Page 278 - Engineering Electromagnetics, 8th Edition

260 ENGINEERING ELECTROMAGNETICS

which is the same in both steel and air, we may find the mmf required for the gap,
6
−4
V m,air = (6 × 10 )(2.65 × 10 ) = 1590 A·t
ReferringtoFigure8.11,amagneticfieldstrengthof200A · t/misrequiredtoproduce
a flux density of1Tinthe steel. Thus,

H steel = 200 A·t
V m,steel = H steel d steel = 200 × 0.30π
= 188 A·t
The total mmf is therefore 1778 A·t, and a coil current of 3.56 A is required.

We have made several approximations in obtaining this answer. We have already
mentioned the lack of a completely uniform cross section, or cylindrical symmetry;
the path of every flux line is not of the same length. The choice of a “mean” path
length can help compensate for this error in problems in which it may be more
important than it is in our example. Fringing flux in the air gap is another source of
error, and formulas are available by which we may calculate an effective length and
cross-sectional area for the gap which will yield more accurate results. There is also
a leakage flux between the turns of wire, and in devices containing coils concentrated
on one section of the core, a few flux lines bridge the interior of the toroid. Fringing
and leakage are problems that seldom arise in the electric circuit because the ratio
of the conductivities of air and the conductive or resistive materials used is so high.
In contrast, the magnetization curve for silicon steel shows that the ratio of H to B
in the steel is about 200 up to the “knee” of the magnetization curve; this compares
with a ratio in air of about 800, 000. Thus, although flux prefers steel to air by the
commanding ratio of 4000 to 1, this is not very close to the ratio of conductivities of,
say, 10 15 for a good conductor and a fair insulator.

EXAMPLE 8.8
As a last example, let us consider the reverse problem. Given a coil current of 4 A in
the magnetic circuit of Example 8.7, what will the flux density be?
Solution. First let us try to linearize the magnetization curve by a straight line from
the origin to B = 1, H = 200. We then have B = H/200 in steel and B = µ 0 H in air.
6
6
The two reluctances are found to be 0.314×10 for the steel path and 2.65×10 for the
6
air gap, or 2.96×10 A · t/Wb total. Since V m is 2000 A · t, the flux is 6.76×10 −4 Wb,
and B = 1.13 T. A more accurate solution may be obtained by assuming several values
of B and calculating the necessary mmf. Plotting the results enables us to determine
the true value of B by interpolation. With this method we obtain B = 1.10 T. The good
accuracy of the linear model results from the fact that the reluctance of the air gap
in a magnetic circuit is often much greater than the reluctance of the ferromagnetic
portion of the circuit. A relatively poor approximation for the iron or steel can thus
be tolerated.