C H A T E R
Magnetostatics Overview 5-1
Magnetic Forcesand Torques
5-2
The Biot-Savart Law
5-3
Magnetic Force betweenTwo Parallel conductors
5-4
Maxwell's MagnetostaticEquations
5-5
Vector Magnetic Potential
5-6
Magnetic Propertiesof Materials
5-7
Magnetic Boundary Conditions
5-8
Inductance
5-9
Magnetic EnergY
YERVIEW us chapteron magnetostatics parallelsthe preced_ : --hapter on electrostatics. Stationarychargesproduce c electricfields, and steady(nontimevarying)cur_ s producemagneticfields.For 0/0t - 0, the mag_ r: fieldsin a mediumwith magneticpermeabirity 1t : ioyernedby the secondpair of Maxwell'sequations, :s givenby Eqs.(4.3aandb):
V.B:0,
(5.1a)
V x H: J,
(5.1b)
rents and the magnetic fields B and H for various types of current distributions and in various types of media and to introducea numberof relatedquantitiis, suchas the magnetic vectorpotentialA, the magneticenergydensity ur,n, and the inductanceof a conducting structure,z. The parallelism betweenthesemagnetostaticquantities and their electrostaticcounterpartsis encapsulatedin Thble 5-1.
5-1 Magnetic Forces andTorques
r: J is the current density. The magnetic flux den_ B and the magnetic field intensity H are related by
B: pH.
anunderstanding of therelationships between sready cur-
(s.2)
we examined electric fields in a dielectric mern Chapter 4, we noted that the relation D : eE rrrrd only when the medium is linear and isotropic. E$cproperties, which are true for most materials. alLi to treat the permittivity r as a constant scalar n'. independentof both the magnitude and direc_ :i E. A similar statementapplies to the relation bl' Eq. (5.2).With the exception of ferromagneric pls, for which the relationship between B and H urnear, most materials are charactenzel by con_ nagnetic permeabilities. Furthermore, p &o : srdielectrics and metals (excluding ferromagnetic
u/s/. Our objectivein this chapteris to develop
Lt9
The electric field E at a point in spacehasbeendefined as the electric force F" per unit chargeacting on a test charge when placed at that point. we now define the magnetic flux density B at a point in spacein terms of the magnetic force F that would be eierted on a charged particre moving with a velocity u were it to bepassingthrough that point. Based on experimentsconducied to;etermine the motion of chargedparticles moving in magnetic fields, it was establishedthat the magnetic force F* acting on a particle of chargeq canbe cast in the form
F'n-quxB
(N).
(5.3)
Accordingly, the strength of B is measured in newtons(c'm/s), which arso is called the tesla (T) in SI units. For a positively charged particle, the direction of F,n is in the direction of ih. ",o* product u x B, which is perpendicularto the prane "ontuining
CHAPTER 5 MAGNETOST
Thble 5-1: Auributes of electrostaticsand magnetostatics.
Electrostatics
Attribute Sources
Stationarycharges Steadycurrents
Fietds
EandD
HandB
Constitutive paremeter(s)
e ando
p
Governing equations o Differential form
V.D-pv VxE:0
r Integral form
/r"
Potential
ds=s
V.B:0 VxH-J
fr".ds:o
-o fr"-dt
-r fr".dt
Scalar V, with E=-VV
Vector A, with B:VxA
Energydensity
u)e: it4'
u)m: *.pH,
charge4
Fs : qE
Fm : qu x B
Circuit element(s)
C and R
L
&olot
u and B and governed by the righrhand rule. If q is negative,the direction of F* is reversed,as illustrated in Fig. 5- l. The magnitudeof Fn is given by f; - quB sinO,
(s.4)
where 0 is the angle betweenu and B. We note that F* is maximum when u is perpendicularto B (0 : 90o), and it is zero when u is parallel to B (A - 0 or 180"). If a chargedparticle is in the presenceof both an electric field E and a magnetic field B, then the total electromagnetic force acting on it is
F:F"*F*
Magnetostatics
-qE*quxB:
q(E+uxB). (5.5)
The force expressedby Eq. (5.5) is known as the force. Electric and magnetic forces exhibit a nu important differences: l. Whereasthe electric force is alwavs in the of the electric field, the magnetic force is perpendicularto the magnetic field. 2. Whereasthe electric force actson a charged whetheror not it is moving, the magneticforce on it only when it is in motion. 3. Whereas the electric force expends energy in placing a chargedparticle, the magnetic force no work when a particle is displaced. Our last statementrequires further elaboration. the magnetic force Fn' is always perpendicular b
: MAGNETIC FORCESAND TORQI.JES
ExERclsE5.1 An erectronmoving in the positive -r-directionperpendicularto a magneticfield experiences a deflectionin the negativez-direction. What is the di_ rectionof the magneticfield? Ans. Positivey-direction. (See O) ExERclsE 5.2 A protonmovingwith a speedof 2 x 106 m/s througha magneticfield with magneticflux denof a magneticforceof magnitude lity 2.7T experiences 4 x 10-13N. what is the angrebetweenthe magnetic field andtheproton'svelocity? Ans. 0 : 30oor 150o. (See$) ExERclsE 5.3 A charged particlewith velocityu is moving in a mediumcontaininguniform fieldsE : iE and B - iB. what shouldu be so rhat the particleexperiencesno netforceon it? Ans. u : 2E/8. [u may also havean arbitraryy_ component uy7. (SeeS)
:di is :d force gv ln
re 5-1: The direction of the magnetic force exerted r charged particle moving in a magnetic field is (a) -ndicular to both B and u and (b) depends on the polarity (positive or negative).
u = 0. Hence, the work performed when a particle ,;laced by a differential distancedl - u dr is dW - Fm .dl : (F* .u) dt - 0.
(5.6)
orce no work is done, a magnetic field cannot change etic energyof a chargedparticle ; the magnet i c e1d fi "lnge the direction of motion of a charged,particle, rlar to
r .;nnot change its speed.
5-1.1 Magnetic Force 0na Gurrent-Garrying Conductor A current flowing through a conducting wire consists of chargedparticles drift i ng through the material of the wire. consequently,when a current-carrying wire is placed in a magneticfield, it will experiencea force equal to the sum of the magnetic forces acting on the charged particles moving within it. Consider, for example, th" u,,ung"_ ment shown in Fig. 5-2 in which a vertical wire oriented along the z-direction is placed in a magnetic field B (producedby a magnet)oriented along the -i-direction (into the page). With no current flowing in the wire, Fm : 0 and the wire maintains its vertical orientation, as shown in Fig. 5-2(a), but when a current is introduced in the wire, the wire deflects to the left (-i-direction) if the current's direction is upward (+2-Oirection), and it de_ flects to the right (*9-direction) if the current's direction
CHAPTER5 MAGNETOSI] chargedensity Pve:
-N"€, where N" is the n
movingelectronsper unit volume,thenthetotal B
@ @
@ @
@ o
@ @
@ @
@ @ @
of moving charge contained in an elemental the wire is dQ : preA dl - -N"eA dI,,
@
/=0 (a)
(b)
and the corresponding magnetic force acting on the presenceof a magneticfield B is dFtn : dQ rreX B : -N"eA dl ue x B, where u" is the drift velocity of the electrons.Si direction of a currentis definedas the directionof positive charges,the electron drift velocity u" is to dl, but oppositein direction.Thus,dlu" - -dl Eq. (5.8a)becomes
(c) Figure 5-2: When a slightly flexible verticalwire is placed in a magnetic field directed into the page (as denoted by the crosses),it is (a) not deflected when the current through it is zero, (b) deflected to the left when I is upward, and (c) deflectedto the right when / is downward.
dFrn - N"eAu" dl x B. From Eqs. (4.11) and (4.12), the current I through a cross-sectionalarea A due to electrons density pu" : -N"€, moving with velocity pr"(-u")A - (-Nre)(-u")AI: N"eAu". Eq. (5.8b) may be written in the compactform dF'.n: I dlxB
is downward(-2-direction). The directionsof thesedeflectionsare in accordancewith the crossproduct given
byEq.(s.3). To quantify the relationshipbetweenF* and the current / flowing in the wire, let us considera small segmentof the wire of cross-sectionalareaA anddifferential lengthdl, with the direction of dl denotingthe direction of the current.Without loss of generality,we assumethat the chargecarriers constituting the current I are exclusively electrons,which is always a valid assumptionfor a good conductor.If the wire contains a free-electron
(N).
For a closedcircuit of contour C canying a current total magneticforce is
F m :r
f
arrn
(N).
(5.10)
We will now examine the application of Eq. (5. eachof two specialsituations.
:-x MAGNETIC FORCESAND TORQIJES
123 ClosedCircuitin a UniformB Field
rmber
Considera closedwire carryinga current/ andplaced in a uniform externalmagneticfield B, as shown in Fig. 5-3(a).SinceB is constant,it can be takenoutside the integralin Eq. (5.10),in which casewe have
ilunrc
Fm: / ndQ
(f,^)
(s.lr)
xB:Q.
This result, which is a consequenceof the fact that the vector sum of the displacementvectorsdl over a closed path is equal to zero, states that the total magnettcforce on any closed curcent loop in a unifurrn magneticfield is zero.
Sincc (a)
ri s 'dlu*
Curved Wire in a UniformB Field If we are interested in the magnetic force exerted on a wire segment,such as that shown in Fig. 5-3(b), when placedin a uniform field B, then Eq. (5.10)becomes
F* : ' (1"' dr) x B: Itx B,
(s.12)
where I is the vector directed from a to b, as shown in Fig. 5-3(b). The integral of dl from a to b has the same value irrespectiveof the path taken betweena andD. For a closed loop, points a and b become the samepoint, in whichcasel:0andF*:0.
I ro{ls
I lc. 1
Exqmple 5-l ' Force ona Semicircular Conduclor
*q;rcr I
(b) Fetsre 5-3: In a uniform magnetic field, (a) the net force n r closed current loop is zero because the integral of h rsplacement vector dl over a closed contour is zero, : r the force on a line segment is proportional to the r ltErr betweenthe end point (Fm : Il x B).
The semicircularconductor shown in Fig. 5-4 lies in the;-y planeandcarriesa current /. The closedcircuit is exposedto a uniform magnetic field B : j Bs. Determine (a) themagneticforceF1 on the straightsectionof the wire and (b) the force F2 on the curved section. Solution: (a) The straight section of the circuit is of length 2r, andthe current flowing through it is along the *,r-direction. Application of Eq. (5.12) with I : *.2r gives
Fr : *,(21r)x iBo : 221rBo
N).
CHAPTERs MAGNETOST dueto verticallyupward?(Hinc Theacceleration -29.8m/s2.) isgAns. B - i0.49 T. (See O;
REVIEW OUESTIONS Q5.1 What are the major differences between havior of the electric force F" and the behavior magneticforce F*?
a Q5.2 TheendsofalO-cm-longwirecarrying Figure 5-4: Semicircular conductor in a uniform field (Example5-1).
(b) Let us considera segmentof differential length dl on the curvedpart of the circle. The direction of dl is chosen to coincide with the direction of the current. Since dl andB areboth in the;r-y plane,their crossproduct dlxB points in the negativez-direction, and the magnitudeof dl x B is proportional to sin rp, where @ is the angle betweendlandB. Moreover,the magnitudeof dl is dl r df.Hence, flr
F2-t I drxB J 6:O - -iI ["
J 6:o
rBs sinO d.O- -i2lrBs
(N).
We note that F2 - -F1, and consequentlythe net force on the closedloop is zero. I
current I areanchoredat two points on the x-axis,l and x : 6 cm. If the wire lies in the x-y plane presentin a magnetic field B - iBo, which of lowing arrangementsproduces a greater magnetic on the wire: (a) wire is V-shapedwith cornersat
(3,4), and(6,0), (b) wire lookslike an open at (0,0),(0,2),(6,2),and(6,0). withcorners
Torque 0na Gurrent'Carry 5-1.2 Magnetic Loop When a force is applied on a rigid body pivoted fixed axis, the body will reactby rotating about The strengthof the reactiondependson the cross of the applied force vector F and the distance measuredfrom a point on the rotation axis (such perpendicularto the axis) to the point of appli (Fig. 5-5).The lengthof d is calledthe moment the crossproduct is called the torque: T-dxF
5.4 A horizontal wire with a mass per unit EXERCISE of 0.2 kg/m caries a culrent of 4 A in the length *.r-direction. If the wire is placedin a uniform magnetic flux density B, what should the direction and minimum magnitudeof B be in order to magneticallylift the wire
(N.m).
