chapter magnetic field in materialsecee.colorado.edu/~mcleod/teaching/eandm3400/lab...

CHAPTER 13: MAGNETIC FIELD IN MATERIALS 113

13. Magnetic Field in Materials

. In magnetic terms, atoms and molecules inside matter resemble tiny current loops. If apiece of matter is situated in a magnetic field, the moment of magnetic forces partly alignsthese loops, and we say that the substance is magnetized. The magnetic field produced bythe substance is due to these aligned current loops, known as Ampere's currents. A substan~ein the magnetic field can therefore be visualized as a large set of oriented elementary currentloops situated in a vacuum. These oriented loops can be replaced by equivalent macroscopiccurrents situated in a vacuum, known as the magnetization currents.

. An elementary current loop is first characterized by a magnetic moment, rn = I S, whereI is the loop current and S its vector area. Next the magnetization vector, M, is defined, as

(L:rn)in dv = NrnM = - dv (Aim), (13.1)

where N is the number of Ampere's currents per unit volume.

. The Ampere currents can be considered to be situated in a vacuum. Consequently, theycan be incorporated in Ampere's law (which is valid for currents in a vacuum):

i B. dl = lLo (lJ . dS + i M. dl) ,(13.2)

or

i H .dl =l J .dS,(13.3)

where

H = B IlLo- M (Aim) (13.4)

is known as the magnetic field intensity vector. For linear magnetic materials

M=XmH (Xm - dimensionless, M - Aim), (13.5)

where Xm is the magnetic susceptibility. Thus, for linear materials,

B = lLo(1 + Xm)H = lLolLrH =ILH (ILr - dimensionless, IL - Him). (13.6)

The constant ILr is the relative permeability, and IL the permeability of the material. If Eq.(13.6) holds, the material is linear, else it is nonlinear. If IL is the same at all points, the

114 PART 2: TIME-INVARIANT MAGNETIC FIELD

material is said to be homogeneous, otherwise it is inhomogeneous. Linear magnetic materialscan be diamagnetic (Xm < 0, i.e., J.lr < 1), or paramagnetic (Xm > 0, i.e., J.lr > 1).

. Ampere's law in Eq. (13.3) can be transformed into a differential equation, i.e., its differ-ential form,

curlH = V'x H = J, (13.7)

valid for time-invariant currents.

. The volume density of magnetization current is given by

Jm =V' x M (A/m2). (13.8)

It is zero in homogeneous materials with no macroscopic currents. On a boundary betweentwo magnetized materials, the surface magnetization current density is given by

Jms =n X (MI - M2) (A/m), (13.9)

where n is the unit vector normal to the boundary, directed into medium 1.

. The most important boundary conditions at the interface of two magnetic materials are

H Hang = H 2tang, (13.10)

(no surface currents on the boundary), and

BInormal = B2normal. (13.11)

In linear media these equations become

BItang B2tang-= ,J.lI J.l2

J.lI H Inormal = J.l2 H 2norma.l' (13.12)

. Diamagnetic and paramagnetic materials are linear, and have J.lr ~ 1 (less than one fordiamagnetic materials, greater than one for paramagnetic materials). The most importantmagnetic materials in electrical engineering are known as ferromagnetic materials, nonlinearmaterials with very large value of relative permeability. In ferromagnetic materials, groupsof atoms (Weiss' domains) are formed as small saturated magnets. Magnetization of ferro-magnetic materials is obtained by aligning these domains, which is accompanied by hysteresislosses. Above the Curie temperature, a ferromagnetic material becomes paramagnetic. Fer-rites are materials that have neighboring Weiss domains of different sizes, oriented in oppositedirections. Therefore they have less pronounced magnetic properties than ferromagnetic ma-terials, but many of them have high electrical resistivity, which makes them important forhigh-frequency applications. (High resistivity makes so-called skin effect less pronounced andred uces eddy-current losses at high frequencies.)


. Ferromagnetic materials are characterized by magnetization curves, obtained by measure-ments. These are curves of B versus H. If the material was previously unmagnetized, theinitial magnetization curve is obtained. If H is periodic, a hysteresis loop is obtained instead.The curve connecting the tips of hysteresis loops corresponding to different amplitudes of His the normal magnetization curve.

. Ferromagnetic materials are characterized by several "permeabilities". The ratio B IHalong an initial magnetization curve at H = 0 is the initial permeability, and that along anormal magnetization curve is the normal permeability. Also used are complex permeability,differential permeability, etc. .

. Magnetic circuits are structures designed to channel the magnetic flux, and are used in awide variety of devices (cores of transformers, motors, generators, relays, etc.) An approximatesolution of magnetic circuits is based on assuming them to be thin (with respect to the lengthof the branches), and linear. Such magnetic circuits are analyzed using Kirchhoff's laws for.magnetic circuits, which are analogous to these laws for electric circuits. In these equations,the electromotive force is replaced by the magnetomotive force (the product N I of the number

.:of turns in a coil and its current), the current by the magnetic flux, and the resistance bythe "magnetic resistance", or reluctance. For a branch of length l, cross-section area Sandpermeability J.L,the reluctance is given by

1

Rm= J.LS(l/H). (13.13)

Real magnetic circuits are both nonlinear and are not thin. Their analysis is quite com-plicated, especially if the circuit is not quite simple.

