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P5 SOLUTIONS TO PROBLEMS
MAGNETOSTATIC FIELD IN MATERIAL
MEDIA
Section 5.3 Magnetization Volume and Surface
Current Densities
PROBLEM 5.1 Nonuniformly magnetized parallelepiped. (a) Combining Eqs.(5.28) and (4.81), the volume magnetization current density vector inside the parallelepiped is obtained to be
Jm = ∇ × M = −
∂My ∂z ˆx −
∂Mz ∂x ˆy −
∂Mx ∂y ˆz
= −πM 0
c
cos
πz c
ˆx +
a
cos
πx a
ˆy +
b
cos
πy b
ˆz
. (P5.1)
(b) Eq.(5.32) tells us that the surface magnetization current density vector over the back (x = 0) and front (x = a) sides of the parallelepiped (assuming that its position in space is as in Fig.5.7) are
Jms1 = M × nˆ = M(0+, y, z) ×(− ˆx) = M 0 sin
πz c
yˆ ×(− ˆx) = M 0 sin
πz c
ˆz ,
and Jms2 = M(a−, y, z) × ˆx = M 0 sin πz c
yˆ × ˆx = −M 0 sin πz c
ˆz , (P5.2)
respectively. On the left- (y = 0) and right-hand (y = b) sides of the parallelepiped,
Jms3 = M(x, 0 +, z) ×(− ˆy) = M 0 sin
πx a
ˆz ×(− ˆy) = M 0 sin
πx a
ˆx ,
and Jms4 = M(x, b−, z) × ˆy = M 0 sin πx a
ˆz × yˆ = −M 0 sin πx a
ˆx , (P5.3)
and on the bottom (z = 0) and top (z = c) sides,
Jms5 = M(x, y, 0 +) ×(− ˆz) = M 0 sin
πy b
ˆx ×(− ˆz) = M 0 sin
πy b
ˆy ,
and Jms6 = M(x, y, c−) × ˆz = M 0 sin
πy b ˆx × ˆz = −M 0 sin
πy b ˆy. (P5.4)
PROBLEM 5.2 Hollow cylindrical bar magnet. (a) According to Eq.(5.29), there is no magnetization volume current inside the magnet, Jm = 0. In the cylin- drical coordinate system adopted such that its z-axis is along the magnet axis, Eq.(5.32) gives the following magnetization surface current densities on the inner (r = a) and outer (r = b) lateral (cylindrical) sides of the magnet, similarly to Eq.(5.36) and Fig.5.8:
Jmsa = M ˆz × (− ˆr) = −M φˆ and Jmsb = M ˆz × ˆr = M φˆ , (P5.5)
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P5. Solutions to Problems: Magnetostatic Field in Material Media 141
whereas Jms = 0 on both the upper and the lower bases of the bar.
(b) Similarly to the analysis in Example 5.3, the magnetic flux density vector along the axis of the magnet can be found as B due to two l long cylindrical current sheets of radii a and b in a vacuum with the surface current densities in Eqs.(P5.5), or due to two corresponding solenoids in Fig.4.10. So, using Eq.(4.36), twice, with N I/l substituted, based on Eq.(4.30), by Jmsa = −M and Jmsb = M , respectively, we obtain
B = Ba + Bb =
μ 0 Jmsa 2
(sin θ 2 − sin θ 1 ) ˆz +
μ 0 Jmsb 2
(sin θ 2 ′ − sin θ 1 ′) ˆz
μ 0 M 2
(− sin θ 2 + sin θ 1 + sin θ′ 2 − sin θ′ 1 ) ˆz , (P5.6)
where the angles θ 1 and θ 2 are defined exactly as those in Fig.4.10(b), while θ′ 1 and θ′ 2 are defined in the same way but for the solenoid radius b.
