0}

, these are the requested expressions. Substituting them into the two

R_{2}

equations gives a two-by-two linear system for

a_{1}

and

b_{3}

Solved by gpt-5.6-sol high.

 Ancestors (11)