The unit for T is the sameas that for work or e torque does not representwork or energy.The applied on the disk shown in Fig. 5-5 lies in plane and makesan angle 0 with d. Hence,
T - irF sin9,
STA
MAGNETTCFORCESAND TORQT.JES
r25 v
t I
--l'X
a
ris. r
Fs-ure5-5: The forceF actingon a circulardisk pivoted rrog tbc z-axisgenerates a torqueT : d x F thatcauses mc-isk to rotate.
Pivota*is/'
me lf the
l€uc ;at(
dl _ r, the radiusof the disk, and F _ lFl. Eq.(5.14)we seethata torquealongthepositive 7
im-tion correspondsto a tendency for the cylinder to m a counterclockwisedirection and, conversely,a i e torquecorrespondsto clockwise rotation.These F*f,{rsaregovernedby the followingright-hand rule: :;:e thumb of the right hand is pointed along the * -,nof the torque,the four fingers indicate the diBd
r tha ss t \ *t
:atia ta
+ I
Ol*"
rnttr ilutt the torque is trying to rotate the body. Te rilf now consider the magnetic torqae exerted on
ing loop underthe influenceof magneticforces. gln *'ith the simple casewherethe magneticfield B is :*eneof theloop, andthenwe will extendthe analysis rrre general casewhere B makes an angle d with normalof the loop.
(b) Figure 5-6: Rectangular loop pivoted along the y-axis: (a) front view and (b) bottom view. The combination of forces F1 and F3 on the loop generatesa torque that tends to rotate the loop in a clockwise direction as shown in (b).
Field in the Plane of the Loop mangular conductingloop shown in Fig. 5-6(a) is
:r ngid wire carryinga current1. The loop lies :-.r planeand is pivotedaboutthe axis shown. =e influenceof an externallygenerateduniform : field B - iBo, anns I and 3 of the loop are to forcesF1 and F3, r€spoctively,with
F i : I ( - 9 , b )x ( i B 6 )- i l b B s ,
(5.15a)
and
F::
I(9,b)x (iBe) : -ilbBo.
(s.15b)
Theseresultsare basedon the applicationof Eq. (5.12). No magneticforce is exertedon either arm2or4 because B is parallelto the direction of the currentflowing in those arms.
CHAPTER5 MAGNETOST The endview of the loop, depictedin Fig. 5-6(b), shows that forcesF1 andF3 producea torque about the origin O, causing thc toop to rotate in a cloclnrr'isedirection. The momentarm is a/2 far both forces, but d1 and d3 are in oppositedirections, resulting in a total magnetic torque of
T : d r x F r*dl x F:
: (-ut) x (ilunil * (*;)
" (ilun,)
:jIabBs - jr ABs,
(s.r6)
where A - ab is the area of the loop. The right-hand rule tells us that the senseof rotation is clockwise. The resultgivenbVEq. (5.16)is valid only when the magnetic field B is parallel to the plane of the loop. As soon as the loop startstorotate, the torque T startsto decrease,and at the end of one quarter of a complete rotation, the torque becomeszeto, as discussednext. MagneticField Perpendicularto the Axts of a RectangularLoop For the situation represented by Fig. 5-7, where B - iBe, the field is still perpendicular to the loop's axis of rotation, but its direction may be at any angle 0 with respectto the loop's surfacenormal ff, we may now have nonzeroforces on all four anns of the rectangular loop. However,forces F2 and Fa are equal in magnitude and oppositein direction and are along the rotation axis; hence,the net torque contributed by their combination is zeto. The directions of the currents in arms I and 3 are always perpendicularto B regardlessof the magnitude of 0. Hence,F1 and F3 have the sameexpressionsgiven previouslyby Eqs. (5.15aand b), and their moment arm is (a/2) sin 0, as illustratedin Fig. 5-7(b). Consequently, the magnitudeof the net torque exerted by the magnetic field about the axis of rotation is the same as that given by Eq. (5.16),but modifiedby sin 0: T : I ABosin9.
(5.17)
According to Eq. (5.17), the torque ls maxlmum the magnetic field is parallel to the plane of the
127
TI{E BIOT-SAVART LAW 11r= 90o)and is zero when the field is perpendicular u fre planeof the loop (0 :0). If the loop consistsof I rurns,eachcontributinga torquegivenby Eq. (5.17), m thetotal torqueis T - NIABssin0.
(5.18)
are quantity NIA is called the magnetic moment m r Se loop, and it may be regarded as a vector m with ur:;don ff, where ff is the surface normal of the loop rnd governedby the following right-hand rule: when no ' ;ur fngers of the right hand advancein the direction w i.e current I, the direction of the thumb specifiesthe s,'i :ion of fi.That is,
m a ffNrA
(A.m2), (5.19)
: m tenns of m, the torque vector T can be written as
T-mxB
(N.m). (5.20)
though the derivation leading to Eq. (5.20) was obfor B being perpendicular to the axis of rotation r rEctangularloop, the expressionis valid for any orin of B and for a loop of any shape.
0uEsTr0lrs Euc
hcular
lco
3 How is the direction of the magnetic moment of rq'defined? 4 [f one of two wires of equal lengths is formed . closed square loop and the other is formed into a
EXERCISE 5.5 A squarecoil of 100turns and0.5-m-long sides is in a region with a uniform magnetic flux density of 0.2 T. If the maximum magnetic torque exertedon the coil is 4 x l0-2 (N.m), what is the current flowing in the coil? Ans. /:8mA.
(SeeO)
Law 5-2 TheBiot-Savart In the preceding section, we elected to use the magnetic flux density B to denotethe presenceof a magnetic field in a given region of space.We will now work with the magnetic field intensity H instead. We do this in part to remind the reader that B and H are linearly related for most materials through B - pH, and therefore knowledge of one is synonymouswith knowledgeof the other (assumingthat p is known). Through his experiments on the deflection of compassneedlesby current-carrying wires, Hans Oerstedestablished that currents induce magnetic fields that form closedloops aroundthe wires [seeSection l-3.3]. Building upon Oersted's results, Jean Biot and Felix Savart arrived at an expressionthat relatesthe magnetic field H at any point in spaceto the current 1 that generatesH. tTneBiot-Savart law statesthat the differential magnetic field dH generatedby a steadycurrent / flowing through a differential length dl is given by
I dH- 4tr
dlxfr, R2
(A/m),
(5.21)
circularloop, andif bothwiresarecarryingequal ens and both loops have their planes parallel to a i:wn magneticfield, which loop would experiencea torque?
where R : RR is the distancevector betweendl and the observationpoint P shownin Fig. 5-8. The SI unit for H is ampere.mlmz - (A/m). It is important to remember
CHAPTER 5 MAGNETOST
(dII out of the page)
'p9 d H ,/R
'@ aw
(a) Volume current density J in (A/m2)
(dtl into the page) Figure 5-8: Magnetic field dH generated by a current element / dl. The direction of the field induced at point P /. is opposite that induced at point P
that the direction of the magnetic field is defined such thatdl is along the direction of the current / and the unit vector fr, points/rom the current element to the observation point. According to F4. 6.2t), dH varies as R-2, which is similar to the distance dependenceof the electric field induced by an electric charge. However, unlike the electric field vector E, whose direction is along the distancevector R joining the charge to the observation point, the magnetic field H is orthogonal to the plane containing the direction of the current element dl and the distancevector R. At point P in Fig. 5-8, the direction of dH is out of the page,whereasat point P' the direction of dH is into the page. To determine the total magnetic field H due to a conductor of finite size, we needto sum up the contributions due to all the current elementsmaking up the conductor. Hence,the Biot-Savart law becomes I f dl x R H r ! -I { - - r - p z 4n J,
(A/m),
(5.22)
(b) Surface current density J. in (A/m) Figure 5-9: (a) The total current crossing the cross section S of the cylinder is 1 : /s J .ds. (b) The total currenl flowing acrossthe surfaceof theconductoris f : It Jtdl.
andVol Fielddueto Surface 5-2.1 Magnetic Distributions Current The Biot-Savart law may also be expressedin terrm
distributedcurrentsources(Fig. 5-9) suchasthe current density J, measuredin (A/m2), or the su current density Jr, measuredin (A/m). The surface rent density Js applies to currents that flow on the faces of conductors in the form of sheets of effecti zero thickness. When the current sources are speci in terms of J, over a surface S or in terms of J volume y, we can use the equivalencegiven by Idl-Jrds:Jdv
where/ is the line path along which / exists.
:
r29
TTM BIOT-SAVART LAW
E crpressthe Biot-Savart law as follows:
H= *Il#*
(for a surfacecurrent), (5.24a)
H : * 1 " # *(for a volumecurrent). (s.24b) Magnetic Fieldof a LinearConduclor r i.inearconductorof length/ andcarryinga current1 alongthee-axisasshownin Fig. 5-10. rurced u letermine the magnetic flux density B at a point P at a distance r in the.r-y plane in free space. rul,l;l,;ad rrrution: From Fig. 5-10, current element dl -
2 dz
imu.:| x ft,: dz (2 x R) - frsin0 dz, wheref ir tt" ;nr: -Cr direction and 0 is the angle between dl and fi.. {rmce.applicationof Eq. (5.22) gives
t =
1
p z : t / 2d l x n
G lr:-,,,
R,
:o
^ I
fI/2sinl
+o l-,,, nz
dz' (5'251 (b)
ilr :r-rnvenience, we will convert the integration variable trnm- to 0 by using the transformations
(5.26a)
R:rcsc0, z: -r cot?,
(s.26b)
dz:rcsc2e d0.
(5.26c)
lmn rnsertingEqs. (5.26a) and (5.26c) into Eq. (5.25), lm Tii[-.e
":a*L:
Figure 5-10: Linear conductor of length / carrying a current f. (a) The field dH at point P due to incremental current element dl. (b) Limiting angles 0r and 02, each measured between vector I dl and the vector connecting the end of the conductor associated with that angle to point P (Example 5-2).
E@EI
sing rcsc2e d0
4n Jer
where 91 and 02 are the limiting angles at e : -l /2 and e :_ I f 2, respectively. From the right triangle in Fig. 5-10(b),
r2csc20
:Q7 I 1 usinodo +o, 1r, :Q
^ J
- cos0z), +nr(cos01
(s.27)
cos91:
t/2
lF\sJry'
(5.28a)
CHAPTER 5 MAGNETDS)
cos0z: - cos0t :
-t /2
(5.28b)
+ Q/D2
Hence,
B - PoIt: Q '
PoIl
2nrJ4r2 + 12
(r).
(s.ze)
For an infinitely long wire suchthat / D r,, Eq. (5.29) reducesto
Solution: For the straightsegmentsOA and OC, magnetic field at O is identically zero. This is
for all poin9 alongtheseTgrynts, dl is parallelor parallelto R and hencedl x R, : Q.For scgment a it perpendicularto f,' and dt x fr' : i dl - 2a F4. 6.22) gives Consequently,
? o a b' ^' IP ' H -=aG fJ 7 : ' m whered is in radians. r
^ LLII
B : A#
'' Hfifi$i0ffi"
(infinitelylong wire). I
(5.30) LooP Fieldof a Girculat Magnetic
Loop Fieldofa Pie'shaped Magnetic
Determine the magnetic field at the apex O of the pie-shapedloop shown in Fig. 5-11. Ignore the contributions to the field due to the current in the small arcs near O.
