Memory AidBoundary Conditions for Mag FieldTangential HTangential H (cont)Normal BBoundary Conditions for a ConductorHall EffectHall Effect: Vertical Internal FieldHall Effect: Density of Carriers (I)Hall Effect: Density of Carriers (II)Forces on Current LoopsTorques on Current Loops (I)Torques on Current Loops (II)Electric MotorsFaraday's Big DiscoveryFaraday's Law in Differential FormExample: TransformersGenerating Sparks!Vector PotentialMore on Vector PotentialIs the vector potential real?The Reality of the Vector PotentialSolenoid TransformerVector Potential Outside of SolenoidCircuit Application of TransformersVoltage TransformerPower TransmissionSummary So Far...Are These Equations Complete?Maxwell's Displacement CurrentMagnetic Field of a CapacitorDisplacement Current of a CapacitorEECS 117Lecture 17: Magnetic Forces/Torque, Faraday’s LawProf. NiknejadUniversity of California, BerkeleyUniversity of California, Berkeley EECS 117 Lecture 17 – p. 1/??Memory AidThe following table is a useful way to remember theequations in magnetics. We can draw a very goodanalogy between the fieldsaE H ρ JD B V Aǫ µ−1· ×P M × ·aI personally don’t like this choice since to me E and B are “real” and so theequations should be arranged to magnify this analogy. Unfortunately the equationsare not organized this way (partly due to choice of units) so we’ll stick with conven-tionUniversity of California, Berkeley EECS 117 Lecture 17 – p. 2/??Boundary Conditions for Mag FieldWe have now established the following equations for astatic magnetic field∇ × H = J∇ · B = 0ICH · dℓ =ZSJ · dS = IISB · dS = 0And for linear materials, we find that H = µ−1BUniversity of California, Berkeley EECS 117 Lecture 17 – p. 3/??Tangential HThe appropriate boundary conditions followimmediately from our previously established techniquesµ1µ2CTake a small loop intersecting with the boundary andtake the limit as the loop becomes tinyZCH · dℓ = (Ht1− Ht2)dℓ = 0University of California, Berkeley EECS 117 Lecture 17 – p. 4/??Tangential H (cont)So the tangential component of H is continuousHt1= Ht2µ−11Bt1= µ−12Bt2Note that B is discontinuous because there is aneffective surface current due to the change inpermeability. Since B is “real”, it reflects this changeIf, in addition, a surface current is flowing in betweenthe regions, then we need to include it in the abovecalculationUniversity of California, Berkeley EECS 117 Lecture 17 – p. 5/??Normal Bµ1µ2SˆnConsider a pillbox cylinder enclosing the boundarybetween the layersIn the limit that the pillbox becomes small, we haveIB · dS = (B1n− B2n)dS = 0And thus the normal component of B is continuousB1n= B2nUniversity of California, Berkeley EECS 117 Lecture 17 – p. 6/??Boundary Conditions for a ConductorIf a material is a very good conductor, then we’ll showthat it can only support current at the surface of theconductor.In fact, for an ideal conductor, the current lies entirelyon the surface and it’s a true surface currentIn such a case the current enclosed by even aninfinitesimal loop is finiteHCH · dℓ = (Ht1− Ht2)dℓ = Jsdℓ Ht1− Ht2= JsThis can be expressed compactly asˆn × (H1− H2) = JsBut for a perfect conductor, we’ll see that H2= 0, soH1t= Jsˆn × H1= JsUniversity of California, Berkeley EECS 117 Lecture 17 – p. 7/??Hall EffectJB0V0When current is traveling through a conductor, at anyinstant it experiences a force given by the LorentzequationF = qE + qv × BThe force qE leads to conduction along the length of thebar (due to momentum relaxation) with average speedvdbut the magnetic field causes a downward deflectionF = qˆxE0− qˆyvdB0University of California, Berkeley EECS 117 Lecture 17 – p. 8/??Hall Effect: Vertical Internal Field+ + + + + + ++ + + + + + ++In steady-state, the movement of charge down (orelectrons up) creates an internal electric field whichmust balance the downward pullThus we expect a “Hall” voltage to develop across thetop and bottom faces of the conducting barVH= Eyd = vdB0dUniversity of California, Berkeley EECS 117 Lecture 17 – p. 9/??Hall Effect: Density of Carriers (I)JB0VHJSince vd= µEx, and Jx= σEx, we can write vd= µJx/σVH=µJxσB0dRecall that the conductivity of a material is given byσ = qNµ, where q is the unit chargeSince vd= µEx, and Jx= σEx, we can write vd= µJx/σVH=µJxσB0dUniversity of California, Berkeley EECS 117 Lecture 17 – p. 10/??Hall Effect: Density of Carriers (II)Recall that the conductivity of a material is given byσ = qNµ, where q is the unit charge, N is the density ofmobile charge carriers, and µ is the mobility of thecarriersVH=JxB0dqNN =JxB0dqVH=IB0dAqVHNotice that all the quantities on the RHS are eitherknown or easily measured. Thus the density of carrierscan be measured indirectly through measuring the HallVoltageUniversity of California, Berkeley EECS 117 Lecture 17 – p. 11/??Forces on Current LoopsFF1F2BBˆxˆyˆzabI1I2FSince the field is notuniform, the net forceis not zero. Note theforce on the ⊥ sidescancel outF1= −ˆyµI1I22πadF2= +ˆyµI1I22π(a + b)dF = F1+ F2= −ˆyµI1I2d2π1a−1bUniversity of California, Berkeley EECS 117 Lecture 17 – p. 12/??Torques on Current Loops (I)B0B0free rotation about this axisforce downforce upIn a uniform field, thenet force on the cur-rent loop is zero. Butthe net torque is notzero. Thus the loopwill tend to rotate.T = r × FF1= −I1B0dˆxF2= +I1B0dˆxF1+ F2= 0University of California, Berkeley EECS 117 Lecture 17 – p. 13/??Torques on Current Loops (II)F1F2IIB0θθˆxˆyˆzT1= −(b/2)I1B0d sin θˆzT2= −(b/2)I1B0d sin θˆzT = T1+ T2= −ˆzB0momentz }| {I × b × d|{z}Area of loopsin θIn general the torque can be expressed asT = m × Bwhere the moment is defined as m = I × AreaUniversity of California, Berkeley EECS 117 Lecture 17 – p. 14/??Electric MotorsIB0A DC electric mo-tor operates on thisprinciple. A uniformstrong magnetic fieldcuts across a currentloop causing it to ro-tate.When the loop is || to the field, the torque drops to zerobut the rotational inertia of the loop keeps it rotating.Simultaneously, the direction of the current is reversedas the loop flips around and cuts into the field. Thisgenerates a new torque that favors the continuousrotation.University of California, Berkeley EECS 117 Lecture 17 – p. 15/??Faraday’s Big DiscoveryIn electrostatics we learned thatHE · dℓ = 0Let’s use the analogy between B and D (and