CHAPTER 18
PROBLEM SOLUTIONS
18.1 A coil of wire wire 0.20 m long and having having 200 turns carries a current of 10 A. (a) What is the magnitude of the magnetic field strength H? (b) Compute the flux density B if the coil is in a vacuum. (c) Compute the flux density inside a bar of titanium that is positioned positioned within the coil. The susceptibility susceptibility for titanium is given in Table Table 18.2. (d) Compute the magnitude of the magnetization magnetization M. Solution (a) We may calculate the magnetic field field strength generated by this coil using Equation 18.1 as
H
=
(200 turns)( turns)(10 10
NI =
A)
0.20 m
l
= 10,000 A-turns/ A-turns/m m
(b) In a vacuum, the flux flux density is determined determin ed from from Equation 18.3. Thus, B = 0
= (1.257
H
0
10-6 H/m H/m)(10,000 A-turns/m) A-turns/m) = 1.257
10-2 tesla
(c) When a bar of titanium is positioned positioned within the coil, we we must use an expression that is a combination combination of Equations 18.5 and 18.6 in order to compute compute the flux density given given the magnetic susceptibility. Inasmuch as = m
1.81 10-4 (Table 18.2), then B =
= (1.257
H +
0
M =
0
H +
0
0
H =
m
(
H 1
0
+
10-6 H/m H/m) (10,000 A-turns/m)(1 + 1.81 10-2 tesla
m
)
10-4 )
which is essentially the same result as part (b). This is to say that the influence of the titanium bar within the coil makes an imperceptible difference in the magnitude of the B field. (d) The magnetization is computed from Equation 18.6:
M =
H =
m
10-4 )(10,000 A-turns/m) = 1.81 A/m
(1.81
18.6 The magnetization within a bar of some metal alloy is 3.2
5
10
A/m at an H field of 50 A/m. Compute the
following: (a) the magnetic susceptibility, (b) the permeability, and (c) the magnetic flux density within this material. (d) What type(s) of magnetism would you suggest as being displayed by this material? Why? Solution (a) This portion of the problem calls for us to compute the magnetic susceptibility within a bar of some metal alloy when M = 3.2 105 A/m and H = 50 A/m. This requires that we solve for m from Equation 18.6 as
m
=
M
5
3.2
=
10 A/m
= 6400
50 A/m
H
(b) In order to calculate the permeability we must first of all employ a altered form of Equation 18.7 as
r
=(
m
+ 1)
In addition, Equation 18.4 may be written as = r
By substituting the expression for
r
0
from the first equation above into the second equation and incorporating
values for the parameters given in the problem statement or determined in part (a) yields
=
= (6400 + 1)(1.257
r
0
= (
m
+ 1)
0
10 -6 H/m) = 8.05
10-3 H/m
(c) The magnetic flux density may be determined using Equation 18.2 as
B = H =
(8.05
10-3 H/m) (50 A/m) = 0.40 tesla
(d) This metal alloy would exhibit ferromagnetic behavior on the basis of the magnitude of its m (6400), which is considerably larger than the m values for diamagnetic and paramagnetic materials listed in Table 18.2.
18.13 The chemical formula for manganese ferrite may be written as (MnFe 2O4 )8 because there are eight formula units per unit cell. If this material has a saturation magnetization of 5.6
10
5
A/m and a density of 5.00
g/cm3 , estimate the number of Bohr magnetons associated with each Mn2+ ion. Solution We want to compute the number of Bohr magnetons per Mn 2+ ion in (MnFe 2O4)8. Let nB represent the number of Bohr magnetons per Mn 2+ ion; then, using Equation 18.9, we have
M s = nB
B N
in which N is the number of Mn 2+ ions per cubic meter of material. But, from Equation 18.10 N A
N =
A
Here A is the molecular weight of MnFe 2O4 (230.64 g/mol). Thus, combining the previous two equations
M s =
nB
B
N A
A
or, upon rearrangement (and expressing the density in units of grams per meter cubed),
nB =
M s A B
=
(5.6
(9.27
N A
105 A/m) (230.64 g/mol)
10 24 A-m2 /BM)( 5.00
106 g/m3 )(6.022
1023 ions/mol)
= 4.6 Bohr magnetons/Mn2+ ion
18.17 A coil of wire 0.1 m long and having 15 turns carries a current of 1.0 A. (a) Compute the flux density if the coil is within a vacuum. (b) A bar of an iron – silicon alloy, the B-H behavior for which is shown in Figure 18.29, is positioned within the coil. What is the flux density within this bar?
(c) Suppose that a bar of molybdenum is now situated within the coil. What current must be used to produce the same B field in the Mo as was produced in the iron – silicon alloy [part (b)] using 1.0 A? Solution (a) This portion of the problem asks that we compute the flux density in a coil of wire 0.1 m long, having 15 turns, and carrying a current of 1.0 A, and that is situated in a vacuum. Combining Equations 18.1 and 18.3, and solving for B yields B0 =
=
(1.257 10
6
0 H =
0 NI
l
H/m) (15 turns)(1.0 A) 0.1 m
= 1.89
10 -4 tesla
(b) Now we are to compute the flux density with a bar of the iron-silicon alloy, the B-H behavior for which is shown in Figure 18.29. It is necessary to determine the value of H using Equation 18.1 as H =
NI
(15 turns)(1.0 A)
l
0.1 m
=
= 150 A-turns/m
Using the curve in Figure 18.29, B = 1.65 tesla at H = 150 A-turns/m, as demonstrated below. 18.19 The following data are for a transformer steel: B H
(A /m )
(tesl as )
B H
(A/ m )
(tesl as )
0
0
200
1.04
10
0.03
400
1.28
20
0.07
600
1.36
50
0.23
800
1.39
100
0.70
1000
1.41
150
0.92
(a) Construct a graph of B versus H. (b) What are the values of the initial permeability and initial relative permeability? (c) What is the value of the maximum permeability? (d) At about what H field does this maximum permeability occur? (e) To what magnetic susceptibility does this maximum permeability correspond? Solution
(a) The B- H data for the transformer steel provided in the problem statement are plotted as follows:
(b) The first four data points are plotted below.
The slope of the initial portion of the curve is i (as shown), is
i
=
B H
=
(0.15 (50
0) tesla 0) A/m
= 3.0
´
10-3 H/m
Also, the initial relative permeability, ri, (Equation 18.4) is just
ri =
i
0
=
3.0 1.257
10 3 H/m 10 6 H/m
= 2387
(c) The maximum permeability is the tangent to the B- H curve having the greatest slope; it is drawn on the plot below, and designated as (max).
The value of (max) is (modifying Equation 18.2)
(max) =
B H
=
(1.3
0.3) tesla
(160
45) A-m
= 8.70
´
10-3 H/m
(d) The H field at which (max) occurs is approximately 80 A/m [as taken from the plot shown in part (c)]. (e) We are asked for the maximum susceptibility, (max). Combining modified forms of Equations 18.7 and 18.4 yields
(max) =
(max)
r
1
(max)
=
0
=
8.70 1.257
10
3
H/m
10
6
H/m
1 = 6920
1