A circular loop of radius a carries a steady Determine the magnetic field H at a point on the the loop. Solution: Let us placethe loop in the x-y plane,as
for in Fig. 5-12.Our taskis to obtainanexpression pointP(0,0, z). We startby notingthat anyelementdl on the ci loop is perpendicular to the distance vector R, and all elementsaround the loop are at the samedi
the FromEq.(5-21), fromP, withp - 'rffi{. given by is dl element to nitudeof dEdue I
^
d H - f u l d r x R :l W ,
IdI
and the direction of dH is perpendicular to the containingR and dl. As shown in Fig. 5-l2,dH is r-e plane, and thereforeit hascomponentsdll' If we considerelementdl', locateddiametrically to dl, we observethat the z-componentsof the fields due to dl and dl' addbecausethey are in the direction, but their r-components cancel becaus are in opposite directions. Hence, the net magnetic is along z only. That is, Figure 5-11: Pie-shapedloop of radius a carrying a current.I (Example5-3).
e : \ffi dH- i dHz-- i dHcos
at.
f,8
THE BIOT-SAVART LAW
131 At the center of the loop (z : 0), Eq. (5.34) reducesto
H-re
(atz :0),
(s.3s)
and at points vgrl far away from the loop such that z2 > a2,Eq.(534) canbe approximated as
H-r#
(at lzl ) a). I
(5.36)
5-2.2 Magnetic Fieldofa Magnetic Dipole In view of the definition given by Eq.(5.19) for the magnetic moment m of a current loop, a loop with a single turn situated in the x-y plane, such as the one shownin Fig. s-lz,has a magnetic moment rl : im with m : Ina2. Consequently,Eq. (5.36) may be expressed in the form
ss for
n H = ='
cl
and
r fixed point P(0,0, z) on the axis of the loop, all itiesin Eq. (5.32) areconstant,exceptfor dl.Hence,
m
z, kp
@tlzl ) a)'
(5.37)
This expressionapplies for a point P very far away from the loop, but on the axis of the loop. Had the problem been solved to find H at any distant point p,(R,,0, , O,) in a sphericalcoordinate system,where R, is the distance betweenthe center of the loop and point p,, we would have obtained the expression
gratingW. 6.32) over a circle of radiusa gives
/ " , o t e oo , n - e4n(az * z f
^ I cosd (2r a). 4r (a2+ z2)
:7.-
2(azl rz1tlz ll. {
#(ft,2cos
o' +6sing') (s.38)
(s.33)
usingthe relation cos0 : a/(a2+r21r/z , we obtain
tnal expression:
H-
for R' )) a.A currentloop with dimensionsmuch smaller than the distance between the loop and the observation point is called a magnetic dipole. This is becausethe pattern of its magnetic field is similar to that of a permanent magnet,as well asto the pattern of the electric field of the electricdipole [seeExample4-7].The similarity is evident from the patternsshownin Fig. 5-13.
CHAPTER5
(c) Bar magnet
(a) Electric dipole
Figure 5-13: pattemsof (a) the electric field of an electric dipole, (b) the magneticficld of a magncticdipole' and (c) thc magneticfield of a bar magnet.Far awayftom the souces,the field patternsare similar in all threec{ses.
5.7 A wirecarryinga currentof 4 A is EXERCISE HEVTEW 0UEST|0ils Q5.5 Two infinitely long parallel wires carry currents of equal magnitude.What is the resultant magnetic field due to the two wires at a point midway betweenthe wires, compared with the magnetic field due to one of them alone, if the currents are (a) in the same direction and (b) in oppositedirections? Q5.6 Devise a right-hand rule for the direction of the magneticfield due to a linear current-carrying conductor. Q5.7 What is a magneticdipole? Describe its magnetic field distribution.
5.6 A semiinfinite linear conductor extends EXERCISE betweenz : 0and z : oo along the z-axis. Ifthe current / in the conductorflows along the positive e-direction, find H at a point in the.r-y plane at a radial distance r from the conductor. ^ I (A/m).(See($) Ans. H:d, +Tr
into a circular loop. If the magnetic field at the center the loop is 20 A"/m, what is the radius of the loop if loop has (a) only one turn and (b) 10 turns? Ans.
(a) a:
10 cm, (b) a : I m.
(See @)
5.8 A wire is formedinto a squareloop EXERCISE placedin the r-y planewith its centerat the origin each of its sides parallel to either the x- or y-axes. side is 40 cm in length, and the wire carries a culrent 5 A whosedirection is clockwise when the loop is vi from above.Calculate the magnetic field at the center the loop. 4I Ans. H - -|fri
- -i1r.25Nm. (See$)
Two Force between 5-3 Magnetic Gonductors Parallel In Section5- l. I we examinedthe
actson a current-carryingconductor tor is placedin an externalmagnetic
force Fm thl conducuneil
ATICS
1-4 MAXWELL'StrlecmEtOSTAnC EQUATIONS
r33 TheforceF2exertedon alength/ of wne l2dueto its prcsencein field 81 ma! be obtainedby applyingEq. (5.12):
Fz: IzI2 xBr : Izl2x (-i) ' 2g n d
= _^ 1ti fsI1I2l r,
(s.40
and the corresponding force per unit length is n/
ft2:
F2
T:
- v^ Pohlz z"a
(s.41)
c)tu A similar analysisperformed for the force per unit length exertedon the wire carrying /1 leads to f,
Frgure5-14:Magneticforcesonparallelcurrent-carrying :,:oductors.
n/
I t L: Vr -
Snter
^ PyI1I2 Z n d
(s.42)
pif
)op gtn 5.
rred ri entcr
Ot
r aneconductor, however, also generatesits own magmr-- field. Hence, if two current-carrying conductorsare rerd in eachother's vicinity, eachwill exert a magnetic lnmr on the other. I-et us consider two very long (or efribstcly infinitely long), straight, parallel wires in free rrce. separatedby a distance d and carrying currents 12in the same direction, as shown in Fig. 5-14. ; @mJ Imnt /1 is located at y - -d /2 and 12 is located at " = .i '2, and both point in the z-direction.We denoteB1 t?emagneticfield due to current /1, definedatthe loca"rf the wire carrying the current 12and, conversely,Bz ril :ne field due to 12 at the location of the wire carrying -nt 11.From Eq. (5.30), with I - Ir,t | : d, and - -i at the location of 12,the field 81 is
B,' :
-uttolt 2nd
(s.3e)
Thus, Fi : -Vl which meansthat the two wires attract each other with equal forces. If the currents are in opposite directions, the wires would repel each other with equal forces.
@s
5-4 Maxwell's Magnetostatic Equations Thus far, we have defined what we mean by a magnetic flux density B and the associatedmagnetic field H, we introduced the formulation provided by the Biot-Savart law for finding the fields B and H due to any specified distribution of electric currents, and we examined how magnetic fields can exert magnetic forces on moving chargedparticles and on current-carrying conductors.We will now examine two additional important properties of magnetostaticfields, those described mathematically by Eqs.(5.1aand b).
CHAPTER5 MAGNETOSTA
LawforMagnetism 5-4.1 Gauss's In Chaptcr 4 we learned that the net outward flux of the electric flux density D through a closed surfaceenclosing a net charge Q is equal to Q.We referred to this property as Gauss'slaw (for electricity), and we expressedit mathematicallyin differential and integral forms as
V . D: p v < +
$ O.ds:e.
(s.43)
/s
Conversionfrom differential to integral form was accomplished by applying the divergence theorem to a volume y enclosed by a surface S and containing charge O Iu pu dv [Section 4-4]. The magnetic analogueto a point charge is a magnetic pole, but whereas electric charges can exist in isolation, magnetic poles do not. Magnetic poles always occur in pairs; no matter how many times a pennanentmagnet is subdivided, eachnew piece will always havea north and a south pole, evenif the processwere to be continued down to the atomic level. Thus, there is no magnetic equivalenceto a charge Q or a chargedensity pu, and it is therefore not surprising that Gauss'slaw for magnetism is given by
V.B: 0 €=+
(s.44) {".ds:0.
The differential form is one of Maxwell's four equations, and the integral form is obtained with the help of the divergence theorem. Formally, the name "Gauss's law" refers to the electric case, even when no specific referenceto electricity is indicated. Theproperty describedby Eq. (5.44)hasbeencalled "the law of nonexistertce of isoIated monopoles,""the law of conservationof magnetic "Gatlss's law for magnetism,"among others. flux," and We prefer the last of the three cited names becauseit reminds us of the parallelism, as well as the differences, between the electric and magnetic properties of nature. The differencebetweenGauss'slaw forelectricity and its counterpartfor magnetismmay be viewed in terms of
(a) Electric dipole Figure 5-15: Whereas (a) the net electric flux through I closed surface surrounding a charge is not zero, (b) thc net magnetic flux through a closed surface surrounding one of the poles of a magnet is zero.
field lines.Electricfield linesoriginatefrom positive tric charges and terminate on negative electric Hence, for the electric field lines of the electric di shown in Fig. 5-15(a),the electric flux through a surfacesurroundingone of the chargesis not zero. In trast, magneticfield lines alwaysform continuous loops. As we saw in Section 5-2, the magnctic field due to currentsdo not begin or end at any point; this is for the linear conductorof Fig. 5-10 andthe circular of Fig. 5-12, as well as for any distribution of cu is also true for a magnet,asillustrated in Fig. 5- 15O) bar magnet.Becausethe magnetic field lines form c loops, the net magnetic flux through the closed surrounding the south pole of the magnet (or any other closed surface) is always zeto, regardless the shapeof that surface.
5-4.2 Ampire's Law We will now examine the property represented E q .( 5 . 1 b ) :
VxH:J,
(5,
:r
/
H
\
l l t
MAXWELL'S MRGNETOSTAIICEQUATIONS
135
.o""g
is the secondof Maxwell'spair of equationscharizing the magnetostatic fields, B and H. The integral
of Eq. (5.45)is calledAmpdre'scircuital bw (or Ampdre'slaw) under magnetostaticconditions mady currents).It is obtainedby integratingboth sides i Eq.(5.45)overan opensurfaceS,
7 , / t
)1' :
>\
pet ough r ft) ttE unding
/,o
x H ). d s :
lg.o",
: then invoking Stokes'stheoremgiven by Eq. (3.39)
-,buinthe result
f
H . d tr:
(Ampbre'slaw),
(5.47) (c)
C is the closed contour bounding the surface ,S tive c
ic di ac l. In uscl
ield h
ris is ular me,nttr.
xul mc
I rdless
(b)
(a)
(5.46)
Figure 5-16: Ampdre's law states that the line integral of H around a closed contour C is equal to the current traversing the surface bounded by the contour. This is true for contours (a) and (b), but the line integral of H is zero for the contour in (c) becausethe current / (denoted by the symbol O) is not enclosed by the contour C.
: I : I l.as is the total currentflowingthroughS. -v
sign conventionfor the direction of C is taken so r.-- / and H satisfi the right-hand rule defined earlier .'onnection with the Biot-Savart law. That is, if the rtion of / is aligned with the direction of the thumb ' tre right hand, then the direction of the contour C be chosen to be along the direction of the other fingers.In words,Ampire's circuital law statesthat line integral of H around a closed path is equnl to :urrent traversing the sudace boundedby that path. n'ay of illustration, for both configurations shown in . 5-16(a) and (b), the line integral of H is equal to --urrent.f, even though the paths have very different and the magnitude of H is not uniform along the I of configuration (b). By the same token, because : (c) in Fig. 5-16 doesnot enclosethe current /, its integral of H is identically zero, even though H is zero along the path. ihen we examinedGauss'slaw in Section4-4. we disthat in practice its usefulnessfor calculating the
*ric flux densityD is limited to chargedistributions possessa certaindegreeof symmetryand that the :ulation procedure is subject to proper choice of the
Gaussiansurface enclosing the charges.A similar statement applies to Ampbre's law: its usefulnessis limited to symmetric current distributions that allow the choice of convenientAmpirian contours around them, as illustrated by Examples5-5 to 5-7.