QUESTIONS

Q13.1. Are any conventions implicit in the definition of the magnetic moment of a currentloop? - (a) There are no conventions at all. (b) The direction around the loop is adopted inthe direction of the current. (c) In addition to (b), the unit vector normal to the surface andthe direction around the loop are connected through the right-hand rule.

Q13.2. A magnetized body is introduced into a uniform magnetic field. Is there a force onthe body? Is there a moment of magnetic forces on the body? Explain. - (a) There is aforce in the direction of vector B, there is a moment of magnetic forces. (b) There is no force,there is a moment. (c) There is a force in the direction opposite to that of vector B, there isa moment.

S Q13.3. A small body made of soft iron is placed on a table. Also on the table is a permanentmagnet. If the body is pushed toward the magnet, ultimately the magnet will pull the bodytoward itself, so that the body will acquire certain kinetic energy before it hits the magnet.Where did this energy come from? - (a) From our pushing the body. (b) From the magnetitself. (c) The small body had a potential energy in the field of the permanent magnet.

Answer. A small body had a potential energy in the field of the permanent magnet. This potentialenergy was used to accelerate the body.

Q13.4. Prove that the units for Band J.LoM are the same. - Hint: use the Biot-Savart lawto express the tesla in terms of the unit for J.LO.


Q13.5. The source of the magnetic field is a permanent magnet of magnetization M. Whatis the line integral of the vector H around a contour which passes through the magnet? -(a) Depends on the shape of the magnet and its magnetization. (b) Depends on the directionof the path with respect to the magnetization vector. (c) Zero (explain).

S Q13.6. The magnetic core of a thin toroidal coil is magnetized to saturation, and then thecurrent in the coil is switched off. The remanent (i.e., remaining) flux density in the core isBr. Determine the magnetization vector and the magnetic field strength vector in the core.- (a) M = Br. (b) M =J.1.oBr.(c) M = Br/J.1.o.

Answer. Since B = /JoH + M, and H = 0 in the core (remanent flux density exists for H = 0),M=Br.

Q13. 7. Is there a magnetic field in the air around the core in question Q13.6? Explain. -(a) Certainly, because the core is magnetized. (b) No, since there is no leakage flux. (c) Yes,

..but only if the permeability of the core is not high.

.Q13.8. Why is the equation

l V' x H.dS= l J.dS

valid for any contour C and any surface bounded by C? - (a) Because we adopt that, at allpoints, curl H = J. (b) It is valid for any C and any surface only under certain conditions(which ones'?). (c) Because it was derived using the Stokes's theorem.

Q13.9. Why is the reference direction of the vector Jms in Fig. Q13.9 into the paper? -Hint: recall the conventions implicit in Eq. (13.9).

M

z

Llh-O

Fig. Q13.9. Boundary of magnetized material. Fig. Q13.11. Surface magnetization currents.

Q13.10. Suppose that all atomic currents contained in the page you are reading can beoriented so that m is toward you. What is then their macroscopic resultant, and what is(qualitatively) the magnetic field of such a "magnetic sheet"? - (a) As that of two lay-ers of current on the two sheet sides. (b) As that of a circulating current around the sheetcircumference. (c) The sheet is thin, and there is no magnetic field.

Q13.11. What are the macroscopic resultants of the microscopic currents of a short, circular,cylindrical piece of magnetized matter with uniform magnetization M at all points, if: (1) M

I


is parallel to the cylinder axis, or (2) M is perpendicular to the axis? - (a) In case 1, surfacecurrents on the two cylinder bases. In case 2 the same, but of different distribution. (b) In bothcases, a volume distribution of magnetization currents. (c) In case 1, a current sheet aroundthe curved surface, equivalent to a short solenoid. In case 2, surface magnetization currentsare like those in Fig. Q13.11.

Q13.12. Sketch roughly the lines of vectors M, B, and H in the two cases in question Q13.11.- Hint: discuss if the sketches in Pig. Q13.12 are correct.

.M B/ I-Lo

(a).M B/I-'-O

(b)

H

H

Fig. Q13.12. Sketch of lines of vectors M, B, and H for the two cases of magnetized short cylinderdescribed in question Q13.11.

Q13.13. A ferromagnetic cube is magnetized uniformly over its volume. The magnetizationvector is perpendicular to two sides of the cube. What is this cube equivalent to in terms ofthe magnetic field it produces? - (a) To a short solenoid of square cross section. (b) To twocurrent sheets (over the two mentioned cube sides) . (c) To a volume distribution of currents.

Q13.14. Is the north magnetic pole of the earth close to its geographical North Pole? Explain.(See Chapter 17.) - Hint: recall that the magnetic needle turns its north pole towards theNorth.