PROBLEM 5.3 Uniformly magnetized square ferromagnetic plate. This is similar to the analysis in Example 5.2 (Fig.5.8), the only difference being the shape of the equivalent current loop; namely, the ferromagnetic plate in Fig.5.36 is replaced by a square loop, with edge length a and the current intensity [Eq.(5.37)]
Im = Jmsd = M 0 d , (P5.7)
assumed to be in a vacuum. The magnetic flux density vector at the center of a square current loop is computed in Example 4.3, whereas B at an arbitrary point along the axis (z-axis) of a rectangular loop, and thus along the z-axis in Fig.5.36, is found in Problem 4.1, and the result for a = b (square loop) and I = Im = M 0 d is B =
2 μ 0 M 0 a^2 d π(4z^2 + a^2 )
2 z^2 + a^2
ˆz. (P5.8)
PROBLEM 5.4 Magnetization parallel to plate faces. Now we have two magnetization current sheets of square shapes and densities
Jms1 = M × ˆn = M 0 ˆy × ˆz = M 0 ˆx and Jms2 = M 0 ˆy × (− ˆz) = −M 0 ˆx , (P5.9)
on the upper and lower faces, respectively, of the ferromagnetic plate in Fig.5.36. We neglect end effects, and thus disregard magnetization surface currents on the two lateral sides (thin strips) of the plate perpendicular to the x-axis and, moreover, find the magnetic flux density vector (B) at the plate center (point O) as that due to two infinite planar sheets with surface currents of the same density (Js) running in opposite directions. To this end, we can employ Eq.(4.47) for a single infinite planar current sheet and the superposition principle or use the result of Problem 4.14, part (b), to obtain
B = B 1 + B 2 = 2B 1 = μ 0 Js ˆy = μ 0 Jms1 ˆy = μ 0 M 0 ˆy = μ 0 M. (P5.10)
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P5. Solutions to Problems: Magnetostatic Field in Material Media 143
−z^2
∫ (^) a
r=
dR R^2
ˆz = − μ 0 M 0 d a^2
a^2 + z^2 − |z| + z^2 √ a^2 + z^2
z^2 |z|
ˆz. (P5.15)
Finally, we add the B field due to the equivalent circular loop with radius a and current Im(a) = Jmsd = M 0 d, representing the surface current in Eq.(P5.12), and this field is given in Eq.(5.38), where we replace M by M 0 , so that the total B amounts to
Btot = −
μ 0 M 0 d a^2
[
a^2 + z^2 − |z| +
z^2 √ a^2 + z^2
z^2 |z|
a^4 2 (z^2 + a^2 )^3 /^2
]
ˆz. (P5.16)
PROBLEM 5.6 Infinite cylinder with circular magnetization. (a) By means of Eqs.(5.28) and (4.84), the volume magnetization current density vector in the ferromagnetic cylinder comes out to be
Jm = ∇ × M =
r
∂r
[rMφ(r)] ˆz =
2 M 0
a
ˆz. (P5.17)
Note that this distribution of the magnetization current, axial and uniform through- out the cylinder (Jm = Jm ˆz and Jm = const), has the same form as the distribution of the conduction current in the cylindrical copper conductor in Fig.4.15.
(b) The surface magnetization current density vector, Eq.(5.32), on the cylinder surface is Jms = Mφ(a−) φˆ × ˆr = −M 0 ˆz. (P5.18)
(c) Because of symmetry, the B field in the cylinder (due to its magnetization currents assumed to reside in a vacuum) is circular (magnetic-field lines are circles centered at the cylinder axis), and of the form in Eq.(4.53). To find this field, we apply Amp`ere’s law, Eq.(4.48), as if the magnetization currents found in (a) and (b) were conduction currents in a nonmagnetic medium, to the circular contour C of radius r, as in Fig.4.15(a),
B 2 πr = μ 0 Jmπr^2 −→ B =
μ 0 Jmr 2
φˆ = μ^0 M^0 r a
φˆ = μ 0 M (0 ≤ r ≤ a). (P5.19)
(d) For the observation point outside the cylinder, the right-hand side of Amp`ere’s law includes the surface magnetization current of density Jms, Eq.(P5.18), as well, and this current amounts, using Eq.(3.13), to Jms times the circumference of the cylinder. Hence, B outside the ferromagnetic cylinder is computed as follows:
B 2 πr = μ 0
Jmπa^2 + Jms 2 πa
−→ B =
μ 0 a r
Jma 2
μ 0 a r
(M 0 − M 0 ) = 0 (a < r < ∞). (P5.20)
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144 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
Section 5.4 Generalized Amp`ere’s Law
PROBLEM 5.7 Magnetic field intensity vector. (a) The magnetic flux density vector, B, along the axis (z-axis) of the magnetized disk in Fig.5.8 is given in Eq.(5.38), from which the magnetic field intensity vector, H, along the axis is found using Eq.(5.50) for points in the disk and Eq.(5.61) for the part of the z-axis in air, as follows:
H =
B
μ 0
− M =
B
μ 0 − M ˆz = M
[
d a^2 2 (z^2 + a^2 )^3 /^2
]
ˆz (|z| ≤ d/2)
and H =
B
μ 0
M d a^2 2 (z^2 + a^2 )^3 /^2
ˆz (|z| > d/2). (P5.21)
(b) By means of Eqs.(5.50) and (5.39), the magnetic field intensity vector at the center of the cylindrical bar magnet (from Example 5.3) amounts to
H =
B
μ 0
− M =
l √ l^2 + 4a^2
M. (P5.22)
(c) The magnetic field intensity is zero both inside and outside the nonuniformly magnetized cylinder in Fig.5.9, which is obtained from Eq.(5.41) for points inside the cylinder and the fact that B = 0 outside it, respectively,
H =
B
μ 0
− M = M − M = 0 (r ≤ a) and H =
B
μ 0
= 0 (r > a). (P5.23)
PROBLEM 5.8 Total (conduction plus magnetization) current density. (a) A combination of Eqs.(5.52), (5.28), and (5.50) yields that the total (conduction plus magnetization) volume current density in a ferromagnetic material can, indeed, be expressed in terms of the magnetic flux density vector, B, in the material only, as follows:
Jtot = J + Jm = ∇ × H + ∇ × M = ∇ × (H + M) = ∇ ×
B
μ 0
μ 0
∇ × B. (P5.24)
(b) For the given function B(x, y, z), we use the formula for the curl in Cartesian coordinates, Eq.(4.81), to obtain
Jtot =
μ 0
∇ × B =
μ 0
[(
∂Bz ∂y
∂By ∂z
ˆx +
∂Bx ∂z
∂Bz ∂x
ˆy +
∂By ∂x
∂Bx ∂y
ˆz
]
μ 0
[
2 xˆ +
y^2
y ˆ +
2(2x + z) y^3
ˆz
]
(A/m^2 ) (x, y, z in m). (P5.25)
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146 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
where μr − 1 can be taken out of the integral sign because μr = const (homogeneous magnetic material).
PROBLEM 5.13 Thin toroidal coil with a linear ferromagnetic core. (a) From the generalized Amp`ere’s law, Eq.(5.49), the circulation of the magnetic field intensity vector through the core, along its mean length (contour C), amounts to (see Fig.5.11) (^) ∮
C
H · dl = N I. (P5.31)
(b) By means of Eqs.(5.60) and (P5.31), the circulation along C of the magnetic flux density vector is ∮
C
B · dl =
C
μH · dl = μ
C
H · dl = μN I. (P5.32)
(c) Finally, with the use Eqs.(5.68) and (P5.31), the circulation of the magnetization vector through the core turns out to be ∮
C
M · dl =
C
(μr − 1) H · dl = (μr − 1)
C
H · dl =
μ − μ 0 μ 0
N I. (P5.33)
PROBLEM 5.14 Solenoidal coil with a linear ferromagnetic core. (a) The magnetic field intensity vector inside the solenoid, H, is axial, while it is zero outside it. Adopting a cylindrical coordinate system with the z-axis along the solenoid axis, as in Fig.4.10(b), and applying the generalized Amp`ere’s law, Eq.(5.49), to the same rectangular contour C as in Fig.4.19, we obtain, for an arbitrary point in the ferromagnetic core, Hl = N ′Il −→ H = N ′I ˆz. (P5.34)
(b)-(c) Using Eqs.(5.60) and (5.68), the magnetic flux density and magnetization vectors in the core are
B = μH = μN ′I ˆz and M = (μr − 1) H = μ − μ 0 μ 0
N ′I ˆz , (P5.35)
respectively.
