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Electromagnetism 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 r × 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 r × A = fr × c − c × rf = −c × rf (2) since r × 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 × rf · dS = −c · rf × dS; (3) but this is true for any c and so I Z f dl = − rf × dS: (4) Now let us choose f = r^ · r0 and take the gradient and integral around a closed loop in the primed coordinates: I Z Z (r^ · r0) dl0 = − r0(r^ · r0) × dS0 = −r^ × dS0 (5) which works because r0(r^ · r0) = r^. In summary: I Z r^ · r0 dl0 = dS0 × r^: (6) We will use this to show that the magnetic vector potential from a magnetic dipole is given by µ A = 0 m × r^; (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 φ^; (8) 4πr2 and so 1 @ 1 @ B = r × A(r) = (r sin θ A )r^ − (r sin θ A )θ^; (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 r^ + θ^ ; (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 atoms and m is the magnetic dipole moment of each atom. • The potential from a box of magnetic dipoles is as if there is a volume (bound) current density J b = −r × M and surface (bound) current density Kb = M × n^. • The magnetic field strength H is given in terms of the magnetic flux density B by B = µ0(H + M): This implies that r × H = J f , where J f is the free current density. • For a linear material, M = χmH, where χm is the magnetic susceptibility. 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, r × H = 0 and we can write H = −rφm and (for a linear material) 2 B = −µoµrrφm. Away from interfaces, r·H = 0 and so r 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 transformers, motors etc. The figure above shows a hysteresis loop with the coercive field Hc, remanent magnetization −1 Hard magnetic materials: Hc ∼ 10 MA m . Mr, and saturation 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 magnets. M = (µr − 1)H then µr = µr(H). SJB Michaelmas 2019.

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