S Q13.15. If a high-velocity charged elementary particle pierces a toroidal core in which thereis only remanent flux density, and H = 0, will the particle be deflected by the magnetic field?


Explain. - (a) No, because in the core H ==O. (b) No, because in the core B = O. (c) Yes,because in the core B is not zero, although His.

Answer. It will be deflected, because the force on the particle is F= Qv x B, and B #- O.

Q13.16. Sketch the initial magnetization curve corresponding to a change of H from zero to-Hm. - Hint: assume that H starts from zero and becomes negative, and think about whatthe variation of B must be like.

Q13.17. Suppose that the magnetization of a thin toroidal core corresponds to the point Br(remanent flux density). The coil around the core is removed, and the magnetic flux densityuniformly decreased to zero by some appropriate mechanical or thermal treatment. How doesthe point in the B-H plane go to zero? - (a) Backwards along the magnetization curve. (b)Continues to a point on the -H axis, and then along that axis to the origin. (c) Along the Baxis to the origin.

SQ13.18. The initial magnetization curve of a certain ferromagnetic material is determined fora thin toroidal core. Explain the process of determining the magnetic flux in the core of the~same material, but of the form shown in Fig. P13.5a. - Hint: note that the magnetic fieldintensity differs from one point of such a core to another, as H(r) = Nlj(27rr).

Answer. The magnetic field intensity differs from one point of the core to another, as H(r) =Nlj(27rr). We need to take the corresponding value of B from the magnetization curve, to multiply itby h dr (the area of a small rectangular surface of the core cross section), and to integrate from a tob.

Q13.19. Make a rough sketch of the curve in the B-H plane obtained if H is increased toHm, then decreased to zero, then again increased to Hm and decreased to zero, and so on. -(a) A siraight line segment. (b) A flat hysteresis loop. (c) Even a rough sketch is not possiblewithout measurements.

Q13.20. Is it possible to obtain a higher remanent flux density than that obtained whensaturation is attained and then H reduced to zero? Explain. - (a) Yes (how'?). (b) No(why'?). (c) For some materials (describe).

S Q13.21. What do you expect would happen if a thin slic~ is cut out of a ferromagnetic toroidwith remanent flux density in it? Explain. - (a) Nothing. (b) The operating point will movedownwards along the B axis. (c) The operating point will move to the left and downward.

Answer. The simplest way to visualize the situation is to imagine that the slice was not cut out, butthat, instead, a surface current of equal magnitude to the magnetization surface current, but oppositeto it, is "glued" over the slice. This amounts to the same, but is easier to understand. The workingpoint in the H - B diagram would move from the point (0, Br) to the left and downward, because wehave a "coil" with negative current.

..

Q13.22. A rod of ferromagnetic material can be magnetized in various ways. If a magnetizedrod attracts most ferromagnetic powder (e.g., iron filings) near its ends, and very little in itsmiddle region, how is it magnetized? - (a) Not enough data for a conclusion. (b) Normal toits axis. (c) Along its axis.

Q13.23. If a small diamagnetic body is close to a strong permanent magnet, does the magnetattract or repel it? Explain. - (a) It repels it. (b) It attracts it. (c) It does not exert a forceon it.

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Q13.24. Answer question Q13.23 for a small paramagnetic body. - The answers are thesame as for the preceding question.

Q13.25. While the core in question Q13.17 is still magnetized, if just one part of the core isheated above the Curie temperature, will there be a magnetic field in the air? If you thinkthere will be, what happens when the heated part has cooled down? - (a) No. (b) Yes, butwill practically disappear when it cools down (why?). (c) It depends how large the heated part~s.

S Q13.26. Assuming that you use a large number of small current loops, explain how you can'make a model of (1) a paramagnetic material, and (2) a ferromagnetic material? - (a) It isnot possible to make models in this way. (b) Just orient the loops in both cases. (c) In case 1,suspend the loops on springs individually, and in random directions. In case 2, make groups ofmany oriented current loops (with the magnetic moment, m, of loops in a group in the samedirections). Suspend these groups of loops on springs, and allow these groups of loops to turn

'. with friction.

Answer. (1) Suspend the loops individually, and in random directions, on springs that allow them to. turn only to some extent in response to the applied magnetic field. (2) Suspend many large groups ofstrictly oriented current loops in random directions, on springs that allow them to turn with frictionin response to the applied magnetic field.

Q13.27. If the current in the coil wound around a ferromagnetic core is sinusoidal, is themagnetic flux in the core also sinusoidal? . Explain. - (a) Depends on the amplitude ofcurrent. (b) Yes. (c) No.

Q13.28. Analyze similarities and differences between Kirchhoff's laws for dc electric circuitsand magnetic circuits. - Hint: analyze possible approximations in the two sets of laws.

Q13.29. How do you determine the direction ofthe magnetomotive force in a magnetic circuit?- (a) The magnetomotive force does not have a direction. (b) According to the right-hand rulewith respect to the current in the winding. (c) According to the left-hand rule.