(d)-(e) According to Eq.(5.29), since M = const, the volume magnetization current density vector in the core is Jm = 0. By means of Eq.(5.32), the surface magneti- zation current density vector over the surface of the core equals
Jms = M × ˆn = M ˆz × ˆr = M φˆ = μ − μ 0 μ 0
N ′I φˆ. (P5.36)
This magnetization current flows in the same way as the conduction current through the solenoid turns.
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P5. Solutions to Problems: Magnetostatic Field in Material Media 147
PROBLEM 5.15 Ferromagnetic cylinder with a conduction current. (a) In the cylindrical coordinate system adopted as in Fig.4.15(a), the magnetic field intensity vector (H) in the ferromagnetic cylinder and outside it is circular and of the form as in Eq.(4.53). From the generalized Amp`ere’s law, Eq.(5.51), applied as in Fig.4.15(a), we obtain
H(r) 2πr = Jπr^2 −→ H(r) = Jr 2
φˆ , for r ≤ a ;
H(r) 2πr = Jπa^2 −→ H(r) =
Ja^2 2 r
φˆ , for r > a. (P5.37)
(b) With the use of Eq.(5.60) for points in the cylinder and Eq.(5.61) for those outside it, the magnetic flux density vector in the two regions turns out to be
B(r) = μH(r) =
μrμ 0 Jr 2
φˆ (r ≤ a) ; B(r) = μ 0 H(r) = μ^0 Ja
2 2 r
φˆ (r > a). (P5.38)
(c) The magnetization vector inside the cylinder is found invoking Eq.(5.68), while it is zero in air, and hence
M(r) = (μr − 1) H(r) =
(μr − 1)Jr 2
φˆ (r ≤ a) ; M(r) = 0 (r > a). (P5.39)
(d) There is no volume magnetization current inside the cylinder [Eq.(5.29)]. By means of Eq.(5.32), the surface magnetization current density vector on the cylinder surface (r = a) amounts to
Jms = M(a−) × ˆn =
(μr − 1)Ja 2
φˆ × ˆr = − (μr^ −^ 1)Ja 2
ˆz. (P5.40)
Section 5.6 Maxwell’s Equations and Boundary
Conditions for the Magnetostatic Field
PROBLEM 5.16 Magnetic-magnetic boundary conditions. (a) From the boundary condition in Eq.(5.75), as there is no surface conduction current on the boundary (Js = 0), the tangential components of H 1 and H 2 , namely, their x- and y-components, must be the same on the two sides of the boundary (plane z = 0), so H 2 x = H 1 x = 5 A/m and H 2 y = H 1 y = −3 A/m. (P5.41) Eq.(5.76) gives the following for the normal component (z-component) of H 2 for z = 0−:
μ 1 H1n = μ 2 H2n −→ H 2 z = H2n =
μ 1 μ 2
H1n =
μr μr
H 1 z = 4.8 A/m , (P5.42)
with which the magnetic field intensity vector in medium 2 near the boundary equals
H 2 = (5 xˆ − 3 ˆy + 4. 8 ˆz) A/m (Js = 0). (P5.43)
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P5. Solutions to Problems: Magnetostatic Field in Material Media 149
where the partial forces are computed based on Eq.(4.166). As can be seen in Fig.P5.2, the resultant force is along the line connecting wires 1 and 4, that is, along the symmetry line in the cross section of the original system in Fig.5.38, and its p.u.l. magnitude is
F (^) m1′ = F (^) m21′ cos 45◦^ + F (^) m31′ cos 45◦^ + F (^) m41′ = 2
μ 0 I^2 2 π(2h)
μ 0 I^2 2 π(2h
2 μ 0 I^2 8 πh
(P5.49)
PROBLEM 5.18 Uniformly magnetized hollow disk on a PMC plane. In the cylindrical coordinate system adopted such that its z-axis is the one in Fig.5.39, Eq.(5.32) tells us that magnetization surface currents flow circumferentially along the the inner and outer lateral surfaces of the hollow ferromagnetic disk, and their densities are, respectively,