$
Magnetic Fieldofa LongWire
A long (practically infinite) straight wire of radius a carries a steady current / that is uniformly distributed overthe crosssectionof the wire. Determinethe magnetic field H at a distance r from the axis of the wire both (a) inside the wire (r < c) and (b) outside the wire (r ,. a). Solution: (a) We choose1 to be along the *z-direction, as shown in Fig. 5-17(a). To determineH1 at a distance rl 1 a, we choosethe Ampbrian contour C1 to be a cir-
CHAPTER5 MAG
t36
cular path of radius l'1 &s shown in Fig' 5-17(b)' In case,Ampbre's law takes the form
f,,
II1dlr : Ir,
a
Czl .
where /1 is the fraction of the total current f
P
a
.
t
through C1. From symmetry,Ht trlust be constas magnitudeandparallelto the contourat any point alo
I I
-t--u
f . , {
/\
I
'-- -/
the path. Furthermore, to satisfy the right-hand rule t given that 1 is along the z-direction, H1 must be along *d-direction in a cylindrical coordinate system. Hen Hr : |Hr. With dlr : 0rr dQ,tke left-hand si&
\ z '
(a) CYlindricalwire
Ampbre'slaw gives f
: I
z
f,,u,'a',
o
A
A
u r ( Q. 0 ) u d Q: 2 n r 1 H 1 .
JO
The current /r flowing through the area enclosed by is equal to the total current / multiplied by the rati the area enclosedby Cr to the total cross-sectional of the wire:
I' :(#)':(?)" (b) Wire crosssection
Equating both sides of Ampere's law and then for H1 leads to
H(r)
Hr : 6H, - r#,
(for11' a).
(b) For rz ) a, we choose path C2, which enclosec the current I. Hence,
f,,
H2' dl2 - 2rr2H2 - L,
(c) Figure 5-l7z Infinitely long wire of radius a carrying a uniform current I along the *e-direction: (a) gen(b) cral configuration showing contours Ct and Cz; (Exr versus H plot of (c) a cross-secti,onalview; and ample 5-5).
and
,r'z: Onr: A#
(for12, a).
(
If we ignore the subscnpt2, we observethat Eq' (! providis the sameexpressionfor B : &oH asEq' (5 which we had derived previously on the basis of Biot-Savart law.
MAXWELL'S MRGNETOSTAnCEQUATIONS
r37
Thevariationof the magnitudeof H asa function of r r plottedin Fig. 5-17(c);II increaseslinearly between - = 0 and r : a (inside the conductor),and then it &creasesas l/r outsidethe conductor. I
therewill be no currentflowing throughthe surfaceof the contour,and thereforeH - 0 for r < a. Similarly, for an contourwith radiusr 7 b,the nel current flowing throughits surfaceis zerobecausean equal numberof currentcoils crossthe surfacein both directions. Hence,H : 0 in the regionexteriorto the toroidal coil. For the regioninsidethe core,we constructa pathof radiusr, asshownin Fig.5-18.Foreachloop,we know from Example 5-4 that the field H at the centerof the looppointsalongtheaxisof theloop,whichin this case is the @-direction,and in view of the directionof the currentf shownin Fig. 5-18,the right-handrule tells us thatH mustbe in the -rp-direction.Hence,H - -irn. Thetotal currentcrossingthesurfaceof thecontourwith radius r is N/ and its direction is into the page.Accordingto the right-handrule associated with Am$re's law,the currentis positiveif it crossesthe surfaceof the contourin the directionof the four fingersof the right handwhenthe thumbis pointingalongthe directionof
H TATICS
Etant nt
rule ilong
.H I sidc
Hr.
dbv r ratio nal
ilEnClSE5.9 Current / flows in the inner conductor of ! I*Jngcoaxial cable and returns through the outer cono.u.1or.What is the magnetic field in the region outside ne coaxialcable and why? (See O; {rLs. H - 0 outside the coaxial cable becausethe net lurent enclosed by the Amperian contour is zero. 5.10 The metalniobium becomesa supercon:IERCISE rucror with zero electrical resistancewhen it is cooled to m.ow 9 K, but its superconductivebehavior ceaseswhen m magneticflux density at its surfaceexceeds0.12 T. lecrmine the maximum currentthat a 0.1-mm-diameter m*:biumwire can carry and remain superconductive. {,m. 1 - 30 A.
bmple 5-6
(See (F)
MagnelicFieldinsldea Toroidal Goil
A toroidal coil (also called a torus or toroid) is a mghnut-shaped structure (called its core) with closely irmuedturns of wire wrapped around it, as shown in ig 5-18. For clarity, we show the turns in the figure u spaced apart, but in practice they are wound in a *-xely spacedarrangementto form approximately cirruar loops. The toroid is used to magnetically couple nlriple circuits and to measurethe magnetic properties .r raterials, as illustratedlater in Fig. 5-30. For a toroid r'f,.r Y turnscarrying a current /, determinethe magnetic n*J H in each of the following three regions: r I a, t < r < b, and r ) b, all in the azimuthal plane of the mr:ld.
h. (5, q .r 5 is of
{rhtion: From symmetry, it is clear that H has uniform nm:certies in azimuth. If we constructa circular Amperian rmrrrour with center at the origin and of radius r 1 a,
Figure 5-18: Toroidal coil with inner radius a and outer radius b. The wire loops usually are much more closely spacedthan shown in the figure (Example 5-6).
CHAPTER 5 MAGNETOSTA To evaluatethe line integralin Ampere'slaw,we a rectangularAmpbrianpath aroundthe sheet,with mensionsI andur,8sshownin Fig. 5-19.Recalling reprcsentscurrentper unit length along the the total currentcrossingthe surfaceof the loop is I : .Irl. Hence,applying AmpBre'slaw
.l
,
the loop, while noting that II is zero along the pathc length ur, we have
f,
H' dl : ZHl : Jrl,
Ftgure 5-1,9:A thin currentsheetin the.r-y planecarrying a surfacecurrentdensityJs = iJs (out of the page) (Example5-7).
from which we obtainthe result
":[-r*, I e+,
forz > 0, forz<0.
thecontourC. Hence,in thepresentsituationthecurrent is -N/. Applicationof Am@re'slaw thengives
il: lo'"<-irn>.0,d0 : f,H
-2nrH: -N/.
We note that the magnetic field is uniform and where parallel to the current sheet. I
Hence,H - NI/(2rr)and
H- -0n:-i.y
z,rr
(foro
Magnetic Field of anInfiniteGurrent Sheet The x-y plane of Fig. 5-19 contains an infinite current sheet with surface current density J, - iJr. Find the magnetic field H. Solution: From symmetry considerations and the right-hand rule, H must be in the direction shown in the figure. That is,
H_ \ \\ \
-i,H, I I in'
fore > 0, forz < 0.
REVIEW OUESTIONS Q5.8 What is a fundamental difference between tric field lines and magnetic field lines? Q5.9
If the line integral of H over a closed
zgro,doesit follow that H : 0 at everypoint on contour?If not,whatthendoesit imply? Q5.10 Comparethe utility of applying the Biot law versus applying Ampdre's law for computing magnetic field due to current-carrying conductors. Q5.11 What is a toroid? What is the magnetic field points outside the toroid?
VECTORMAGNETIC POTE}ITIAL
r39
Vector Magnetic Potential
whereJ is the currentdensitydueto freechargesin motion. If we substitute Eq. (5.53)into Eq. (5.54),we have
cu treatmentof electnostaticfields in Chapter4, we Vx(VxA):pJ.
the electrostatic potential V and defined it 'dms of the line integral of the electric field E. In
form, we found that V and E are relatedby - -V V . This relationshipprovedusefulnot only in the electricfield distributionin circuit elements :h as resistorsand capacitors)to the voltagediffer:ss acrossthem, but it also proved more convenientin :ein casesto first determine the potential V due to a charge distribution and then apply the relationship
- -VV to find E thanto useCoulomb'slaw directly :nd E. We will now explorea similar approachin ion with the magneticflux densityB. tccordingto Eq. (5.M), V . B - 0. Wewishto define ! tennsof a magneticpotentialwith the constraint
I el'
;uch a definition guaranteesthat the divergenceof B rr*'ays equal to zero. This can be realized by taking sntageof the vector identity given by Eq. (3.38b), '-h statesthat, for any vector A,
V.(VxA)-0.
(s.s2)
, by defining the vector magnetic potential A such
en a
ntorr Ion
B:VxA
(Wb/m2), (5.53)
ire guaranteed that V .B = 0. The SI unit for B is =sla(T). An equivalentunit usedin the literatureis per squaremeter (Wb/m2). Consequently,the SI rr tor A is (Wb/m). rlth B : FH, the differential form of Ampbre's law by Eq. (5.45) can be written as
VxB-pJ,
(s.54)
(5.ss)
For anyvectorA, the Laplacianof A obeysthe vector identitygivenby Eq.(3.45),thatis, v2A- v(v.A) -v x (v x A),
(5.56)
where, by definition, V2A in Cartesian coordinatesis given by V2A -
(#.#.#)^ ivz A, + jvz Ay + 2v2Az.
(5.57)
CombiningEq. (5.55)with Eq. (5.56) gives
V(V .A) - VzA: LtJ.
(s.58)
When we introduced the defining equation for the vector magnetic potential A, given by Eq. (5.53), the only constraint we placed on A is that its definition satisfy the condition V .B - 0. Equation (5.58) containsa tenn involving V .A. It turns out from vector calculus that we have a fair amount of latitude in specifying a value or a functional form for V .A, without conflicting with the requirementrepresentedby Eq. (5.52). The simplest among theseallowed specificationsis
V.A:0.
(s.se)
Using this choice in Eq. (5.58) leads to the vector Poisson's equation given by
v2A : -pJ.
(5.60)
CHAPTER 5 MAGNETOST
r40 Using the definitionfor V2A givenby Eq' (5'57)' the vector Foisson'sequationcan be broken up into three scalarPoisson'sequations: (5.61a)
YzA, : -ltrJt, Y'Ar: -pJy,
(5.6lb)
Y'Ar:-PJz.
(5.61c)
In electrostatics,Poisson'sequation for the scalarpotential V is given bY Eq. (4.60) as
vzv - -g: , t
(s.62)
and its solution for a volume charge distribution pv occupying a volumev' was givenby Eq' (4'61) as
v' -
1
t 9:dr,
4ne Ju, R'
(5.63)
Poisson'sequationsfor Ar, Ay' and A, arc mathematically identical in form to Eq.(5.62). Hence,for a current density J with r-component J' distributed over a volume v', the solutionfor Eq. (5.61a)is IL f J, . A': fr J",* o''
(wb/m)' (5.64)
and similar solutions can be written for A, in terms of ./, and A. in terms of Jr. The three solutions can be combined into a vector equation of the form
A-h 1,,*o'' (Wb/m).
(5.65)
In view of Eq. (5.23\,if the culrentdistribution is given in the form of a surfaceculrent density J, over a surfaceS', then J dv' should be replacedwith J, dt' and v' should be replacedwith s'; and, similarly, for a line distribution, J dr-, should be replaced with I dl' and the integration should be performed over the associatedline path I'.
In addition to the Biot-Savart law and law, the vector magnetic potential provides a third
proachfor computingtbe magneticfield due to carryingconductors.For a specifiedcurrentdist Eq. (5.65) can be usedto find A, and then Eq' canbe usedto find B. Exceptfor simplecurrentd
tions with symmetrical geometries that lend to the application of Ampbre's law, the choice is between the approachesprovided by the Bio law and the vector magnetic potential, and among two the latter often is easier to apply becauseit is
to perform the integrationgiven by Eq. (5'65) performthe integrationgivenby Eq. (5.22)' T-bemagneticflux O linking a surfaceS is
the total magnetic flux density passingthrough S, c
o-
lrn.a"
ffb).