S Q13.30. A thin magnetic circuit is made of a ferromagnetic material with an initial magneti-zation curve that can be approximated by the expression B(H) = BoH/(Ho + H), where Boand Ho are constants. If the magnetic field strength, H, in the circuit is much smaller than Ho,can the circuit be considered as linear? What in that case is the permeability of the material?What is the physical meaning of the constant Bo? - (a) No, it cannot be considered as linear.(b) Depends on the values of the parameters. (c) Yes, and the permeability is IL= Bo/ Ho.

Answer. It can, for then B(H) =BoH/(Ho + H) ==(Bo/Ho)H. The permeability of the material isthus 11= Bo/ Ho. If H ~ Ho, then B(H) ==Bo, so that Bo is the magnetic flux density correspondingto saturation.

Q13.31. Why can't we have a magnetic circuit with no leakage flux (stray field in the airsurrounding the magnetic circuit)? - (a) Because IL cannot be large enough. (b) Because ILOis too large. (c) A combination of the two preceding answers.

Q13.32. Is it possible to construct a magnetic circuit closely analogous to a dc electric circuit,if the latter is situated (1) in a vacuum, or (2) in an imperfect dielectric? Explain. - (a) No,in both cases. (b) Yes in case 1, no in case 2. (c) No in case 1, yes in case 2.


Q13.33. One half of the length of a thin toroidal coil is filled with a ferromagnetic material, andthe other half with some paramagnetic material. Can the problem be analyzed as a magneticcircuit? Explain. - (a) No. (b) Yes. (c) Depends on the permeability of the paramagneticmaterial.

PROBLEMS

S P13.1. The magnetic moment of the earth is about 8 . 1022Am2. Imagine that there is agiant loop around the earth's equator. How large does the current in the loop have to be toresult in the same magnetic moment? Would it be theoretically possible to cancel the magneticfield of the earth with such a current loop (1) on its surface, or (2) at far points? The radiusof the earth is approximately 6370km. - (a) I = 6.28.106 A. (b) I = 6.28.107 A. (c)I = 6.28 . 108 A.

Solution. The magnitude of the magnetic moment of the contour is given by

m = brR2,

where R is the radius of the earth, so that the loop current has to be as large as I = 6.28 . 108 A! Withsuch a loop it could not be possible, not even theoretically, to cancel the magnetic field of the earthon its surface, because this field is due to currents distributed inside the earth volume, and not on theearth equator. At far points, however, the magnetic field of an arbitrary current distribution dependson the magnetic moment of the equivalent contour only, and the cancellation of the magnetic field ofthe earth is theoretically possible.

P13.2. The number of iron atoms in one cubic centimeter is approximately 8.4. 1022, andthe product of J.Loand the maximum possible magnetization (corresponding to "saturation")is J.LoMsat= 2.15T. Calculate the magnetic moment of an iron atom. - (a) m = 1.04.10-21 Am2. (b) m =4.04 .10-20 Am2. (c) m =2.04.10-23 Am2.

P13.3. A thin toroid is uniformly magnetized along its length with a magnetization vectorof magnitude M. No free currents are present. Noting that the lines of M, B, and H insidethe toroid are circles by symmetry, determine the magnitude of B and prove that H =O. -Hint: use Ampere's law to prove that H = O.

S P13.4. A straight, long copper conductor of radius a is covered with a layer of iron of thicknessd. A current of intensity I exists in this composite wire. Assuming that the iron permeabilityis J.L,determine the magnetic field, the magnetic flux density, and the magnetization in copperand iron parts of the wire. Note that the current density in the copper and iron parts of thewire is not the same. - Hint: assume that O'Cuis the conductivity of copper, and O'Fethat ofiron. Note that the electric field intensities in copper and iron are the same (due to continuityof Etang on the boundary).

Solution. Let O"Cube the conductivity of copper, and O"Fethat of iron. The electric field intensitiesin copper and iron are the same (due to continuity of Etang on the boundary). Hence we have that

JCu = O"Cu.JFe O"Fe

The total current in the wire is given by


1= JCu1ra2 + JFe1r(d2 + 2ad),

so that JCu = II[1ra2 + 1r(d2+ 2ad)O'FeIO'CuJand JFe = JCuO'FeIO'Cu'

Due to symmetry, vector H has a circular component only, and its intensity depends only to thedistance r from the wire axis. By applying the generalized Ampere's law to the contour in the form ofa circle of radius r, we obtain '.

H= JCur2 (in copper) , H = JCua2 + JFe(r2 - a2)

2r (in iron) .

In copper, B =J.1.0H and M =O. In iron, B =J.1.Hand M = (J.1.1J.1.0- l)H.

b B

h

H

(a) (b)

Fig. P13.5. (a) A ferromagnetic core, and (b) its idealized initial magnetization curve.