Jmsa = M ˆz × (− ˆr) = −M φˆ (r = a) and Jmsb = M ˆz × ˆr = M φˆ (r = b). (P5.50) As d ≪ a, b, these circumferential sheets of magnetization current can be regarded as equivalent circular current loops with radii a and b and current intensities [Eq.(5.37)]
∓Im = ∓Jmsb d = ∓M d (P5.51)
(currents in the two loops flow in opposite directions). By the image theory for the magnetic field (Fig.5.15), the current loops right on the plane (at z = 0+) and their positive images right below the plane (at z = 0−) add to each other, so the currents intensities ∓Im in fact double, and are considered to reside in a vacuum. Using Eq.(4.19) or (5.38), with current intensities − 2 Im and 2Im, and radii a and b, respectively, the magnetic flux density vector along the z-axis above the PMC plane in Fig.5.39 comes out to be
B = Ba + Bb =
μ 0 (− 2 Im) a^2 2 (z^2 + a^2 )^3 /^2
ˆz +
μ 0 (2Im) b^2 2 (z^2 + b^2 )^3 /^2
ˆz
= −μ 0 M d
[
a^2 (z^2 + a^2 )^3 /^2
b^2 2 (z^2 + b^2 )^3 /^2
]
ˆz (0 < z < ∞). (P5.52)
PROBLEM 5.19 Magnetized cylinder between two PMC planes. We can replace the ferromagnetic cylinder in Fig.5.40 by a cylindrical magnetization- current sheet of radius a, hight (length) h, and surface current density in Eq.(5.36), where M = M 0 , in a vacuum. By multiple applications of the image theory for the magnetic field (Fig.5.15), much like those in Fig.5.17, the cylindrical current sheet of length h becomes infinite, and thus equivalent to an infinitely long solenoid, Fig.4.19. This means that the magnetic field outside it (for r > a), so in air in Fig.5.40, is zero, while that inside it is axial, given by Eq.(4.37), which we modify according to Eq.(4.30). Hence the magnetic flux density vector in the space between the PMC planes in Fig.5.40, inside and outside the cylinder,
B = μ 0 N ′I ˆz = μ 0 Jms ˆz = μ 0 M 0 ˆz = μ 0 M (r < a , 0 < z < h)
and B = 0 (r > a , 0 < z < h). (P5.53)
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150 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
Section 5.9 Magnetic Circuits – Basic Assumptions
for the Analysis
PROBLEM 5.20 Magnetic flux through a thick toroid. (a) The magnetic field value corresponding to the “knee” point between the two parts of the piece- wise linear magnetization curve in Fig.5.30(b) is Hk = 1000 A/m, the same as in Fig.5.27(b). Therefore, the radial distance c in Fig.5.27(a) representing the bound- ary between the two parts of the core, one in which the second (farther) part of the curve in Fig.5.30(b) applies and the other with B(H) given by the first part (containing the point B = 0) of the curve, is the same as in Eq.(5.91), so c = 3.2 cm. In specific, from Eq.(5.103), the magnetic flux density in the first part of the core is
B 1 (r) = 0.4+2. 5 × 10 −^5 [H(r)−1000] (T) =
7. 96 × 10 −^4
r
T (a ≤ r ≤ c) (P5.54) [H(r) is given in Eq.(5.83)], whereas we observe in Fig.5.30(b) that B in the second part can be expressed as
B 2 (r) = 4 × 10 −^4 H(r) (T) =
1. 27 × 10 −^2
r T (c < r ≤ b). (P5.55)
Hence, in place of Eq.(5.92), the magnetic flux through the core amounts to
∫ (^) b
r=a
B(r) (^) ︸︷︷︸h dr dS
= h
∫ (^) c
a
B 1 (r) dr + h
∫ (^) b
c
B 2 (r) dr = 77. 1 μWb. (P5.56)
(b) The assumption of a uniform field distribution in the core results in the following approximate value of the magnetic flux through the core [see Eq.(5.93)]:
Φapprox = B 1 (rmean) (b − a)h ︸ ︷︷ ︸ S
= 80. 3 μWb
[
rmean = a + b 2
= 3 cm
]
, (P5.57)
and the relative error is now δΦ = 4.2%.