If we insertEq. (5.53)into Eq. (5.66)andthen Stokes'stheorem,we have
where C is the contour bounding the surface S' Q can be determined by either Eq. (5.66) or Eq' whichever is easier to integrate for the specific under consideration.
ofMaterials Properties 5-6 Magnetic According to our precedingdiscussions,becausethc tern of the magnetic field generated by a current is similar to that exhibited by a permanent magnet'' loop is regardedas a magnetic dipole with a north *O u south pole [see Section 5-2.2 and Fig' 5-13]' magnetic moment m of a loop of area A has a m : /A, and the direction of m is normal to the the loop in accordancewith the right-hand rule' tization in a material substanceis associatedwith current loops generatedby two principal mechani
56 MAGNETIC PROPERTIESOF MATERIALS
third t
tn
h ( Etn t
ot Dng lis i) thil
h r S .a
- , orbital motions of the electrons around the nucleus d similar motions of the protons around each other in t nucleus and (2) elecfion spin. The magnetic moment r an electron is due to the combination of its orbital mion and its spinning motion about its own axis. As re n'ill see later, the magnetic moment of the nucleus s rnuch smaller than that of an electron, and therefore u total magnetic moment of an atom is dominated by t sum of the magnetic moments of its elecftons. The rrsnetic behavior of a material is governed by the interrE rn of the magnetic dipole moments of its atoms with ro :rternal magnetic field. This behavior,which depends ,r fre crystalline stmcture of the material, is used as a -rs for classifying materials as diamagnetic, param,ryetic, or ferromagnetic. The atoms of a diamagnetic unrerial haveno pennanentmagneticdipole moments.In i:urerast,both paramagnetic and ferromagnetic materials r.E itofils with permanent magnetic dipole moments, ,- as will be explained later, the atoms of materials rcr--ngingto these two classeshave very different organerrional structures.
$4.1 OrtitalandSpinMagnetic Moments h heepthe following presentation simple, we will be-
ur discussionwith a classicalmodel of the atom, rhich the electrons are assumedto move in circular
aboutthe nucleus,and we will then extendthe by incorporatingthe predictionsprovidedby the
ials
correct quantum-mechanical model of matter. An with chargeof -e moving with a constantveloc-
scfu
e in acircularorbitofradiusr [Fig.5-20(a)]completes revolutionin time T : 2nr /u.\\is circularmotion
lTent
agn€t. Dord i-131. El
e.I itt
t
electron constitutes a tiny current loop with cur/ given by I - -'
T
:
eu
(5.68)
2nr rnagnitudeof the associatedorbital magnetic mo-
m" is
,to: IA:(#)@rr)
t4l
?:::::-.fi (a) Orbiting electron
(b) Spinning electron
Figure 5-20: An electron generates(a) an orbital magnetic moment mo as it rotates around the nucleus and (b) a spin magnetic moment ms, as it spins about its orvn axis.
eur
whereLe : meur is the angularmomentumof the electron andnr, is its mass.Accordingto quantumphysics, the orbital angularmomentumis quanttz.ed; specifically, f,. is alwayssomeintegermultiple of h : h/2n , wlrere ft is Planck's constant. Thatis, Z" : 0, h,2F1,....Consequently,the smallestnonzeromagnitudeof the orbital magneticmomentof an electronis tlto :
eh
2m"
(5.70)
Despite the fact that all substancescontain electronsand the electronsexhibit magneticdipole moments,most substancesare effectively nonmagnetic. This is because,in the absenceof an external magnetic field, the atoms of most materialsareoriented randomly, asa result of which the net magnetic moment generatedby their electrons is either zero or very small.
TECHNOLOGY BRIEF: ELECTROMAGNETSA}'ID MAGNETIC
Relays andMagnetic Electromagnets WilliamSturgeondevelopedthe firstpracticalelectromagnetin the 1820s.Today,the principleof the ebctrjmagnetis usedin motors,relayswitchesin forharddisksandtapedrives,loud read/writeheads speakers,magneticlevitation,and manyotherapplications.
the wirecoiledaroundthe centralcore,it i tfm a magneticfield with field lines resembling strengfl The generatedby a bar magnet(A1). to the current'I the magneticfieldis proportional permeability magnetic the and turns, numbeiof ferromagnetic a the core material.By using the field strengthcan be increasedby several ders of magnitude,dependingon the purityof iron material.Whensubiectedto a magnetic materials,suchas ironor nickel, ferromagnetic andact likemagnetsthemselves. magnetized
PrinciPle Basic Electromagnetscan be constructedin various the linearsolenoiddescribedin shapes,inLtuOing section5-8.1.when an electriccurrentgenerated by a powersource,suchas a batteryflowsthrough
Relays Magnetic breaker A magneticrelayis a switchor circuit "OFF" "ON" and can be activatedintothe
OLOCYBRXET: EIffiOMACNETS
I ngilt efil,
ttility lc efd
7d lc fGl,
AND MAGNE:TIC RELAYS
One exampleis the low-powerreed usedin telephoneequipment, whichconsists 'm flat nickel-ironbladesseparatedby a small {B).The bladesare shapedin sucha waythat fu absenceof an externalforce,they r€main and unconnected(OFF position).Electrical betweenthe blades(ONposition)is realized ptying a magneticfieldalongtheirlength.The irducedby a currentflowingin the wirecoiled theglassenvelope,causesthetwobladesto oppositemagneticpolarities,therebyforcftem to attracttogetherand closeout the gap.
CHAPTER 5
t44 its In additionto the magneticmoment producedby magnetic a spin generates orbital motion, an electron monentn,duetoitsspinningmotionaboutitsown axistFig.5-20(b)l.Themagnitudeofmrpredictedby quantumtheorYis eh nls:
2m"
,
(5.71)
mowhich is equal to the minimum orbital magnetic number even an ment mo.The electronsof an atom with of of electronsusually exist in pairs, with the members cancela pair having opposite spin directions, thereby number the If ini eachothers'-ipin magnetic moments. spin of-"lr.t ons is od-d,the atom will have a nonzero magneticmoment due to its unpaired electron' According to Eq. (5.71), the spin magnetic moment electron of an electron is inversely proportional to the massm".Tbenucleus of an atom also exhibits a spinning that of motion, but becauseits massis much greaterthan 10-3 of order the on is an electron,its magnetic moment of that of an electron.
quantitycalledthemaX whereXmis a dimensionless ,urrrptibility otthe material.For diamagneticand tougntti. materials,Im is a constantat a givente ftr; resultingin a linear relationshipbetweenM This is not the casefor ferromagnetic relationship between M and H not only is nt 'history" of the also dependson the previous explained in the next section. While keeping this fact in mind, let us combinc (5.12) and (5.73)to get
B : po(H+ XmH): &o(1* Xo')H, or B: &H, where;r.,the magneticpermeabilityof the givenin termsof 1* bY tt :
B : I.roH* poM : Po(H + M),
(5'72)
where the first term represents the contribution of the contriexternal field and the second term represents the a general, In material. the bution of the magnetizationof external the materialbecomesmagnetizedin responseto field H. Hence,M can be exPressedas
M - x.H,
(5.73)
GVm).
(5'76)
Often it is convenientto define the magnetic of a material in terms of the relative permeability
5-6.2 MagneticPermeabilitY "t\e magnetization vector M of a material is defined as of the the vector sum of the magnetic dipole moments The atoms contained in a unit volume of the material. : &oM' magneticflux density correspondingto M is B* H, field magnetic applied In the presenceof anexternally is material the toial magnetic flux densrty in the
tto(l * Xm)
rtrr:P &o
-l*xo,,
6-77)
where /.ro is the permeability of free space' A
paramagnetig, usuallyis classifiedasdiamagnetic, rornagneticon the basisof the value of its Xm,8ss Table 5-2. Diamagnetic materials have negative
tibilities and paramagneticmaterialshavepositivc ceptibilities.How"y"r, the absolutemagnitude-ofI on tt. order of 10-5 for both classesof materials, allows us to ignore 1,n relative to 1 in Eq' (5'77r' - Po for diamagnetic an'd1 givestrt, ry | or p magneticsubstances,which include dielectric ord ^ott metals.In contrasr,lpl'l S) l forfern materials; llt,l of purified iron, for example, order of 2 x 10s.Ferromagneticmaterials are next.
F H I
I I
:
o
STAN
MAGNETIC PROPERTIESOF MATERIALS
Table5-2: Propertiesof magneticmaterials.
and
M and nces; inear, ateriaL bine
Permenent magnetic dipole moment Primary magnetizadon mechanism Direction of induced magneticfield (relativeto external field) Commonsubstances
Ilpicat value of X. Ilpical value of p,
\ ticI E
Diamagnetism
Paramagnetism
Ferrumagnetism
No
Yes,but weak
Yes,andsEong
Electronorbital magneticmoment
Electronspin magneticmoment
Magnetized domains
Opposite
Same
Hysteresis [seeFig. 5-221
Bismuth,copper,diamond, Aluminum,calcium, gold, lead,mercury silver, chromium,magnesium, silicon niobium,platinum, tungsten -10-s ry ry 10-5 lXtnl> ryl a;l lptl )
5.11 The magnetic vector M is the vector il 6e magnetic momentsof all the atoms contained mrt volume (1m3). If a certain type of iron with r lrP atoms/m3contributes one electron per atom is spin magnetic moment along the direction of Efr€dfield, find (a) the spin magnetic moment of b electron,given that m" - 9.1 x 10-31(tg) anO - ".6 x 10-34(J.s),and (b) the magnitudeof M. t t) ms : 9.3 x l0-u (A.m2), (b) M : 7 .9 x 105 (S€e O)
liitt
lof
Ir [7- r*
Hagnetic Hysteresis ofFerromagnetic ilaterials tic materials, which include iron, nickel, and crhibit strong magnetic properties due to the fact magnetic moments tend to align readily along ion of an external magnetic field. Moreover, s remain partially magnetized even after
kon, nickel, cobalt l andhysteretic I andhysteretic
the removal of the external field. Becauseof this property, ferromagnetic materials are used in the fabrication of permanentmagnets. Common to all ferromagnetic materials is a characteristic featuredescribedby magnetizeddomains. A magnetized domain of a material is a microscopic region (on the order of 10-lo m3; within which the magnetic moments of all its atoms (typically on the order of l0le atoms) are aligned parallel to each other. This permanent alignment is attributed to strong coupling forces between the magnetic dipole momentsconstituting an individual domain. In the absenceof an external magnetic field, the domains take on random orientations relative to each other, as shownin Fig. 5-21(a),resultingin a net magnetizationof z&ro.The domain walls forming the boundariesbetween adjacentdomainsconsistof thin transition regions.When an unmagnetized sample of a ferromagnetic material is placed in an external magnetic field, the domains will align partially with the external field, as illustrated in Fig. 5-21(b). A quantitativeunderstandingof how these domains form and how they behaveunder the influence
5 CHAPTER
(a) Unmagnetizeddomains
pable of measuring B and H. The unmagnetized B (b) Magnetizeddomains and (b) Figure 5-21:Comparisonof (a) unmagnetized material' domainsin a ferromagnetic ma-gnetized
of of an external magnetic field requires a heavy dose the of quantum mechanics, which is outside the scope presenttreatment.Hence, we will confine our discussion process io a qualitative description of the magnetization and its implications. The magnetization behavior of a ferromagnetic material is describedin termsof its B-H magnetization curve , where f;I is the amplitude of the externally applied magnetic field, H, and I is the amplitude of the total magnetic to flux density B present within the material. According Eq. (5.72), ttrehux B consistsof a contribution p'sH due to the external field and a contribution psM due to the magnetizationfield induced in the material. Supposethat we start with an unmagnetized sample of iron, and let uSassumethat we have an experimental zurangementca-
denotedby point O inFig. 5-22.As we startto H continuously,B increasesalso,and the responsc lowsthecurvefrom point O to point A1,Ltwhichu all the domains have become aligned with H' Poi representsa saturation condition' If we now start b "r"ur" H from its value at point 41 back to zero, magnetizationcurve follows the path from A 1 to A point 42, the externalfield H is zero,but the flux dens
in the materialis not zero.This valueof B is residualflux densityBr. The iron materialis now netized and can serve as a permanent magnet the fact that a large fraction of its magnetization have remained aligned. Reversing the direction of
increasingits intensitycausesB to decreasefrom point Az to zetoatpointA 3, oodif theintensityof H
creasedfurther (while maintaining its negativedi the magnetizationcurve movesto the saturation at point A+. Finally, as H is made to return to zero then increasedagain in the positive direction, the follows the path from Aa to Ar. This processis magnetic hysteresis.The term hysteresismeans behind." T\e hysteresis loop shows that the tion processin ferromagnetic materials dependsnot on the externalmagnetic field H, but on the magneuc
DI{GNETIC BOUNDARY CONDITIONS
147 Q5.14 What doesthe magnetizationcurve describe? What is the differencebetrveenthe magnetizationcurves of hard and soft ferrornagneticmaterials?
