Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Chapter 32 Maxwell’s equations; Magnetism in matterIn this chapter we will discuss the following topics:-Gauss’ law for magnetism -The missing term from Ampere’s law added by Maxwell -The magnetic field of the earth -Orbital and spin magnetic moment of the electron -Diamagnetic materials -Paramagnetic materials -Ferromagnetic materials(32 – 1)The magnetic flux through each of five faces of a die (singular of ''dice'') is given by ΦB = ±N Wb, where N (= 1 to 5) is the number of spots on the face. The flux is positive (outward) for N even and negative (inward) for N odd. What is the flux (in Wb) through the sixth face of the die? A.1 B.2 C.3 D.4 E.5Fig.aFig.bIn electrostatics we saw that positive and negative chargescan be separated. This is not the case with magnetic poles,as is shown in the figure. In fig.a we have a pGauss' Law for the magnetic fieldermanent barmagnet with well defined north and south poles. If we attempt to cut the magnet into pieces as is shown in fig.bwe do not get isolated north and south poles. Instead newpole faces appear on the newly cut faces of the pieces and the net result is that we end up with three smaller magnets,each of which is a i.e. it has a north and asouth pole. This result can be exprmagnetic dipoleessed as follows:The simplest magnetic structure that can exist is a magnetic dipole.Magnetic monopoles do not exists as far as we know. (32 – 2)iBrˆinifΔAi1 2 3The magnetic flux through a closed surfaceis determined as follows: First we dividethe surface into area element with areas, , ,...,nnA A A AD D D DBMagnetic Flux ΦFor each element we calculate the magnetic flux through it: cosˆHere is the angle between the normal and the magnetic field vectorsat the position of the i-th element. The indei i i ii i iB dAn BffDF =r1 1x runs from 1 to n We then form the sum cosFinally, we take the limit of the sum as The limit of the sum becomes the integral: cosn ni i i ii iBiB dAnBdA B dAff= =DF =� �F = = �� �� �SI magnetic flux unrr� �2 T m known as the "Weber" (Wb)�it :(32 – 3)BB dAF = ��rr�Gauss' law for the magnetic field can be expressed mathematically as follows: For any closed surface Contrast this with Gauss' law for costhe electric field: 0BencEoBdA B dAqE dAefF =F = �� ===� ��rrrr��� Gauss' law for the magnetic field expresses the fact that there is no such a thing as a" ". The flux of either the electric or the magnetic field through a surface is proportionalFmagnetic charge to thenet number of electric or magnetic field lines that either enter or exit the surface. Gauss' law for the magnetic fieldexpresses the fact that the magnetic field lines are closed. The number of magnetic field lines that enter any closedsurface is exactly equal to the number of lines that exit thesurface. Thus 0.BF =(32 – 4)0 BB dAF = � =�rr�Faraday's law states that: This law describeshow a changing magnetic field generates (induces) an electricfield. Ampere's law in its original form reads: BdE dSdtF� =-�Induced magnetic fieldsrrr�. Maxwell using an elegant symmetryargument guessed that a similar term exists in Ampere's law. The new term is written in red : This term, also known as "Eo oo enco encB dS iB dS iddtmemm� =� =F+��rrr��M "desrcibes how a changing electric field can generate a magnetic field. The electric field between the plates of thecapacitor in the figure changes with time . Thus the eletaxwell's law of inductionEctricflux through the red circle is also changing with and a non-vanishing magnetic field is predicted by Maxwell's lawof induction. Experimentaly it was verified thatthe predicted magnetic fieldtF exists. (32 – 5),Ampere's complete law has the form: We define the displacement currentUsing Ampere's law takes the form:In t he eEd oEo enc o odo enc o d encdB dS idtiB dS i ididtm mememFF� = +� ==+��The displacement currentrrrr��xample of the figure we can show that between the capacitor plates is equal to thecurrent that flows through the wires whichcharge the capacitor plates.diiEo enc o odB dS idtm meF� = +�rr�The electric flux through the capacitor plates . 1The displacement current Eo oEd o oo o oqAE Ad q qi idtse ee ee e eF = = =F= = = =(32 – 6),o enc o d encB dS i im m� = +�rr�,Consider the capacitor withcircular plates of radius In the space between the capacitorplates the term is equal to zeroThus Ampere's law becomes:We will use Ampere's law to determo d encB dSRiim� =�rr�ine the magnetic field. The calculation is identical to that of a magnetic field generated by a long wireof radius . This calculation was carried out in chapter 29 for a point P at a distance from the wire center. We wRr( )( )ill repeat the calculation for points outside as well as inside the capacitor plates. In this example is the distance of the point P from the capacitor center C. r Rr R r><(32 – 7)�rBdirRCrdSPWe choose an Amperian loop that reflects the cylindrical symmetry of the problem.The loop is a circle of radius that has its center at the capacitor platrMagnetic field outside the capacitor plates :,e center C. The magnetic field is tangent to the loop and has a constant magnitude . cos 0 22o do d enc o diBB ds Bds B ds rB i i Brp mmmp� = = = = = � =� � �rr� � �(32 – 8)�rBdirRrdSPCWe assume that the distribution of within the cross-section of the capacitor plate is uniform.We choose an Amperian loop is a circle of radius ( ) thatdirr R<Magnetic field inside the capacitor plates,2 2,2 2222 has its center at C. The magnetic field is tangent to the loop and has a constant magnitude . cos 0222do encd enc d do do dBB ds Bds B ds rB ir ri i iR RrrB i B rRiRp mpmppp m� �=� �� �� = = = == == �� � �rr� � �R2o diRmprBO(32 – 9)Below we summarize the four equations on which electromagnetic theoryis
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