Section 5.10 Kirchhoff’s Laws for Magnetic Circuits
PROBLEM 5.21 Simple nonlinear magnetic circuit. The equation of the load line for the magnetic circuit in Fig.5.41, Eq.(5.101), for the given numerical data becomes H + 497. 4 B = 2000 (H in A/m ; B in T). (P5.58)
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152 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
and the solution of the system with Eqs.(P5.64), (P5.61), and (P5.62) is H 1 = 375 A/m, H 2 = 875 A/m, and H 3 = 1250 A/m, which is impossible, given the assumption that none of the field intensities is larger than 1000 A/m in Eq.(P5.63). Since N 2 I 2 = 3N 1 I 1 , it is logical to expect that the middle branch in Fig.P5. would first reach saturation, so B 2 = 1.5 T. Assuming that the other two branches are in the linear regime (B 1 = μaH 1 and B 3 = μaH 3 ), Eq.(P5.60) becomes
H 1 + H 3 = 1000 A/m , (P5.65)
and the solution of the new system of three equations is H 1 = 250 A/m, H 2 = 1500 A/m, and H 3 = 750 A/m, where these values are consistent with the new assumption made about the magnetization stages in the branches. The remaining two flux densities in the circuit are B 1 = 0.375 T and B 3 = 1.125 T.
PROBLEM 5.23 Magnetic circuit with a zero flux in one branch. Refer- ring to Fig.P5.4, the equations for the magnetic circuit are (note that S 2 = 2S 1 = 2 S 3 ):
B 1 S 1 − B 2 S 2 + B 3 S 3 = 0 (B 1 = 0) −→ 2 B 2 = B 3 , (P5.66)
−H 1 l 1 + H 3 l 3 = N 1 I 1 (H 1 = 0) −→ H 3 l 3 = N 1 I 1 , (P5.67)
H 2 l 2 + H 3 l 3 = N 2 I 2 , (P5.68)
Bi =
μaHi for Hi ≤ 1000 A/m 1 T for Hi > 1000 A/m , i = 1, 2 , 3 (μa = 0.001 H/m). (P5.69)
N 1 N 2
I 1
I 2
B 1 =
B 3
B 2
N 1
C 1
C 2
Figure P5.4 Finding the current I 1 of the winding in the first branch of the mag- netic circuit in Fig.5.42(a) such that the magnetic flux in that branch is zero.
Assuming, first, that both branches with nonzero fields in Fig.P5.4 are in the linear regime (B 2 = μaH 2 and B 3 = μaH 3 ), Eq.(P5.66) gives 2H 2 = H 3 , which, substituted in Eq.(P5.68), results in H 2 = N 2 I 2 /(l 2 + 2l 3 ) = 714.3 A/m and H 3 = 2 H 2 = 1428.6 A/m > 1000 A/m, which is impossible. Assuming, then, that the third branch is in saturation, B 3 = 1 T, we have, from Eq.(P5.66), that B 2 = 0.5 T, meaning that the second branch is in the linear regime, so, from Eq.(P5.69), H 2 = B 2 /μa = 500 A/m. Eq.(P5.68) then yields
H 3 =
N 2 I 2 − H 2 l 2 l 3 = 1500 A/m , (P5.70)
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P5. Solutions to Problems: Magnetostatic Field in Material Media 153
which is consistent with the new assumption that this branch is in saturation. Fi- nally, the current of the winding in the first branch in Fig.P5.4 is obtained from Eq.(P5.67), I 1 =
H 3 l 3 N 1
= 0.3 A. (P5.71)
PROBLEM 5.24 Nonlinear magnetic circuit with two air gaps. Equations for the magnetic circuit, Fig.P5.5, are [see also Eqs.(5.98) and (5.100) for the air gap]: −B 1 S + B 2 S + B 3 S = 0 (B 3 = 0) −→ B 1 = B 2 , (P5.72)
H 1 (l′ 1 + l 1 ′′ ) + H 0 l 0 + H 2 l 2 = N 1 I 1 , where H 0 =
B 0
μ 0
B 1
μ 0
, (P5.73)
−H 2 l 2 = N 2 I 2 (H 3 = 0). (P5.74)
Assuming that B 1 = B 2 = 1 T (saturation) and H 1 = H 2 , Eq.(P5.73) gives
H 1 =
N 1 I 1 − B 1 l 0 /μ 0 l′ 1 + l′′ 1 + l 2
= 2145.8 A/m , (P5.75)
so that the mmf of the second coil in Fig.P5.5 turns, from Eq.(P5.74), out to be
N 2 I 2 = −H 2 l 2 = −H 1 l 2 = − 107 .3 A turns. (P5.76)
B 3 =
N 1 N 2
l 2
I 1
l 0 l 0
l 1 ’^ l 3 ’^ I 2
l” 1 l” 3
H 1 H 2
B 3
C^ B^2
1 C 2
B 1
H 0
H 3 =
Figure P5.5 Analysis of the nonlinear magnetic circuit in Fig.5.43.