5-7 Magnetic Boundary Gonditions
'izd
tre materialaswell. Thespecificshapeandextent
1 t oi
hlsteresis loop dependon the propertiesof the feric material and on the peak-to-peakrange over H is madeto vary. Materials characterizedby wide loops are called hardferromagnetic materials !13(a)1. These materialscannot be easily demagibl'an externalmagneticfield becausethey havea csidual magnetization .Br.Hard ferromagnetic marre used in the fabrication of permanent magnets
yhich
H. Pci ' start 3o 1 1t o
[x F S IIOTY
ct owi KXr
bn of r from
andgenerators.Softferromagnetic materials Errrowhysteresisloops [Fig. 5-23(b)], and hence m be moreeasilymagnetizedand demagnetized. ize any ferromagnetic material, the material to several hysteresiscycles while gradually g the peak-to-peakrange of the applied field.
yofEi D
f,l
)zero u the ss is }ans ds not
In Section 4-9, we derived a set of boundary conditions that describes how, at the boundary between two dissimilar contiguous media, the electric field quantities D and E in the first medium are related to those in the second medium. We will now derive a similar set of boundary conditions for the magnetic field quantities B and H. By applying Gauss'slaw to a pill box that straddlesthe boundary,we determinedthat the difference betweenthe normal components of the electric flux densities in the two media is equal to the surfacechargedensity pr. That is,
-0 ->
f,o'a'
Dn - D2n: p".
(5.78)
By analogy,application of Gauss'slaw for magnetism, as expressedby Eq.(5.44), would lead to the conclusion that
===+ fr".ds:o t"":"- 1
(s.7e)
This result statesthat the nornutl componentof'B i.r c.ontinttottsacrosstlrc boundarybetweentw,oudjacentmedia. In view of the relations Br : FtH1 and 82 : pzHrz for linear, isotropic media, the boundary condition for H correspondingto Eq. (5.79) is
OUESTIONS I What are the three types of magnetic materials rbat are typical values of their relative magnetic
ilities? i What causesmagnetic hysteresis in ferromagnaterials?
lhHn:
lrzHzn
(5.80)
Comparisonof Eqs. (5.78) and (5.79)tells us rhar, w'hereasthe nonnal contltottent of B is corttitzuousoct-oss the bowtclary, the nonttal contponent of D mav rutt be
CHAPTER 5 MAGNETOST
4
I
I
Figure 5-242Boundary betweenmedium 1 with p.1and medium 2with 1t'2.
(unless ps : 0). A similar reversal applies to the tangential componentsof the electric and magnetic fields E and H: whereasthe tangentialcomponentof E is continuousacrossthe boundary,the tangentialcomponentof H may not be (unlessthe surface current densityJr : 0). To obtain the boundary condition for the tangentialcomponent of H, we follow the samebasic procedurethat we used previously in Section 4-9 to establishthe boundary condition for the tangential component of E. \Mth referenceto Fig. 5-24, if we apply Ampbre's law [Eq. (5.47)] to a closed rectangular path with sides A/ and A/r, and then we let Lh -+ 0, we end up with the result
the loop approachesa thin line of length A/. total current flowing through this thin line is / : where J, is the magnitude of the normal co
the surfacecurrentdensitytraversingthe loop.In Eq. (5.81)becomes theseconsiderations, H21A,l - Hn L,l : /r A/
Hz, - H*:
J,
(A/m).
This result can be generalized to a vector form
r r . d r + ^ r . d r :I , ( s . 8 1 )poratesthe directional relationship defined by I,o fr,.dr: I,u hand rule, where H1 and H2 are the magnetic fields in media I and2, respectively.According to Ampdre's law, / is the net current crossing the surface of the loop in the direction specified by the right-hand rule (/ is in the direction of the thumb when the fingers of the right hand extend in the direction of the loop C). For the directions of Hr and H2 and the direction of the integration contour C indicated in Fig. 5-24, the componentof H2 tangential to the boundary,H21,is parallel to and in the samedirection as dl over segmentab,butthe tangential component of H1 is antiparallel to dl over segmentcd. Furtherrnore, as we let Ah of the loop approach zero, the surface of
ffz x (Hr - Hz) : Jr,
(5.84)
where ffz is the normal unit vector pointing medium 2 (Fig. 5-24). Surface currents can on the surfacesof perfect conductors and tors. Hence, at the interface betweenmedia w conductivities,Js : 0 and
Hr:
Hzt. (5.85)
!
INDUCTANCE
CfSE5.12 With referenceto Fig. 5-24, determine anglebetrveenH1 and ffz - i it H2 : (t3 + 22) 1m),&rr - 2, andpr, - 8, andJ, : 0. 0 :20.6.
( S e eO )
Inductance rnductoris the magnetic analogueof an electrical ca-
:tor. Justasa capacitorcanstoreelectricenergyin the :sic field presentin the mediumbetweenits conductsurfaces,an inductor can store magneticenergyin rolumecomprisingthe inductors.A typicalexample rn inductoris a coil consistingof multiple turns of wound in a helical geomefry around a cylindrical as shown in Fig. 5-25(a). Such a structure is called bnoid. The core may be air filled or may contain a ic material with magnetic permeability 1t,.If the
r6ri y thc
I
l
J
wire carriesa current/ andthe turnsarecloselyspaced, the solenoidcan producea relatively uniform magnetic field within its interior region,andits rnagneticfield pattern resemblesthat of a permanentmagnet,asillustrated by the field linesin Fig. 5-250).
5-8.1 Magnetic Fieldin a Solenoid We precedeour discussionof inductanceby derivingan expressionfor the magneticflux densityB in the interior region of a tightly wound solenoidwith n turns per unit length.Eventhoughthe turns are slightly helical in shape,we will tneatthem ascircular loops,as shownin Fig. 5-26.The solenoidis of length / and radiusa and carriesa currentI.Let us startby consideringthe magnetic flux densityB at point P, locatedon the axis of the solenoid.In Example5-4, we derivedthe following
TI Hs BII ,,
il HH, l#' ( t H I
t H I
18 | Figure 5-262Solenoidcrosssectionshowinggeometry for calculatingH at a point P on the solenoidaxis.
150
Recording Magnetic ValdemarPoulsen,a Danish engineor,inventred magneticrecordingby demonstratingin 1900that spelch can b€ recoded on a thin steel wire with a Magnetictapesweredevelslmpleelectromagnet. mediumto wiresin the 1940s opedas an alternative and becameverypopularfor recordingand playing were intromusicwell into the 1960s.Videotapes motionpictures ducedinthelate1950sforrecording for laterreplayon television.Becausevideosignals occupya muchwider bandwidth,tape speedsfor videoiecording(pastthe magnetichead)haveto be at rateson thi orderof 5 m/s,comparedw1honly 0.3 m/s for audio.CIhertypesof magneticrecording mediaweredevelopedsincethen,includingthe ftJxibleplasticdiskscalledtloppies,'theharddisks drum,and themagnetic madeof glassor aluminum, advantage take the magneticbubblememory.All of the samefundamentalprincipleof beingableto storeelectricalinformationthroughselectivemagnetizationof a magneticmaterial,aswellas theabilwhenso desired. ity to retrieveit (playback)
Processes Record/Read Nlagnetictape recordingof sound is illustrated in (A).Soundwavesincidenton a mischematically to vibrate,creating crophone(1)causea diaphragm time and (2) corresponding with current electric an sound original the of that as amplitudevariations is pattern.The acousticto electricconversion realcacrystal,an electrostatic izedby a piezoelectric types other or field, magnetic pacitor,a coil in a
TECHNOLOGY BRIEF: MAGNETIC
OLOGY BRIEF: MAGNETIC RECORDING
y -nsducers. After amplification(3), the current ryral drivesa recordinghead (4) consistingof an cectromagnet to magnetizethe tape as it is drawn :as the head. The tape (5) is made of a plastic re materialwith a coatingof ferricoxidepowder lLed to its surface.Whenexposedto the magnetic re,ri.the previouslyrandomlyorientedmoleculesof t"ebnomagneticpowderbecomepermanently orir:ed alonga specificdirection,therebyestablishr: a magneticimprintof the originalsoundsignal :- the tape. Replayingis accomplished by reversrnr;[f1sprocess(B). Drawingthe magnetizedtape lE:st€lreproducing headinducesa currenthaving
151 TJT
a vibrationproportional to thaton thetape,whichis thenamplifiedand c-onverted backto soundwaves througha loudspeaker.
CHAPTER5 MAG expressionfor the magneticfield H at a distancez along the axisof a circularloop of radiusa: I'a2
H:i
' 2(a2+ z2)3/2
(5.86)
where f is the current carried by the loop. [f we treat anincrementallengthdz of the solenoidasanequivalent acurrentI' : Indz,then loopof nd.zturnsandcarrying the inducedfield at point P is t
(s.87)
The total field B at P is obtained by integrating the contributions from the entire length of the solenoid. This is facilitated by expressing the variable z in terms of the angle 0. That is,
(5.88a)
atan9,
az + z2 : o2 + a2 tan20 -- a2 secz0,
Upon substitutingthe last two expressionsin Eq. (5.87) and integrating from 0r to 02, wQhave
a3 sec30
: iPll (singz- sin91). 2
(5.8e)
If the solenoid length / is much larger than its radius a, then 0r - -90" and02 - 90", in which caseEq. (5.89) reducesto
B - itrtn, :U#
situated at its end points. How does B at the end compareto B at the midpoint of the solenoid?
Ans. B - 2(pN I /21) at the endpoints,which is aslargeasB at the midpoint. (See C)
5-8.2 Self-inductance From Fq. (5.66),the magneticflux O linking a is given by
o- l,s'a'
a e=7e o' B : iFnI 1u yz Jt,
EXERCISE 5.13 UseEq. (5.89)to obtainan ex for B at a point on the axis of a very long solenoid
(5.88b) (5.88c)
dz: a sec2e do.
2
in another circuit. Usually, when the term used,the intendedreferenceis to self-inductance.
)
! n ' o - - = ,*'' =d, d.B:pdlfl-2 2(a2I 7213/2
z:
it is approximatelyvalid at dl pointsin the interior solenoid,exceptnearthe ends. We now return to a discussionof inductance, includesself-fuducttnce, representingthe magnetic linkageof a coil or circuit with itself, andmutrnl ductance,which involvesthe magneticflux linkage circuit due to the magneticfield generatedby a
(tongsolenoid witht/a D l),
(s.e0) where N : nI is the total number of turns over the length /. Even though the expressiongiven by Eq. (5.90) was derived for the field B at the midpoint of the solenoid,
(wb).
(5.e1)
In a solenoid with an approximately uniform field given by Eq. (5.90), the flux linking a single
,: I,u(-+,)uds: ulrs, where S is the cross-sectionalareaof the loop. M flux linkage A is definedasthe total magneticflux li a given circuit or conducting structure. If the consistsof a single conductorwith multiple loops, the caseof the solenoid, A is equal to the flux linki loops of the structure.For a solenoid with N turns, N2
A-NQ-r-7IS
(wb).
