Page 286 - Engineering Electromagnetics, 8th Edition

268 ENGINEERING ELECTROMAGNETICS

In Chapter 11, we will see that the current distribution in a conductor at high
frequencies tends to be concentrated near the surface. The internal flux is reduced, and
it is usually sufficient to consider only the external inductance. At lower frequencies,
however, internal inductance may become an appreciable part of the total inductance.
We conclude by defining the mutual inductance between circuits 1 and 2, M 12 ,
in terms of mutual flux linkages,

N 2 12
M 12 = (63)
I 1

where 12 signifies the flux produced by I 1 which links the path of the filamentary
current I 2 , and N 2 is the number of turns in circuit 2. The mutual inductance, there-
fore, depends on the magnetic interaction between two currents. With either current
alone, the total energy stored in the magnetic field can be found in terms of a single
inductance, or self-inductance; with both currents having nonzero values, the total
energy is a function of the two self-inductances and the mutual inductance. In terms
of a mutual energy, it can be shown that (63) is equivalent to
1
M 12 = (B 1 · H 2 )dν (64)
I 1 I 2 vol
or
1
M 12 = (µH 1 · H 2 )dν (65)
I 1 I 2 vol
where B 1 is the field resulting from I 1 (with I 2 = 0) and H 2 is the field arising from
I 2 (with I 1 = 0). Interchange of the subscripts does not change the right-hand side of
(65), and therefore

M 12 = M 21 (66)
Mutual inductance is also measured in henrys, and we rely on the context to allow
us to differentiate it from magnetization, also represented by M.

EXAMPLE 8.9
Calculate the self-inductances of and the mutual inductances between two coaxial
solenoids of radius R 1 and R 2 , R 2 > R 1 , carrying currents I 1 and I 2 with n 1 and
n 2 turns/m, respectively.
Solution. We first attack the mutual inductances. From Eq. (15), Chapter 7, we let
n 1 = N/d, and obtain

H 1 = n 1 I 1 a z (0 <ρ < R 1 )
= 0(ρ> R 1 )
and

H 2 = n 2 I 2 a z (0 <ρ < R 2 )
= 0(ρ> R 2 )