P13.5. The ferromagnetic toroidal core sketched in Fig. P13.5a has an idealized initialmagnetization curve as shown in Fig. P13.5b. Determine the magnetic field strength, themagnetic flux density, and the magnetization at all points of the core, if the core is wounduniformly with N = 628 turns of wire with current of intensity (1) 0.5A, (2) 0.75A, and (3)1 A. The core dimensions are a = 5 em, b = 10 em, and h = 5 em, and the constants of themagnetization curve are Ho = 1000 Aim, and Bo = 2 T. For the three cases determine themagnetic flux through the core cross section. Assume that the core was not magnetized priorto turning on the current in the winding. - Hint: use Ampere's law to find H, and notethat in the three cases different parts of the magnetization curve are valid. The magnetic fluxthrough the core cross section for the three cases is: (a) <1>1= 0.466 m Wb, <1>2= 3.658 m Wb,<1>3= 4 m Wb. (b) <1>1= 3.466 m Wb, <1>2= 4.658 m Wb, <1>3= 5 m Wb. (c) <1>1= 6.466 m Wb,<1>2= 2.658 m Wb, <1>3= 2.1 m Wb.

P13.6. A straight conductor of circular cross section of radius a and permeability /l carries acurrent I. A coaxial conducting tube of inner radius b (b > a) and outer radius c, with no cur-rent, also has a permeability /l. Determine the magnetic field intensity, magnetic flux density,and magnetization at all points. Determine the volume and surface densities of macroscopiccurrents equivalent to Ampere currents. - Hint:-use Ampere's law to find H, and henceBand M, and Eqs. {13.B} and {13.9} to find the magnetization currents.

P13.7. Repeat problem P13.6 if the conductor and the tube are of permeability /l(H) =/loHI Ho, where Ho is a constant. - Hint: note that in this case B = /loH2IHo and M =


(HjHo - l)H in the conductor and in the tube, and that H is the same as in the precedingproblem.

P13.8. Repeat problems P13.6 and P13.7 assuming that the tube carries a current -1, sothat the conductor and the tube make a coaxial cable. - Hint: the only difference is that Hin the tube is determined also by the current in the tube.

P13.9. A ferromagnetic sphere of radius a is magnetized uniformly with a magnetization vec-tor M. Determine the density of magnetization surface currents equivalent to the magnetizedsphere. - Hint: use Eq. (13.9). The surface currents are circulating currents, as in Fig.P13.9.

B z

z=r

Fig. P13.9. A magnetized sphere. Fig. P13.10. A magnetized thin disk.

S P13.10. A thin circular ferromagnetic disk of radius a = 2 cm and thickness d = 2 mm isuniformly magnetized normal to its bases. The vector J.LoMis of magnitude 0.1T. Determinethe magnetic flux density vector on the disk axis normal to its bases, at a distance r = 2 cmfrom the center of the disk. - Hint: there are no volume magnetization currents, and surfacemagnetization currents are obtained from Eq. (13.9), Fig. P 13.10.

Solution. Since M is constant (the disk is uniformly magnetized), Jrn =V' X M =0, i.e., there are novolume magnetization currents inside the disk. The distribution of the surface magnetization currentscan be determined on the basis of Eq.(13.16). On the two cylinder bases, Jrns =0, since vectors n andM are parallel. On the cylinder curved surface, however, there are circulating surface magnetizationcurrents, of density Jrns =M and with direction indicated in Fig. P13.10. The magnetized disk beingthin (d ~ a), it can be replaced by a circular loop of radius a, situated in a vacuum. The loop currentIS

I = Jrnsd = Md.

The magnetic flux density at the point specified in the text of the problem is hence (see problem P12.7)

B = J-loMda22(a2 + r2)3/2 = 1.77mT.

The vector B is shown in Fig. P13.10.

I


P13.11. A thin ferromagnetic toroidal core was magnetized to saturation, and then the currentin the winding wound about the core was turned off. The remanent flux density of the corematerial is Br = 1.4 T. Determine the surface current density on the core equivalent to theAmpere currents. If the mean radius of the core is R =5 cm, and the winding has N = 500turns of wire, find the current in the winding corresponding to this equivalent surface current.If the cross section area of the core is S = 1cm2, find the magnetic flux in the core. - Themagnetic flux in the core is (a) ~ =85J.lWb. (b) ~ = 140J.lWb. (c) ~ = 23J.lWb.

P13.12. A round ferrite rod of radius a = 0~5cm and length b = 10 cm is magnetized uniformlyover its volume. The vector POM is in the direction of the rod axis, of magnitude 0.07 T.Determine the magnetic flux density at the center of one of the rod bases. Is it importantwhether the point is inside the rod, outside the rod, or on the very surface of the rod? - (a)B = 15mT. (b) B =25mT. (c) B =35mT.

P13.13. Shown in Fig. P13.13 is a stripline with ferrite dielectric. Since a ~ d, the magnetic"field outside the strips can be neglected. Under this assumption, find the magnetic fieldintensity between the strips if the current in the strips is I. If the space between the stripsis filled with a ferrite of relative permeability J.lr that can be considered constant, determinethe magnetic flux density and magnetization in the ferrite, and the density of surface currentsequivalent to the Ampere currents in the ferrite. - Hint: use the Ampere law to find H,and hence B, M, and Jms. (a) Jms = (J.lr- 1)lj(2a). (b) Jms = (Pr - l)lja. (c) Jms =2(J.lr - 1)1 j a.

a

J.L»J.Lo

Fig. P13.13. A stripline with ferrite dielectric.