(
INDUSTANCE
Radius a
(a) Parallel-wiretransmissionline
:noid rd oc
;h is
153 for thepurposeof determiningtheinductanceof a given structure,andinductanceis of interestprimarilyin thea-c case(i.e.,time-varyingcurrents,voltages,andfields).As we will seelater in Section7-6, the currentflowing in a conductorundera-c conditionsis concenratedwithin a very thin layer on the skin of the conductor.For the parallel-wiretansmissionline, the currentsflow on the outersurfacesof the wires,andfor the coaxialline, the currentflows on the outer surfaceof the inner conductor and on the inner surfaceof the outerconductor(the current-carryingsurfacesarethoseadjacentto the electric andmagneticfieldspresentin theregionbetweenthe conductors). Tlneself-inductanceof anyconductingstructureis defined as the ratio of the magneticflux linkageA to the current/ flowing throughthe structure:
, _ A I
(b) Coaxialtransmissionline mgure 5-272To computethe inductanceper unit length r r twmonductor transmissionline, we needto detere the magneticflux through the area.Sbetweenthe xoductors.
(H).
(5.e4)
The SI unit for inductanceis the henry (H), which is equivalentto webersper irmpere(Wb/A). For a solenoid,useof Eq. (5.93)gives
\:n the other hand, the structure consists of fvto sep.-onductors,as in the case of the parallel-wire and
172
L : U7S
(solenoid), (5.95)
transmission lines shownin Fig. 5-27,the flux ;3 A associatedwith a length / of either line refers frc flux O through the closed surface between the :cnductors, as highlighted by the shaded areas in :-27.In reality, thereis also somemagneticflux that through the conductors themselves,but it may be C by assuming that the currents flow only on the of the conductors, in which case the magnetic :nsidethe conductorsare zero. This assumptionis i*J by the fact that our interest in calculating A is
and for two-conductorconfigurationssimilar to thoseof
Fis.5-27,
L : + : ? : + l d, Bs(5.e6)
TECHNOLOGY BRIEF: INDUCTIVE S
lnductive Sensors Magnetic couplingbetweendifferentcoilsformsthe basisof severaldifferenttypesof inductivesensors. Applicationsincludethe measurementof position (withsubmillimeter resolution)in and displacement proximitydetection processes, of devicefabrication related applications. conductiveobjects,and other
Transformer Variable Diflerential Linear (LVDT) to an An LVDTcomprisesa primarycoilconnected ac source(typicallya sine waveat a frequencyin coils, the 1-10kHzrange)and a pairof secondary (A1). ferromagnetic core The all sharinga common magneticcore servesto couplethe magneticflux generatedby the primarycoil into the two secondaries,therebyinducingan outputvoltageacross
eachof them.The secondarycoilsare so thatwhenthe coreis in opposition, of the LVDT,the individual magnetic center the put signalsof the secondariescanceleach out, producinga null output voltage.The con connectedto the outsideworld via a rod. Whenthe rod movesthe core awayfrom magneticcenter,the magneticfluxesinducedin secondarycoilsare no longerequal,resulting nonzerooutputvoltage.The LVDTis calleda becausethe outputvoltageis a ear' transformer overa wide earfunctionof displacement range. The cutawayviewof the LVDTmodelin (A2) pictsa configuration in whichall threecoilsthe primarystraddledby the secondarieswoundarounda glasstubethatcontainsthe neticcore and attachedrod. Sampleapplica in (A3). are illustrated
LOGY BRIEF: INDUCTTVESENSORS
Gurrent Proximity$ensor
tn
-nsformer principlecan be appliedto build u'uimity sensorin which the outputvoltageof s€condary coil becomesa sensitiveindicatorof :resenceof a conductive objectin its immediate (B).Whenan objectis placedin frontof the coil,themagnetic fieldof thecoilinduces n (circular) currentsin the object,whichgener-ragneticfieldsof their own havinga direction
that opposesthe magneticfield of the secondary coil.The reductionin magneticflux causesa drop in outputvoltage,withthe magnitude of thechange beingdependent properties ontheconductive of the objectand its distancefromthe sensor.
CHAPTER 5 MAG betweenthe conductors,B is everywhere to the surface.Hence, the flux through ,Sis
#'"(:)
o - , l "B udr:,l"'fin,:
orI
Using Eq. (5.96),the inductanceper unit lengthd coa:rialffansmissionline is givenby
ool l L l
ool Outer
oor
Outer conductor
conductor Inner conductor
L,:L -|Q-4mrl\. r I
lI
2n
(s.ee)
\a/
Inductance 5-8.3 Mutual
view of coaxial transmisFigure 5-2E: Cross-sectional sionline (Example5-8).
Magnetic coupling between two different
structuresis describedin termsof the mutualin betrveenthem.For simplicity,let us assumethatwe two closed loops with surfaces51 and 52 and a cu flowing through the first loop, as shown in Fig. 5-29.
Inductance of a Goaxial Transmission Line Develop an expression for the inductance per unit length of acoaxial transmissionline. The conductorshave radii a and b, as shown in Fig. 5-28, and the insulating material has a linear permeability 1t.. Solution: Due to the current / in the innerconductor, the magnetic field generatedin the region with permeability tt betweenthe two conductorsis given by Eq. (5.30) as
Nr turns
C1 ^ ILI
B:0;-
LTT
,
(s.e7)
where r is the radial distance from the axis of the coaxial line. [,et us choose a transmission-linesegmentof length I as shown in Fig. 5-28. Over the planar surface S
Figure 5-29: Magnetic field lines generatedby current in loop I linking surface ,S2of loop 2.
r MAGNETICENERGY
REUIEW OUE$TIONS Q5.15 Whatis the magneticfield like in theinteriorof a long solenoid?
,(: )
(5.
Q5.16 Whatis the differencebetweenself-inductance andmutualinductance?
lgth of
Q5.17 How is the inductanceof a solenoidrelatedto its numberof turnsN? ftgure 5-30: Toroidal coil with two windings used as a ma;former.
5-g Magnetic Energy field Br generatedby 1r results in a flux 012
+h loop 2, givenby
orz:
Ir,u,
or,
(5.100)
: loop 2 consists of N2 turns all coupled by 81 in ) fte sameway, then the total magnetic flux linkage
$ lmp 2 dueto 81 is Atz :
NzQn -
N2
tr,nr.ar-
(s.101)
When we introducedelectrostaticenergyin Section 4-ll, we did so by examining what happensto the energy expended in charging up a capacitor from zero voltage to somefinal voltage V . We will now use a similar analogy by consideringan inductor with inductanceL connected to a current source. Suppose that we were to increase the current i flowing through the inductor from zero to a final value f. From circuit theory we know that the voltage u acrossthe inductor is given by u - L di /dt. Power p is equal to the product of u and i, and the time integral of power is work, or energy. Hence, the total energy in joules (J) expendedin building up the current in the inductor is
t>
\
rurual inductance associatedwith this magnetic
g is givenby
fms
f
f
7l
wm- na,: idi J J , u d t - LJ o
- lrr2
rnductanceis important in transformers wherein rmlings of two or more circuits share a common c core, as illustrated by the toroidal arrangement
m Fig.5-30.
(J).(s.103)
For reasonsthat will become apparent shortly, we call this the magnetic energy stored in the inductor. Let us considerthe solenoidinductor.Its inductanceis given by Eq. (5.95) as ^L : pN2 S//, and the magnitude of the magnetic flux density in its interior is given from Eq.(5.90)byB - 1tNI//.Equivalently,t - Bl/(pN).
CHAPTER5
158 for L and/ in Eq.(5.103)'we If we usetheseexpressions get
l'm: Ir,':i(r+t(#)' -!B'(I$ \-- / = ! p n ' r , 2'
2 t't
(5.104)
where v - /S is the volume of the interiorof the solenoid andH : B /tt.The magnetic energy density ttr* is defined asthe magneticenergy W. Perunit volume,
u)m:
W m l
f
:
(J/m3). (5.105)
z*H'
Even though this expressionwas derived for a solenoid inductor, it is equally valid for any medium with magnetic field H. Furthennore, for any volume v containing a material with permeability tt (including free spacewith permeability po), the total magneticenergy stored in the medium due to the presenceof a magneticfield H is
wmEramBle5-9
(r).
dv ; I"p"H2
(s.106)
MagneticEnergyin a CoaxialCable
Derive an expressionfor the magneticenergy storedin a coaxial cableof length / andinner andouterradii a andb . The insulation material haspermeability pt. Solution: From Eq. (5.97),the magnitudeof the magnetic field in the insulating material is given by
H-
B
I
i:
zor
''
where r is the radial distancefrom the centerof the inner conductor, as shown in Fig. 5-28-The magnetic energy storedin the coaxial cable is then given by
dv:# 1"1 or. wm : I,qLH.
Since II is a function of r only, we choosedv to cylindrical shell of lengthJ, radiusr, and thic alongthc radial diroction.Thus,dv = Znrl dr and
wm-# I"'\.znrta, =*!4hfl) 4tt
( r ) .r
\a /
HIGHLIGHTS CHAPTER o Magnetic force acting on achargedparticleq with a velocity u in a region containing a flux densityB is Frn : qurx B. o The total electromagnetic force, known at Lorentz force, acting on a moving charge i presence of both electric and magnetic F:q(E*uxB). o Magneticforcesactingon currentloopscan magnetictorques. o The magnetic field intensity induced by a c ement is defined by the Biot-Savart law. Gauss'slaw for magnetism statesthat the net netic flux flowing out of any closed surfaceis Ampdre's law statesthat the line integral of E a closedcontour is equal to the net current the surfaceboundedby the contour. o Vector magnetic potential A is related to B-VxA. o Materials are classified as diamagnetic, magnetic, or ferromagnetic, depending on
andbehaviorunderthei crystallinestructure of anexternalmagneticfield. o Diamagnetic and paramagnetic materials linear behavior between B and H, with lL both. o Ferromagnetic materials exhibit a hystereticbehavior betweenB and H and, for their p maybe aslarge as 10spe.
A| I J i I I
I il
At the boundary between two different media, the normal componentof B is continuous,and &e tangential componentsof H are related by Hzt- Hn: "Ir,whereJr is thesurfacecurrentdensityflowingin a directionorthogonalto H1-andH21. of a circuit is definedastheratioof maglnductance neticflux linking the circuit to the currentflowing rhroughit. \{agneticenergydensityis givenby u,. : IpHz.
dv to icknesr r and
@n @ @ @ q
BLEMS leq
5-1:Forces andTorques
a
( \ n as arge ir ic
[e net Ee ls
rl of E int c dto EIIC.
rgm he rn ilse tl-
I'
Anelectronwithaspeedof4x 106m/sisprojected !€ the positive x-direction into a medium containing a rrm magnetic flux density B - (i2 - 23) T. Given ! - 1.6 x lO-le C and the mass of an electron is = 9.1 x 10-31kg, determinethe initial acceleration :r of the electron (at the moment it is projected into redium).
@
@
'*osr.er(s)
availablein Appendix D.
availablein CD-ROM.
@
@
Figure 5-31:Particleof charge4 projectedwith velocityu into a mediumwith a uniform field B perpendicularto u (Problem5.2).
\\tren a particle with chargeq and massm is introi into a medium with a uniform field B such that the . r'elocity of the particle u is perpendicular to B, as n in Fig. 5-31, the magneticforce exertedon the par:ausesit to move in a circle of radiusa.By equating r rhe centripetal force on the particle, determinea in :f.q. m, u, andB. The circuit shown in Fig. 5-32 usestwo identical gs to support a lO-cm-long horizontal wire with a :l 5 g. In the absenceof a magneticfield, the weight i rrecausesthe springsto stretchadistanceof0.2 cm - \\hen a uniform magneticfield is turned on in the :r containingthe horizontal wire, the springsare ob:; to stretchanadditional0.5 cm. What is theintensitv :ragneticflux densityB?
@
4cl
tzv
SprinSsl
o o
o
o
Os
o
Figure 5-32: Configuration of Problem 5.3.
CHAPTER5 MAGNETOSTA 5.4 The rectangularloop shown in Fig. 5-33 consistsof 20 closely wrapped turns and is hinged along the e-axis. The plane ofthc loop makesanangleof 30' with the y-axis, and the current in the windings is 0.5 A. What is the magnitude of the torque exertedon the loop in the presenceof a uniform field B - j 1.2 T? When viewed from above, is the expecteddirection of rotation clockwise orcounterclockwise?
