Phys560Notes-16 - Magnetism B H 4 M H M = magnetization Bi...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Magnetism 4  BH M H M = magnetization Bournday conditions: 0   B  io BB 41 0  cc t D Hj , assuming steady state and no free current  H H Ferromagnetic 0 M for 0 a B (zero applied field); generally nonlinear Paramagnetic a MB ~ constant ~ 1 + O (10 –2 – 10 –6 ) BH Diamagnetic   a ~ constant ~ 1 – O (10 –5 ) Shape anisotropy or demagnetization effects (1) A very large plate perpendicular to the external field: 4 i o BBH MH near the center. HH (assuming M is in the same direction.) The internal field is reduced by demagnetization (because of surface magnetic charge). If 0 oo  (zero external field), 0 i B and 4 i H M   . 2 (2) A long thin needle along the field: 4 i o H HB M B No demagnetization. If 0 0 o (zero external field), 0 i H and 4 i BM . (3) An ellipsoid: 4  H 4  H HM 4  N M 01  N N = demagnetization factor A plate perpendicular to the field: 1 N , , 4 H . o B i B ellipsoid
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
3 A thin needle along the field: 0 N , 4 io BM B , H H . We will ignore demagnetization effects. Or, we will consider the dilute limit, and so M is small, and BH (fields caused by other atoms can be ignored). Microscopic theory of magnetism: Assume a uniform external field. ˆ Bz B 1 2  Ar B Hamiltonian for a set of electrons without a magnetic field:  2 0 1 2  ii i Hp U m r With a field: 2 0 1 2     B i z e H Ug B s mc pA r 0 2, , 2 2 Bi i e g mc  s σ  spin 0 H HH 2 22 2 0 2 8   i i e H gB x y mc LS B (verify the math yourself) where i i Ss (total spin in units of );  i Lr p = total orbital angular momentum in units of 4 2 2 0 4    Bo i i i e H gx y d mc B B B Classically,  E μ B for a fixed dipole in a magnetic field.  dE d μ B . If μ depends on B (for an induced dipole), 0   Ed B μ B . 0 H Vd B MB V M = total dipole moment of system V M (an operator) 2 0 2 4 i i e y mc B Energy of state n (a many e state):   0 n En H V d n B 00 |     n nn n E nHV d n nHV d n V n n nE n nE n Vn n n V n n V n n B M BB B M M B assuming that n is normalized.
Background image of page 2
5 Theorem: n nn MM = average magnetic moment per unit volume for state n =  1 n E V B B (Only the eigenvalue is needed; not the wave function!) Total magnetization: n n E kT n n Ek T n e e M M Define Helmholtz free energy   ln ln     n T n Fk T Z k T e B Can show 1 F
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/21/2012 for the course PHYS 560 taught by Professor Flynn during the Spring '08 term at University of Illinois, Urbana Champaign.

Page1 / 7

Phys560Notes-16 - Magnetism B H 4 M H M = magnetization Bi...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online