S P13.14. The ferromagnetic cube shown in Fig. P13.14a is magnetized in the direction of thez axis so that the magnitude of the magnetization vector is Mz(x) = Moxja. Find the densityof volume currents equivalent to the Ampere currents inside the cube, as well as the surfacedensity of these currents over all cube sides. Follow the surface currents and note that in partthey close through the magnetized material. - Hint: the volume and surface currents in across section of the cube perpendicular to the z axis are as in Fig. P13.14b.

Solution. Having in mind the expression for the curl in rectangular coordinates, the density of volumemagnetization currents inside the cube is given by

8Mz MoJm =V x M= --Uy = --Uy.8z a

According to Eq. (13.16), there are no resultant surface currents on the upper, lower and back sideof the cube, while on the left, right and front cube side, the surface magnetization current density isJmsl =Mo(z/a) Ux, Jms2 = -Mo(z/a) Ux, and Jms3 =Mo iu, respectively.


By analysing the above expressions we can conclude that, indeed, the surface magnetizationcurrents close in part through the cube, by forming the current contour integral with the volumemagnetization currents. This is shown graphically in a cross section of the cube perpendicular to thez axis in Fig. P13.14b.

z y

a Jmsl Jms2

y.a Jms3

x

a.x

(a) (b)

Fig. P13.14. (a) A magnetized ferromagnetic cube, and (b) distribution of volume and surfacemagnetization currents inside and over the cube.

P13.15. Prove that on the boundary surface of two magnetized materials the surface magneti-zation current is given by the expression Jrns =n X (Ml - M2)' Ml and M2 are magnetizationvectors in the two materials at close points on the two sides of the boundary, and n is theunit vector normal to the boundary, directed into medium 1. - Hint: interpret properly theAmpere law in Eq. (13.2), and find the current through the rectangular contour sketched inFig. Q13.9.

P13.16. At a point of a boundary surface between air and a ferromagnetic material of per-meability JL» JLothe lines of vector B are not normal to the boundary surface. Prove thatthe magnitude of the magnetic flux density vector in the ferromagnetic material is then muchgreater than that in air. - Hint: use boundary conditions in Eqs. (13.12).

P13.17. A current loop is in air above a ferromagnetic half-space. Prove that the field in theair due to the Ampere currents in the half-space is very nearly the same as that due to a loopbelow the boundary surface symmetrical to the original loop, carrying the current of the sameintensity and direction, with the magnetic material removed. (This is the image method forferromagnetic materials.) - Hint: prove that the magnetic field of the original loop and itsimage satisfy the boundary conditions on the interface.

P13.18. Inside a uniformly magnetized material, of relative permeability JLr,are two cavities.One is a needlelike cavity in the direction of the vector B. The other is a thin-disk cavity,normal to that vector. Determine the ratio of magnitudes of the magnetic flux density in thetwo cavities and that in the surrounding material. Using these results, estimate the greatestpossible theoretical possibility of "magnetic shielding" from time-invariant external magneticfield. (We shall see that the shielding effect is greatly increased for time-varying fields.) -Hint: use boundary conditions in Eqs. (13.12).

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S P13.19. Sketched in Fig. P13.19 is the normal magnetization curve of a ferromagneticmaterial. Using this diagram, estimate the relative normal and differential permeability, andplot their dependence on the magnetic field strength. What are the initial and maximalpermeabilities of the material? - Hint: check if in the table below the values of normal relativepermeability, fLrn = (B / H) / fLo, and differential relative permeability, fLrd = (dB / dH) / fLO,agree with your estimates from the diagram in Fig. P 13.19. Sketch the functions fLrn(H) andfLrd(H) on the basis of the tabulated values.

1.5

1m]

2.5

2.0

1.0

0.5

00 100 200 300 400 500 600

Fig. P13.19. A normal magnetization curve.

Solution. In the table below are given a few values of the normal relative permeability, JJrn =(B I H) 11100,and the differential relative permeability, JJrd = (dBIdH) I 1100,which are estimated from

the diagram in Fig. P13.19. It is left to the reader to sketch the functions p.rn(H) and JJrd(H) on thebasis of the tabulated values.

H (Aim) 0

2600

100

3800

200

6500

300

5100

400

4100

500

3300p.rn

p.rd 2600 5600 2900 1200 300 60

The initial relative permeability is JJri = 2600, while the maximal normal and differential relativepermeabilities are (JJrn)ma.x =6900 for H =180 Aim and (JJrd)ma.x =11000 for H =150Aim.

P13.20. Approximate the normal magnetization curve in Fig. P13.19 in the range 0 ~ H ~300 A/m by a straight line segment, and estimate the largest deviation of the normal relativepermeability in this range from the relative permeability of such a hypothetical linear material.- (a) about 800 and -500. (b) about 1800 and -2500. (c) about 8000 and -5000.