@ 5.5*
In a cylindrical coordinate system, a 2-m-long straight wire carrying a current of 5 A in the positive e-direction is located at r - 4 cm, Q _ r f2, and -1m < z < 1m. (a) If B - i0.2 cos@ (T), what is the magneticforce acting on the wire? (b) How much work is required to rotate the wire once about the e-axis in the negative {-direction (while maintainingr - 4cm)? (c) At what angle/ is the force a maximum? 5.6 A 20-turn rectangularcoil with sides / -
15 cm
1 = 0 . 5A
Figure 5-33: Hinged rectangularloop of Problem 5.4.
and ur : 5 cm is placedin the y-z planeas shown Fig.5-34. (a) ffthe coil, whichcarriesacurrentf '= 10A, is in presence of a magneticflux density
B:2 x t0-2(i+j,2)
(T)
determine the torque acting on the coil. (b) At what angle rpis the torque z,erc? (c) At what angle f is the torque maximum? De its value. Law Section5-2:Biot-Savart 5.7* An 8 cm x 12 cm rectangularloop of wire is uated in the x-y plane with the center of the loop at origin and its long sides parallel to the -r-axis. The has a current of 25 A flowing clockwise (when vi from above). Determine the magnetic field at the of the loop.
'TA
PROBLEMS
161
shown
\,rsm
Pzkz)
Figure 5-372Circular loop next to a linear current (Problem5.11).
P{z)
l:::='f P(r, Q,z) 'rre ls np at
The n v hec
Figure5-35: Current-carryinglinearconductorof Problem5.8.
;J Use the approachoutlined in Example 5-2 to de:lop an expressionfor the magnetic field H at an arbir,:ry point P due to the linear conductor defined by the s:ometry shown in Fig. 5-35. If the conductor extends retweenZr : 3 m and Z2 :7 m and carries a current - 5 A, find H at P (2,, ,0). 0 :"9' The loop shownin Fig. 5-36 consistsof radial lines ud segmentsof circles whose centers are at point P. letermine the magneticfield H at P in terms of a, b,0, md /.
;.6.
x
Figure 5-36:Configuration of Problem5.9.
5.10 An infinitely long, thin conducting sheetdefined over the space 0 carrying a current with a uniform surfacecurrent density J, : i5 (A/m). Obtain an expressionfor the magneric field at point P(0, 0, e) in Cartesiancoordinates. 5.11* An infinitely long wire carrying a 50-A currentin the positive r-direction is placed along the.r-axis in the vicinity of a lO-turn circular loop located in the x-y plane as shown in Fig. 5-37.If the magneticfield at the center of the loop is zero, what is the direction and magnitude of the current flowing in the loop? 5.12 TWoinfinitely long, parallel wires arecarrying 6-A currents in opposite directions. Determine the magnetic flux density at point P in Fig. 5-38. 5.13* A long, East-West-orientedpowercablecarrying an unknown current / is at a height of 8 m above the Earth's surface.If the magnetic flux density recordedby a magnetic-field meter placed at the surface is 12 peT when the current is flowing through the cable and20 p,T when the current is zero, what is the magnitudeof I?
CHAPTER5 MAG 5.14 TWoparallel,circularloopscarryinga 20 A eacharearrangedas shownin Fig. 5-39.Th€ loop is situatedin ttre .r-y planc with its ccnter et origin, andthe secondloop'scenteris at e : 2 m. If two loopshavethe sameradiusa -- 3 m, determine magnetic field at:
( a )z : o ( b )e : l m ( c )z : 2 m Figure 5-38: Arrangementfor hoblem 5.I2.
5-3:Forces Section between Currents 5.15* The long, straightconductor shown in Fig. lies in the plane of the rectangular loop at a di d : 0 . 1 m . T h e l o o p h a s d i m e n s i o n sa : 0 . 2 m b : 0.5 m, and the currents are It : 10 A 12* 15 A. Determine the net magnetic force acting the loop.
1 b=0.5m
1 x Figure 5-39: Parallel circular loops of Problem 5.14.
Figure 5-40: Current loop next to a conducting (hoblem 5.15).
r63
FROBLEMS curTent ). The
ntc,rd 2 m .I f :rmine
5-f6 In the arrangementshownin Fig. 541, eachof fu rwo long, parallelconductorscarriesa current/, is rydted by 8-cm-longstrings,andhasa rnss per unit hgth of 0.3 glcm.Dueto therepulsiveforceactingonthe croductors,the angle 0 betweenthe supportingstrings u l0o. Determinethe magnitudeof I and the relative dnctions of the currentsin the two conductors.
/r
T
6
h
1d
/
p
T
'
l--w-l
Figure 5-4224linearcurrentsourceabovea currentsheet (Problem5.17).
5.18 Threelong,parallelwiresarerurangedas shown in Fig. 543. Determinethe force per unit lengthacting on the wire carrying/3.
r Fig. adi
) . 2m t0A
I
I I
acting
A.
T I I l I l
/r = l0A
I I
2m
I I
Figure 5-41: Parallel Froblem5.16).
I
I
l-2m->o |
- - J
L = 10A
I I
2m
: i-' An infinitely long, thin conducting sheet of u'ddr u.' along the r-direction lies in the x-y plane and ries a current / in the -y-direction. Determine the r The magnetic field at a point P midway between the edges of the sheet and at a height h above it.
1
I I I I I
@ h= loA I I
Figure 5-43: Three parallel wires of Problem 5.18.
(Fie.s-42)
ngr
'l' The force per unit length exerted on an infinitely S long wire passing through point P and parallel to the sheetif the current through the wire is equal in magnitude but opposite in direction to that carried by the sheet.
5.1,9* A squareloop placedas shownin Fig. 5-44has 2-m sidesand carriesa currentIr : 5 A. If a straight, longconductorcarrying acurrent12- l0 A is introduced andplacedjust abovethemidpointsof two of theloop's sides,determinethe net force actingon the loop.
CHAPTER 5 MAG axis.Obtainan expressionfor the magneticfield E ( a )0 < r = 0 ( b )r > c 5.22 RepeatProblem5.21 for a currentdensity iJse-'. 5.23* In a certain conducting region, the magnetic is given in cylindrical coordinatesby
H - O!lr
Figure5-44:Longwire carryingcurrent/2, just abovea squueloopcarrying/1 (Problem5.19).
(t + zr)e-z'l
Find the currentdensityJ.
Polential 5-5:Magnetic Section
Lawlor Magnetism 5-4:Gauss's andAmpire'sLaw Section 5,20 Cunent / flows along the positive z-direction in the inner conductorof a long coaxial cable and returns throughthe outerconductor.The inner conductor has radiusa, andtheinnerandouterradii of the outer conductor areb andc, respectively. (a) Determinethe magneticfield in each of the followingregions:OS r 1 a, a 1 r < b, b < r < c, andr>c. (b) Plot the magnitudeof H as a function of r over the rangefromr: 0tor : l0cm,giventhat/ : 10A, a = 2 cm,b : 4 cm, and c : 5 cm. 5.21* A longcylindricalconductorwhoseaxis is coincidentwith the z-axis hasa radius a andcarries a current acurrentdensityJ - iJo/r, where "/eis characterizedby a constantandr is the radial distancefrom the cvlinder's
to Fig. 5-10: 5.24 With reference (a) Derive an expressionfor the vector magnetic tial A at a point P located at a distance r from wire in the x-y plane. (b) Derive B from A. Show that your result is i with the expressiongiven by Eq. (5.29), which derived by applying the Biot-Savart law.
s 5.25*
In a given region of space,the vector potential is given by A : i5 cos zy + i(2 * (Wb/m).
sin
(a) DetermineB. (b) UseEq. (5.66)to calculatethe magneticflux througha squareloop with 0.25-m-longedgesif loop is in the x-y plane, its center is at the ori and its edgesare parallel to the x- and y-axes. (c) CalculateQ again using Eq. (5.67).
qLtsLEMS
165
.A,uniform currentdensitygiven by
t-I
J : iJs
is 9.27 x l0-2a (A.m2),how many electronsper atom contributeto the saturatedfield?
(A/m2) Section 5-7:Magnetic Boundary Gonditions
riseto a vertor magneticpotential sg,*r,*
A: _try@, + yr) 4 ,
(Wb/m)
5.30 Ther-y planeseparates two magneticmediawith magneticpermeabilities and h &z,xsshownin Fig. 5-45. If thereis no surfacecurrentat the interfaceandthe magneticfield in mediumI is
{pply the vector Poisson'sequationto confirm the rbove statement.
n, {-se the expressionfor A to find H. --"e the expression for J in conjunction with .{m[ire's law to find H. Compare your result with tat obtained in part (b). A thin current element extending between : = -Llz and e - L/2 carries a current 1 along *2 ltln'..;ifra circular cross-sectionof radius a. r Fnd A at a point P located very far from the origin ilssumeR is so much larger than L that point P may be consideredto be at approximately the same dismncefrom every point along the current element). 'r Determinethe corresponding H.
Hr: frHu*iHry+LHk find: (a) Hz (b) 0r and 02 (c) EvaluateH2,01,and02forHb :3 (A/m), Hb : 0, Hk : 4 (Nm), &r : ps, andpz : 4po 5.31* Given that a current sheet with surface current density Js : *.4 (A/m) exists at y - 0, the interface between two magnetic media, and H1 - i8 (A/m) in medium I (y > 0), determineH2 in medium? (y < 0).
of Malerials luc:oo5-6:MagneticProperties In the model of the hydrogen atom proposed by 5 :r r:i io 1913, the electron moves around the nucleus ,rr- speedof 2 x 106 m/s in a circular orbit of radius 10-ll m. What is the magnitude of the magnetic m r€ot generatedby the electron'smotion?
x-y plane
' j"u'
Iron contains8.5 x 1028atoms/m3.At saturation, -rlrr-dignmentof the electrons' spin magnetic moments : :rn cancontribute 1.5T to the total magneticflux denB. If the spin magneticmoment of a single electron
Figure 5-45: Adjacent magneticmedia (Problem 5.30).
CHAPTER5 MAGNETOST
Figure 5-46: Magnetic media separatedby the plane x-y-l(Problem5.32).
Energy andMagnetic 5-8and5-9:Inductance Sections 5.32 In Fig. 5-46, the plane defined by t - y _ I separatesmedium I of permeabilitypl1 from medium 2 of permeabilitypz. If no surfacecurrent exists on the boundaryand
Br : iZ + 9,3
(T)
find Bz and thenevaluateyour result for p'1 : Sltz. Hint: Start by deriving the equation for the unit vector normal to the given plane. ,S-l
5.33 The plane boundary defined by e - 0 separates air from a block of iron. If 81 - *4 - 96 + e8 in air (e > 0), find 82 in iron (z . 0), given that trr.: 5000p0 for iron. 5.34 Show that if no surfacecurrent densitiesexist at theparallelinterfacesshownin Fig. 5-47,the relationship between0a,and91is independentof pz.
5.35* Obtain an expressionfor the self-induc unit length for the parallel wire transmission Fig.5-27 (a) in termsof a, d, andp, wherea is the of the wires, d is the axis-to-axis distance wires, and p is the permeability of the medium in thev reside. 5.36 A solenoidwith a length of 20 cm and a 5 cm consistsof 400 turns and carriesa culrent of lf z : 0 representsthe midpoint of the solenoid, a plot for lH(z) | as a function of z along the axis solenoid for the range -20 cm < z : 20 cm in steps. 5.37* In termsof the d-c current 1, how much energyis storedin the insulatingmedium of a2air-filled sectionof a coaxial transmissionline, given the radius of the inner conductor is 5 cm and the i radius of the outer conductoris 10 cm?
.tt Therectangularloop shownin Fig. 5-48is coplawith the long, straight wire carrying the current = 2OA. Determinethe rnagneticflux throughtheloop. -+5.41 Additional SolvedProblemson O. tumions
complete
Figure 5-48: Loop and wire zurangement for Problem 5.38.
LCtanoe lon
s the cn|-eer min
ta
nt of
it a\li
m m
tru r l-: LT
d L\r