Page 279 - Engineering Electromagnetics, 8th Edition

CHAPTER 8 Magnetic Forces, Materials, and Inductance 261

Figure 8.13 See Problem D8.9.

D8.9. Given the magnetic circuit of Figure 8.13, assume B = 0.6Tat the
midpoint of the left leg and find: (a) V m,air ;(b) V m,steel ;(c) the current required
in a 1300-turn coil linking the left leg.

Ans. 3980 A · t; 72 A · t; 3.12 A

D8.10. The magnetization curve for material X under normal operating con-
ditions may be approximated by the expression B = (H/160)(0.25 + e −H/320 ),
where H is in A/m and B is in T.Ifa magnetic circuit contains a 12 cm length
of material X,as well as a 0.25-mm air gap, assume a uniform cross section
of 2.5 cm and find the total mmf required to produce a flux of (a)10 µWb;
2
(b) 100 µWb.
Ans. 8.58 A · t; 86.7 A · t

8.9 POTENTIAL ENERGY AND FORCES
ON MAGNETIC MATERIALS
In the electrostatic field we first introduced the point charge and the experimental law
of force between point charges. After defining electric field intensity, electric flux
density, and electric potential, we were able to find an expression for the energy in an
electrostatic field by establishing the work necessary to bring the prerequisite point
charges from infinity to their final resting places. The general expression for energy is
1
W E = D · E dν (45)
2 vol
where a linear relationship between D and E is assumed.
This is not as easily done for the steady magnetic field. It would seem that we
might assume two simple sources, perhaps two current sheets, find the force on one