P13.21. Figs. P13.21a and P13.21b show two hysteresis loops corresponding to sinusoidalvariation of the magnetic field strength in two ferromagnetic cores between -Hm and +Hm.Plot the time dependence of the magnetic flux density in the core. Does the magnetic fluxalso have a sinusoidal time dependence? - Hint: check if Fig. P13.21c shows the dependenceon time you expect of the magnetic field intensity, H(t), and the corresponding magnetic fluxdensities, Ba.(t) and Bb (t).

P13.22. If in Figs. P13.21a and b the magnetic flux density varies sinusoidally in time, sketchthe time dependence of the magnetic field strength in the core. Is it also sinusoidal? If the

B [T]

/

/)

/ H[A


hysteresis loops were obtained by measurements with sinusoidal magnetic field strength, is itabsolutely correct to use such loops in this case? - Hint: check if the variations of H in time,obtained from the hysteresis loops in Figs. P13.21a and P13.21b and sketched in Fig. P13.22,are what you expect.

B B

-Hm -Hm

IIIII

: H

HmHm

(a) (b)

Fig. P13.21. Hysteresis loops for two ferromagnetic materials.

S P13.23. The initial magnetization curve (first part of hysteresis curve) B(H) of a ferromag-netic material used for a transformer was measured and it was found that it can be approxi-mated by a function of the form B(H) = BoH/(Ho+H), where the coefficients are Bo = 1.37Tand Ho = 64 A/m. Then a thin torus with mean radius R = 10 cm and a cross-section ofS = 1cm2 is made out of this ferromagnetic material, f'-pd N = 500 turns are densely woundaround it. Find B( H) and the flux through the magnetk circuit as a function of current in-tensity I through the winding. Find the flux for (1) I = 0.25A, (2) 1= 0.5A, (3) 1= 0.75A,and (4) I = 1A. - Hint: use Ampere's law to find H in the core, then B from the givenmagnetization curve. (a) <1>1= 204JLWb, <1>2= 218JLWb, <1>3= 224JLWb, <1>4= 227 JLWb. (b)<1>1= 94JLWb, <1>2= 98JLWb, <1>3= 94JLWb, <1>4= 97 JLWb. (c) <1>1= 104JLWb, <1>2= 118JLWb,<1>3= 124 JLWb, <1>4= 127 JLWb.

.

B,H --"'"/ \\ B (t),y b,

, \, \

B,H

~'/~B(t)

Ij / \ \

'-;/

Hb (t)

Fig. P13.21c. B(t) for sinusoidal H(t). Fig. P13.22. H(t) for sinusoidal B(t).

Solution. The magnetic field intensity in the core is given by H =NI/(27rR) (according to Ampere'slaw). The magnetic flux density can then be determined by the function approximating the initial

I


magnetization curve, B = B(H), and, finally, the magnetic flux through the magnetic circuit foundas 4>= BS. For the four specified values of I, we obtain: (1) 4>= 104J.1Wb, (2) 4>= 118J.1Wb, (3)4>= 124J.1Wb, and (4) 4>= 127 J.lWb.

P13.24. Assume that for the ferromagnetic material in problem P13.23 you did not have ameasured hysteresis curve, but you had one data point: for H = 1000 Aim, B was measuredto be B = 2T. From that, you can find an ~quivalent permeability and solve the cir~uitapproximately, assuming that it is linear. Repeat the calculations from the preceding problemand calculate the error due to this approximation for the four current values given in problem13.23. - Let 6 be the relative error. (a) q)1 = 40jLWb (6 = 61.5%), q)2= 80jLWb (6 =32.2%), q)3 = 119jLWb (6 = 4%), q)4 = 159JLWb (6 = 25.2%). (b) q)1 = 20jLWb (6 =31.5%), q)2 = 40JLWb (6 = 12.2%), q)3 = 60jLWb (6 = 2%), q)4 = 80jLWb (6 = 15.2%).(c) q)1 = 60jLWb (6 = 31.5%), q)2 = 120jLWb (6 = 52.2%), q)3 = 160jLWb (6 = 45%),q)4 = 180 jLWb (6 = 55.2 %).

"P13.25. The thick toroidal core sketched in Fig. P13.25 is made out of the ferromagneticmaterial from problem PI3.23. There are N = 200 turns wound around the core, and the

.=Coredimensions are a = 3cm, b = 6cm, and h = 3 cm. Find the magnetic flux through thecore for 1 = 0.2 A and 1 = 1 A in two different ways: (1) using the mean radius; and (2) bydividing the core into 5 layers and finding the mean magnetic field in each of the layers. - (a)q)1 = 0.85 (1.13) m Wb, the same for case 2. (b) q)1= 0.54 (0.78) m Wb, q)2 = 0.62 (0.88) m Wb.(c) q)1 = 0.27 (0.43)mWb, q)2 = 0.57 (0.76)mWb.

ra

Nt

12h 11-LO NZ,b

a SI

Fig. P13.25. A thick toroidal coil. Fig. P13.26. A magnetic circuit with air gap.

