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.

A converging magnetic field is
often used as a magnetic mirror. Consider a symmetric converging field with
∂B_{z}/∂z
= f(z).
Show that the radial component of **B** in cylindrical coordinates, namely B_{ρ},
where **ρ** = x**i** + y**j** is given by B_{ρ} = -(ρ/2)f(z).

Solution:

- Concepts:

Gauss's law for**B** - Reasoning:

**∇**∙**B**= 0, ∂B_{z}/∂z = f(z) is given, the field has no φ dependence, we can solve for B_{ρ}. - Details of the calculation:

**∇**∙**B**= 0, ∂B_{x}/∂x + ∂B_{y}/∂y = -∂B_{z}/∂z = -f(z).

In cylindrical coordinates: (1/ρ)∂(ρB_{ρ})/∂ρ + (1/ρ) ∂B_{φ}/∂φ = -f(z).

For the symmetric converging field B_{φ}= 0, so ∂(ρB_{ρ})/∂ρ = - ρf(z),

ρB_{ρ}= - ρ^{2}f(z)/2, B_{ρ}= - ρf(z)/2.

At the interface between one linear magnetic material and another the
magnetic field lines bend. Show that tanθ_{2}/tanθ_{1}
= μ_{2}/μ_{1},
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**B**_{1}makes an angle θ_{1}with the normal to the interface in the medium with m_{1}.

B_{1x}/B_{1y}= tanθ_{1}.

B

_{y}is continuous across the boundary. H_{x}is continuous across the boundary. In the medium with μ_{1}we have H_{1x}= B_{1x}/μ_{1}. In the medium with μ_{2}we have H_{2x}= B_{2x}/μ_{2}. H_{1x}= H_{2x}. B_{1x}μ_{2}/μ_{1}= B_{2x}.

B_{2x}/B_{2y}= (B_{1x}μ_{2})/(B_{1y}μ_{1}) = tanθ_{2}= (μ_{2}/μ_{1})tanθ_{1}.

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.

(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**= ω**k**.

The rotating surface charge density produces a surface current density,**k**_{m}= σ**ω**x**R**,

k_{m}= σωRsinθ.

For r >> R we will have a dipole potential,

**A**(**r**) = (μ_{0}/(4π))**m**x**r/**r^{3}, or A_{φ}= (μ_{0}/4π)msinθ/r^{2}.

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 πR^{2}sin^{2}θ = R^{4}σωπ sin^{3}θdθ.

Integrating dm from θ = 0 to θ = π we find m = (4/3)R^{4}σωπ,**m**= m**k**.

[∫sin^{3}xdx = -(1/3)cosx (sin^{2}x + 2)]

For r >> R we have A_{φ}= (μ_{0}/3)R^{4}σωsinθ/r^{2}.

**∇**^{2}**A**(**r**) = -μ_{0}**j**(**r**),**A**(**r**) = (μ_{0}/(4π)) ∫_{V}dV'_{'}**j**(**r'**)/|**r**-**r**'|. (SI units).

Assume that**A**(**r**) will only has φ component.

Assume that the 1/r^{2}term is the only term in the expansion of**A**in powers of 1/r^{n}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 A_{φ}= Csinθr = (μ_{0}/3)Rσωsinθr inside the sphere.

If this solution fulfills the boundary conditions,

(**B**_{2}-**B**_{1})·**n**_{2}= 0, (**B**_{2}-**B**_{1})·**t**= μ_{0}**k**·**n**,

we have the only solution.

**B**=**∇**×**A**,

B_{in_r}= (1/)rsinθ))∂(sinθA_{φ_in})/∂θ = (2μ_{0}/3)Rσω cosθ, = (μ_{0}/4π)2m cosθ/R^{3}.

B_{in_θ}= -(1/r)∂(rA_{φ_in})/∂r = -(2μ_{0}/3)Rσω sinθ = -(μ_{0}/4π)2m sinθ/R^{3}.

B_{out_r}= (1/(rsinθ))∂(sinθA_{φ_out})/∂θ = (2μ_{0}/3)R^{4}σω cosθ/r^{3}= (μ_{0}/4π)2m cosθ/r^{3}.

B_{out_θ}= -(1/r)∂(rA_{φ_out})/∂r = (μ_{0}/3)R^{4}σωsinθ/r^{3}= (μ_{0}/4π)m sinθ/r^{3}.^{}Now check that the boundary conditions are satisfied.

At r = R the radial component of**B**is continuous.

B_{out_θ}- B_{in_θ}= μ_{0}R^{4}σω sinθ/R^{3}= μ_{0}k_{m}.

The boundary conditions are satisfied and we have the only solution.

The field outside the sphere is a pure dipole field,

**B**(**r**) = (μ_{0}/(4π))[3(**m∙r**_{1})**r**_{1}/r_{1}^{5}-**m**/r_{1}^{3}],

and the field inside the sphere is constant,**B**(**r**) =**k**(μ_{0}/4π)2m/R^{3}.

(b) For a uniformly magnetized sphere with radius R and**M**= M**k**we have**m**= (4/3)πR^{3}**M**.

The magnetization current densities are**j**_{m}= 0,**k**_{m}=**M**×**R**/R, k_{m}= Msinθ .

We may write k_{m}= (3m/(4πR^{3})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**) = (μ_{0}/(4π))[3(**m∙r**_{1})**r**_{1}/r_{1}^{5}-**m**/r_{1}^{3}],

and the field inside the sphere is constant,(

B**r**) =**k**(μ_{0}/4π)2m/R^{3}= (μ_{0}/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**) = (μ_{0}/(4π))[3(**m∙r**_{1})**r**_{1}/r_{1}^{5}-**m**/r_{1}^{3},

The dipole moment is**m**=**M**V =**M**4πR^{3}/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**= M**k**, 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

**k**_{m}=**M**×**R**/R, k_{m}= Msinθ .

Any surface distribution of the same form produces the same fields.