Magnetic FluxMagnetic Flux and Vector PotentialFlux LinkageThe Geometry of Flux CalculationsGeometry of Flux (cont)Mutual and Self InductanceSystem of Mutual Inductance EquationsSolenoid Magnetic FieldSolenoid InductanceCoaxial ConductorCoaxial Transmission Line (cont)Magnetization VectorVector PotentialAnother Divergence TheoremAnother Divergence Thm (cont)Vector Potential due to MagnetizationVolume and Surface CurrentsRelative PermeabilityAmpere's Equation for MediaMagnetic MaterialsEECS 117Lecture 16: Magnetic Flux and MagnetizationProf. NiknejadUniversity of California, BerkeleyUniversity of California, Berkeley EECS 117 Lecture 16 – p. 1/21Magnetic FluxMagnetic flux plays an importantrole in many EM problems (inanalogy with electric charge)Ψ =ZSB · dSDue to the absence of magneticchargeΨ =ISB · dS ≡ 0but net flux can certainly crossan open surface.CSUniversity of California, Berkeley EECS 117 Lecture 16 – p. 2/21Magnetic Flux and Vector PotentialS1S2CCMagnetic flux is independent of the surface but onlydepends on the curve bounding the surface. This iseasy to show sinceΨ =ZSB · dS =ZS∇ × A · dS =ICA · dℓΨ =ZS1B · dS =ZS2B · dSUniversity of California, Berkeley EECS 117 Lecture 16 – p. 3/21Flux LinkageI1S2B fieldConsider the flux crossing surface S2due to a currentflowing in loop I1Ψ21=ZS2B1· dSLikewise, the “self”-flux of a loop is defined by the fluxcrossing the surface of a path when a current is flowingin the pathΨ11=ZS1B1· dSUniversity of California, Berkeley EECS 117 Lecture 16 – p. 4/21The Geometry of Flux CalculationsThe flux is linearly proportional to the current andotherwise only a function of the geometry of the pathTo see this, let’s calculate Ψ21for filamental loopsΨ21=IC2A1· dℓ2butA1=14πµ−10IC1I1dℓ1R − R1substituting, we have a double integralΨ21=I14πµ−10IC2IC1dℓ1R − R1· dℓ2University of California, Berkeley EECS 117 Lecture 16 – p. 5/21Geometry of Flux (cont)dℓ1dℓ2R2- R1R2R1We thus have a simple formula that only involves themagnitude of the current and the average distancebetween every two points on the loopsΨ21=I14πµ−10IC2IC1dℓ1· dℓ2R2− R1University of California, Berkeley EECS 117 Lecture 16 – p. 6/21Mutual and Self InductanceSince the flux is proportional to the current by ageometric factor, we may writeΨ21= M21I1We call the factor M21the mutual inductanceM21=Ψ21I1=14πµ−10IC2IC1dℓ1· dℓ2R2− R1The units of M are H since [µ] = H /m.It’s clear that mutual inductance is reciprocal,M21= M12The “self-flux” mutual inductance is simply called theself-inductance and donated by L1= M11University of California, Berkeley EECS 117 Lecture 16 – p. 7/21System of Mutual Inductance EquationsIf we generalize to a system of current loops we have asystem of equationsΨ1= L1I1+ M12I2+ . . . M1NIN...ΨN= MN1I1+ MN2I2+ . . . LNINOr in matrix form ψ = Mi, where M is the inductancematrix.This equation resembles q = Cv, where C is thecapacitance matrix.University of California, Berkeley EECS 117 Lecture 16 – p. 8/21Solenoid Magnetic FieldWe have seen that a tightlywound long long solenoidhas B = 0 outside andBx= 0 inside, so that byAmpère’s lawByℓ = NIµ0where N is the number ofcurrent loops crossing thesurface of the path.The vertical magnetic fieldis therefore constantBy=NIµ0ℓ= µ0InUniversity of California, Berkeley EECS 117 Lecture 16 – p. 9/21Solenoid InductanceThe flux per turn is therefore simply given byΨturn= πa2ByThe total flux through N turns is thusΨ = NΨturn= Nπa2ByΨ = µ0N2πa2ℓIThe solenoid inductance is thusL =ΨI=µ0N2πa2ℓUniversity of California, Berkeley EECS 117 Lecture 16 – p. 10/21Coaxial ConductorSIn transmission line problems, weneed to compute inductance/unitlength. Consider the shaded areafrom r = a to r = bThe magnetic field in the regionbetween conductors if easilycomputedIB · dℓ = Bφ2πr = µ0IThe external flux (excluding the volume of the idealconductors) is given byψ′=ZbaBφdr =µ0I2πZbadrr=µ0I2πlnbaUniversity of California, Berkeley EECS 117 Lecture 16 – p. 11/21Coaxial Transmission Line (cont)The inductance per unit length is thereforeL′=µ02πlnba[H/m]Recall that the product of inductance and capacitanceper unit length is a constantL′C′=1c2where c is the speed of light in the medium. Thus wecan also calculate the capacitance per unit lengthwithout any extra work.University of California, Berkeley EECS 117 Lecture 16 – p. 12/21Magnetization VectorWe’d like to study magnetic fields in magnetic materials.Let’s define the magnetization vector asM = lim∆V →0Pkmk∆Vwhere mkis the magnetic dipole of an atom or moleculeThe vector potential due to these magnetic dipoles isgiven by in a differential volume dv′is given bydA = µ0M × ˆr4πR2dv′soA =µ04πZV′M × ˆrR2dv′University of California, Berkeley EECS 117 Lecture 16 – p. 13/21Vector PotentialUsing∇′1R=ˆrR2A =µ04πZV′M × ∇′1Rdv′Consider the vector identity∇′×MR=1R∇′× M + ∇′1R× MWe can thus break the vector potential into two termsA =µ04πZV′∇′× MRdv′−µ04πZV′∇′×MRdv′University of California, Berkeley EECS 117 Lecture 16 – p. 14/21Another Divergence TheoremConsider the vector u = a × v, where a is an arbitraryconstant. Then∇ · u = ∇·(a × v) = (∇ × a)·v−(∇ × v)·a = − (∇ × v)·aNow apply the Divergence Theorem to ∇ · uZV− (∇ × v) · adV =IS((a × v) · u) · dSRe-ordering the vector triple product−IS(a · v × n) · dSUniversity of California, Berkeley EECS 117 Lecture 16 – p. 15/21Another Divergence Thm (cont)Since the vector a is constant, we can pull it out of theintegralsa ·ZV(−∇ × v) dV = a ·ISr × n · dSThe vector a is arbitrary, so we haveZV(∇ × v) dV = −ISr × n · dSApplying this to the second term of the vector potentialZV′∇′×MRdv′= −ISM × ˆnR· dSUniversity of California, Berkeley EECS 117 Lecture 16 – p. 16/21Vector Potential due to MagnetizationThe vector potential due to magnetization has a volumecomponent and a surface componentA =µ04πZV′∇′× MRdv′+µ04πISM × ˆnR· dSdv′We can thus define an equivalent magnetic volumecurrent densityJm= ∇ × Mand an equivalent magnetic surface current densityJs= M × ˆnUniversity of California, Berkeley EECS 117 Lecture 16 – p. 17/21Volume and Surface CurrentsˆnJsJsJsIn the figure above, we can see that