### Maxwell's equations for magnetostatics, boundary conditions

#### Problem:

Find the magnetic vector potential for the case of a long, straight wire carrying a steady current I.  Let R be the radius of the wire.

#### Problem:

A converging magnetic field is often used as a magnetic mirror.  Consider a symmetric converging field with ∂Bz/∂z = f(z).   Show that the radial component of B in cylindrical coordinates, namely Bρ, where ρ = xi + yj is given by Bρ = -(ρ/2)f(z).

Solution:

• Concepts:
Gauss's law for B
• Reasoning:
B = 0, ∂Bz/∂z = f(z) is given, the field has no φ dependence, we can solve for Bρ.
• Details of the calculation:
B = 0, ∂Bx/∂x + ∂By/∂y = -∂Bz/∂z = -f(z).
In cylindrical coordinates: (1/ρ)∂(ρBρ)/∂ρ + (1/ρ) ∂Bφ/∂φ = -f(z).
For the symmetric converging field Bφ = 0, so ∂(ρBρ)/∂ρ = - ρf(z),
ρBρ = - ρ2f(z)/2, Bρ = - ρf(z)/2.

#### Problem:

At the interface between one linear magnetic material and another the magnetic field lines bend.  Show that tanθ2/tanθ1 = μ21, assuming there are no free currents at the boundary.

Solution:

• Concepts:
Boundary conditions for B and H
• Reasoning:
No free currents are at the interface.  The normal component of B and the tangential component of H therefore have to be continuous at the boundary.
• Details of the calculation:
(a) Assume B1 makes an angle θ1 with the normal to the interface in the medium with m1.
B1x/B1y = tanθ1.

By is continuous across the boundary.  Hx is continuous across the boundary.  In the medium with μ1 we have H1x = B1x1.  In the medium with μ2 we have H2x = B2x2.  H1x = H2x.  B1xμ21 = B2x.
B2x/B2y = (B1xμ2)/(B1y μ1) = tanθ2 = (μ21 )tanθ1.
Comment: This will remind you of refraction, but different boundary conditions for the normal and tangential components of E, D and B, H result in a different formula.

#### Problem:

(a)  Find the magnetic field B of a rotating spherical shell with uniform surface charge density σ.
(b)  Find the magnetic field  B of a uniformly magnetized sphere.

Solution:

• Concepts:
Boundary conditions, the uniqueness theorem, Magnetization M, magnetization current densities
• Reasoning:
We can solve for the magnetic field of a rotating spherical shell with uniform surface charge density σ, by treating the problem as a boundary value problem.  The magnetization of a uniformly magnetized sphere results in a magnetization surface current density like the surface current density of a rotating spherical shell with uniform surface charge density σ.
• Details of the calculation:
(a)  Let the radius of the sphere be R, its center be located at the origin, and let the sphere rotate with angular velocity ω = ωk.
The rotating surface charge density produces a surface current density, km = σω x R
km = σωRsinθ.
For r >> R we will have a dipole potential,
A(r) = = (μ0/(4π))m x r/r3, or Aφ = (μ0/4π)msinθ/r2.
To find the dipole moment m, we consider the rotating sphere to be a stack of current loops.
For a current loop at θ we have dm = IA = Rσω sinθ Rdθ πR2sin2θ = R4σωπ sin3θdθ.
Integrating dm from θ = 0 to θ = π we find m = (4/3)R4σωπ, m = mk.
[∫sin3xdx = -⅓cosx (sin2x + 2)]
For r >> R we have Aφ = (μ0/3)R4σωsinθ/r2.
2A(r) = -μ0j(r),  A(r) = (μ0/(4π)) ∫V'dV' J(r')/|r - r'|.  (SI units).
Assume that A(r) will only has φ component.
Assume that the 1/r2 term is the only term in the expansion of A in powers of 1/rn outside of the sphere, i.e. assume that outside the sphere the potential is a pure dipole potential.
Since A must be continuous at r = R and A must be finite at the origin,
assume that Aφ = Csinθr = (μ0/3)Rσωsinθr  inside the sphere.

If this solution fulfills the boundary conditions,
(B2 - B1)∙n2 = 0,  (B2 - B1)∙t = μ0kn,
we have the only solution.
B = xA,
Bin_r =  (1/)rsinθ))∂(sinθAφ_in)/∂θ = (2μ0/3)Rσω cosθ, = (μ0/4π)2m cosθ/R3.
Bin_θ =  -(1/r)∂(rAφ_in)/∂r = -(2μ0/3)Rσω sinθ = -(μ0/4π)2m sinθ/R3
Bout_r =  (1/(rsinθ))∂(sinθAφ_out)/∂θ = (2μ0/3)R4σω cosθ/r3  = (μ0/4π)2m cosθ/r3
Bout_θ =  -(1/r)∂(rAφ_out)/∂r = (μ0/3)R4σωsinθ/r3 = (μ0/4π)m sinθ/r3
Now check that the boundary conditions are satisfied.
At r = R the radial component of B is continuous.
Bout_θ - Bin_θ = μ0R4σω sinθ/R3 = μ0km.
The boundary conditions are satisfied and we have the only solution.
The field outside the sphere is a pure dipole field,
B(r) =  (μ0/(4π))[3(m∙r1)r1/r15 - m/r13,
and the field inside the sphere is constant,  B(r) = k0/4π)2m/R3.
(b)  For a uniformly magnetized sphere with radius R and M = Mk we have m = (4/3)πR3M
The magnetization current densities are jm = 0, km = M x R/R, km = Msinθ .
We may write km = (3m/(4πR3)sinθ.
Expressed in terms of the magnetic moment m the uniformly magnetized sphere has the same surface current density as the rotating spherical shell with uniform surface charge density σ.
The same currents produce the same fields.
The field outside the uniformly magnetized sphere is a pure dipole field,
B(r) =  (μ0/(4π))[3(m∙r1)r1/r15 - m/r13,
and the field inside the sphere is constant,
B
(r) = k0/4π)2m/R3 = (μ0/4π)(8π/3)M.

Comment:

It is important that you remember the following from this problem:
The magnetic field of a uniform magnetized sphere outside the sphere is a pure dipole field
B(r) =  (μ0/(4π))[3(m∙r1)r1/r15 - m/r13,
The dipole moment is m = MV = M4πR3/3.
The magnetic field inside the sphere is constant (magnitude and direction).

If you place the origin of the coordinate system at the center of the sphere and let M = Mk, then you can find the magnitude of the field inside by matching it to the field outside on the z-axis at z = R.
A uniform magnetized sphere has a magnetization surface current distribution
km = M x R/R, km = Msinθ .
Any surface distribution of the same form produces the same fields.