Page 243 - Electromagnetics

where ω M =|ω M | is the saturation magnetization frequency, we find that

˜ ˜ ω 0 1 ˜
M T + M T × = − ω M × H T , (4.112)
jω jω
where ω 0 = γµ 0 H dc with ω 0 now called the gyromagnetic response frequency. This has
the form v + v × C = A, which has solution (4.80). Substituting into this expression and
remembering that ω 0 is parallel to ω M , we find that
1 ˜ 1 ˜ ˜
− ω M × H T + 2 ω M [ω 0 · H T ] − (ω 0 · ω M )H T
˜ jω ω
M T = .
ω 2
1 − 0
ω 2
˜
˜
If we define the dyadic ¯ω M such that ¯ω M · H T = ω M × H T , then we identify the dyadic
magnetic susceptibility
jω ¯ω M + ω M ω 0 − ω M ω 0 I ¯
˜ ¯ χ (ω) = (4.113)
m 2 2
ω − ω 0
˜
˜
with which we can write M(r,ω) = ¯χ m (ω) · H(r,ω). In rectangular coordinates ¯ω M is
represented by
 
0 −ω Mz ω My
[ ¯ω M ] =  ω Mz 0 −ω Mx  . (4.114)
−ω My ω Mx 0
˜
˜
˜
¯
˜
˜
Finally, using B = µ 0 (H + M) = µ 0 (I + ˜ ¯χ ) · H = ˜ ¯µ · H we find that
m
¯
˜ ¯ µ(ω) = µ 0 [I + ˜ ¯χ (ω)].
m
To examine the properties of the dyadic permeability it is useful to write it in matrix
form. To do this we must choose a coordinate system. We shall assume that H dc is
aligned with the z-axis so that H dc = ˆ zH dc and thus ω M = ˆ zω M and ω 0 = ˆ zω 0 . Then
(4.114) becomes
 
0 −ω M 0
[ ¯ω M ] =  ω M 0 0 
0 0 0
and we can write the susceptibility dyadic (4.113) as
 
−ω 0 − jω 0
ω M
[ ˜ ¯χ (ω)] = 2  jω −ω 0 0  .
m
2
ω − ω
0 0 0 0
The permeability dyadic becomes
 
µ − jκ 0
[ ˜ ¯µ(ω)] =  jκµ 0  (4.115)
0 0 µ 0
where

ω 0 ω M
µ = µ 0 1 − , (4.116)
2
ω − ω 2
0
ωω M
κ = µ 0 2 2 . (4.117)
ω − ω
0
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