S P13.26. Find the number of turns N1 = N2 = N for the magnetic circuit shown in Fig.P13.26 so that the magnetic flux density in the air gap is Eo = 1 T when h = 12 =1 =5A.The core is made of the same ferromagnetic material as in problem PI3.23. Solve the problemin two ways: (1) taking the magnetic resistance ofthe core into account; and (2) neglecting themagnetic resistance of the core. Use the following values: a = 10 cm, b = 6 cm, d1 = d2 = 2 cm,Sl = S2 = S = 4 cm2, and 10= 1 mm. What is the percentage difference between the answersin (1) and (2)? - (a) N = 263, error in neglecting the reluctance of the core is about 5%. (b)N = 163, error in neglecting the reluctance of the core is about 2.5%. (c) N = 363,error inneglecting the reluctance of the core is about 7.5%.

Solution. (1) The magnetic flux density in the central part of the core is B2 = Bo = 1T. Bysymmetry, the magnetic flux densities in the left and right branches of the magnetic circuit are the


same. Using the first Kirchhoff's law for magnetic circuits, we obtain BI = 0.5 T. The magnetic fieldintensity in the core is given by

H = HoBBo - B'

which results in HI = 36.78A/m and Hz = 173A/m. The length of the left (or right) branch isII = 11.14 em, and that of the central branch lz =3.9 em. The second Kirchhoff's law for magneticcircuits finally yields

NI =HIll + Hzlz + Holo,

where Ho =Bo/ J-Io,so that N =163.

. (2) If we neglect the magnetic resistance of the core, i.e., if we neglect HIll +Hzlz with respectto Holo, the result is N =159. The error is about 2.5%.

P13.27. The magnetization curve of a ferromagnetic material used for a magnetic circuit canbe approximated by B(H) = 2HI(400 + H), where B is in T and H is in Aim. The magneticcircuit has a cross-sectional area of S = 2cmz, a mean length of L = 50em, and N = 200turns with I = 2 A flowing through them. The circuit has an air gap Lo = 1 mm long. Findthe magnetic flux density vector in the air gap. - (a) B = 0.433T. (b) B = 0.833T. (c)B = 0.633 T.

P13.28. The magnetic circuit shown in Fig. P13.28 is made out of the same ferromagneticmaterial as the one in the previous problem. The dimensions of the circuit are a =.6 em,b = 4em, d = 1em, SI = Sz = S = 1cmz, NI = 50, Nz = 80, and N3 = 40. Withlz = h = 0, find the value of h needed to produce a magnetic flux of 50 J1.Wb in branch 3 ofthe circuit. - (a) h = 0.66A. (b) h = 1.66A. (c) h =2.66A.

ba

d12

511--

c

'I I- 51 52 1

1351 -- 12

N31b Nl2c

N21a

d c c 2c c c5 c

Fig. P13.28. A magnetic circuit. Fig. P13.29. A linear magnetic circuit.

P13.29. A linear magnetic circuit is shown in Fig. P13.29. The first winding has NI = 100turns, and the second one Nz = 48. Find the magnetic flux in all the branches of the circuit ifthe currents in the windings are (1) h = lOrnA,lz = lOmA;(2) h = 20mA, lz = OmA;(3)h = -10 mA, lz = 10mA. The magnetic material of the core has J1.r= 4000, the dimensions

.-.

I


of the core are a = 4 cm, b = 6 cm, c = 1cm, and the thickness of the core is d = 1 cm. -Hint: designate the left, right and central branches by 1, 2 and 3, and orient them upward,downwardand downward, respectively. Possible results for case 3 are: (a) cI>1= -7.25 J.LWb,cI>2= 2.79J.LWb, cI>3= -14.04J.LWb. (b) cI>1= -3.25J.LWb, cI>2= 3.79J.LWb, cI>3= -9.04J.LWb.(c) cI>1= -5.25J.LWb, cI>2= 1.79J.LWb, cI>3= -7.04J.LWb.

','" ,/' ,/ , \

/ /' '\/ / \ \

I / \ \I I \ I

J '__I I I~ ~L-J , ,

\ \ / ~l-'-r»l\ \ / I\ \ /' /

\ " //\ /''" -'"------

Fig. P13.30. A single loop on toroidal core.

PI3.30. Shown in Fig. P13.30 is a single current loop on a toroidal core (indicated in dashedlines) of very high permeability. Assume that the core can be obtained by a gradual increaseof the number of the Ampere currents, from zero to the final number per unit volume. Followthe process of creating the magnetic field in the core as the core becomes "denser". If yourreasoning is correct, you should come to the answer to an important question: which is thephysical mechanism of channeling the magnetic flux by ferromagnetic cores? - Hint: whenthe number of Ampere currents is small, they are oriented by the magnetic field of the loop(primary field) only. By increasing the number of Ampere currents, the resultant field due tothem (secondary field) is also increased, and the currents are oriented in response to the totalmagnetic field.

chapter magnetic field in materialsecee.colorado.edu/~mcleod/teaching/eandm3400/lab...

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