Page 275 - Engineering Electromagnetics, 8th Edition

CHAPTER 8 Magnetic Forces, Materials, and Inductance 257

magnetic phenomena takes on a slightly different form,

H · dL = I total
for the closed line integral is not zero. Because the total current linked by the path
is usually obtained by allowing a current I to flow through an N-turn coil, we may
express this result as

H · dL = NI (44)

In an electric circuit, the voltage source is a part of the closed path; in the magnetic
circuit, the current-carrying coil will surround or link the magnetic circuit. In tracing
a magnetic circuit, we will not be able to identify a pair of terminals at which the
magnetomotive force is applied. The analogy is closer here to a pair of coupled circuits
in which induced voltages exist (and in which we will see in Chapter 9 that the closed
line integral of E is also not zero).
Let us try out some of these ideas on a simple magnetic circuit. In order to avoid
the complications of ferromagnetic materials at this time, we will assume that we
2
have an air-core toroid with 500 turns, a cross-sectional area of 6 cm ,a mean radius
of 15 cm, and a coil current of 4 A. As we already know, the magnetic field is confined
to the interior of the toroid, and if we consider the closed path of our magnetic circuit
along the mean radius, we link 2000 A · t,
V m, source = 2000 A · t

Although the field in the toroid is not quite uniform, we may assume that it is, for all
practical purposes, and calculate the total reluctance of the circuit as
d 2π(0.15)
9
= = = 1.25 × 10 A·t/Wb
µS 4π10 −7 × 6 × 10 −4
Thus
2000
V m,S −6
= = = 1.6 × 10 Wb
1.25 × 10 9
This value of the total flux is in error by less than 1 4 percent, in comparison with the
value obtained when the exact distribution of flux over the cross section is used.
Hence
1.6 × 10 −6 −3
B = = = 2.67 × 10 T
S 6 × 10 −4
and finally,

B 2.67 × 10 −3
H = = = 2120 A·t/m
µ 4π10 −7
As acheck, we may apply Amp`ere’s circuital lawdirectly in this symmetrical problem,

H φ 2πr = NI