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Handout 3:

3. Magnetization (Lectures 7–9): Here is the proof of a useful vector identity. First recall Stokes’ theorem which states that Z I ∇ × A · dS = A · dl. (1)

Let us suppose that A takes the form A(r) = f(r)c where c 6= c(r) is a constant vector. Hence we have ∇ × A = f∇ × c − c × ∇f = −c × ∇f (2) since ∇ × c = 0 (because c is a constant vector, its derivative vanishes). Stokes’ theorem then gives us I I Z Z A · dl = c · f dl = − c × ∇f · dS = −c · ∇f × dS, (3) but this is true for any c and so I Z f dl = − ∇f × dS. (4)

Now let us choose f = rˆ · r0 and take the gradient and integral around a closed loop in the primed coordinates: I Z Z (rˆ · r0) dl0 = − ∇0(rˆ · r0) × dS0 = −rˆ × dS0 (5) which works because ∇0(rˆ · r0) = rˆ. In summary: I Z  rˆ · r0 dl0 = dS0 × rˆ. (6)

We will use this to show that the magnetic vector potential from a is given by µ A = 0 m × rˆ, (7) 4πr2 where m is the magnetic dipole moment. For example, if we put m parallel to z and use spherical polars, we have that µ A = 0 m sin θ φˆ, (8) 4πr2 and so 1 ∂ 1 ∂ B = ∇ × A(r) = (r sin θ A )rˆ − (r sin θ A )θˆ, (9) r2 sin θ ∂θ φ r sin θ ∂r φ because Ar = Aθ = 0 and hence µ m 2 cos θ sin θ  B = 0 rˆ + θˆ , (10) 4π r3 r3 which is the same form as an electric dipole. • The magnetization M is the magnetic dipole moment per unit volume, so that M = nm where n is the number density of and m is the magnetic dipole moment of each .

• The potential from a box of magnetic dipoles is as if there is a volume (bound) J b = −∇ × M and surface (bound) current density Kb = M × nˆ. • The magnetic field strength H is given in terms of the magnetic flux density B by

B = µ0(H + M).

This implies that ∇ × H = J f , where J f is the free current density.

• For a linear material, M = χmH, where χm is the . B = µ0(H + M) = µ0µrH and so µr = 1 + χm is the relative permeability.

• The boundary conditions are: Hk and B⊥ are continuous.

Magnetic scalar potential φm

If J f = 0 everywhere, ∇ × H = 0 and we can write H = −∇φm and (for a linear material) 2 B = −µoµr∇φm. Away from interfaces, ∇·H = 0 and so ∇ φm = 0 and we can use Laplace’s equation formalism.

Relative permeability µr

substance µr air 1.00000037 platinum 1.000265 iron ∼5000 5 permalloy (Ni0.80Fe0.20) up to 10

Coercive field Hc

−1 Soft magnetic materials: Hc ∼ 0.1 A m . It is easy to move domain walls and these mate- rials are useful for , motors etc. The figure above shows a loop with the coercive field Hc, remanent magnetization −1 Hard magnetic materials: Hc ∼ 10 MA m . Mr, and magnetization Ms. This is It is hard to move domain walls and these ma- obviously non-linear and so if you model it as terials are useful for permanent . M = (µr − 1)H then µr = µr(H).

SJB Michaelmas 2019