PROBLEM 5.25 Reverse problem with a nonlinear magnetic circuit. With reference to the notation in Fig.P5.6, the flux densities in the ferromagnetic section (B 1 ) and in the air gap (B 10 ) of the first branch of the circuit are
B 1 = B 10 =
S
= 0.5 T. (P5.77)
From the magnetization curve, i.e., by solving the magnetization equation given analytically, for the corresponding field intensity in the material,
H 1 = 250 tan B 1 = 137 A/m , (P5.78)
whereas in the gap, H 10 =
B 10
μ 0
= 398 kA/m. (P5.79)
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P5. Solutions to Problems: Magnetostatic Field in Material Media 155
H
B
B m = 1.114 T
H m = 1114 A/m
switch open switch closed
P 2
P 1
Figure P5.7 Operating points for the magnetic circuit in Fig.5.30(a) in two sta- tionary states (switch K closed and open, respectively) for a modified set of numer- ical data.
also shown in Fig.P5.7. The magnetic field intensity in the gap is H 0 = B/μ 0 = Bm/μ 0 = 886.5 kA/m.
PROBLEM 5.27 Linear magnetic circuit with three branches. (a)-(b) Fig.P5.8 shows the equivalent electric circuit for the problem, with reluctances, Eq.(5.97),
R 1 = R 3 =
l 1 ′ + l′′ 1 μrμ 0 S
l′ 3 + l′′ 3 μrμ 0 S
= 7. 64 × 105 H−^1 , R 2 =
l 2 μrμ 0 S
= 1. 59 × 105 H−^1 ,
R 10 = R 30 =
l 0 μ 0 S
= 1. 91 × 106 H−^1. (P5.86)
The figure also shows the adopted loops for the loop analysis of the (electric) cir- cuit (of course, there are multiple other ways to solve this electric circuit). The corresponding loop equations are
(R 10 + R 1 + R 2 ) Φ 1 + (R 10 + R 1 ) Φ 2 = N 1 I 1 , (P5.87)
(R 10 + R 1 ) Φ 1 + (R 10 + R 1 + R 3 + R 30 ) Φ 2 = N 1 I 1 + N 2 I 2 , (P5.88)
F 1 F 2
+ N I^1 1 +
N I 2 2
R 1
R 2
R 3
R 10 R 30
Figure P5.8 Equivalent electric circuit for the loop analysis of a linear magnetic circuit with three branches, two air gaps, and two mmf’s (in Fig.5.43).
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
156 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
and their solution is Φ 1 = 200. 5 μWb and Φ 2 = 199 μWb. The magnetic field intensities in the air gaps in Fig.5.43 come out to be
H 10 =
B 10
μ 0
μ 0 S
= 1.27 MA/m and H 30 =
B 30
μ 0
μ 0 S
= 633 kA/m. (P5.89)
Section 5.11 Maxwell’s Equations for the
Time-Invariant Electromagnetic Field
PROBLEM 5.28 Continuity equation from Amp`ere’s law in integral form. Consider an arbitrary closed surface S in the time-invariant electromagnetic field and a contour C that splits S into two parts, the upper part S 1 and the lower part S 2 (similar to Fig.4.27 for S 1 and S 3 ). Let the surfaces S 1 and S 2 , which are both enclosed by C, be oriented in the same way, that in accordance with the right- hand rule with respect to the orientation of the contour. Applying Maxwell’s second equation in integral form, Eqs.(5.70), to the contour C and either S 1 or S 2 must give the same result, i.e., the fluxes through S 1 and S 2 of the current density vector, J, are the same. This means in turn that [note the similarity with Eq.(4.101)] ∮
S
J · dS = 0 , (P5.90)
which is the time-invariant integral form of the continuity equation, Eqs.(3.59). Note that this derivation parallels the one given in Example 5.19 in differential notation.
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