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Chapter 7. Time-Varying Fields and Maxwell's Equations

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Chapter 7. Time-Varying Fields and Maxwell's Equations Chapter 7. Time-Varying Fields and Maxwell’s Equations Electrostatic & Time Varying Fields • Electrostatic fields ED 0, BH=J0, DE 1 HB • In the electrostatic model, electric field and magnetic fields are not related each other. Faraday’s law • A major advance in EM theory was made by M. Faraday in 1831 who discovered experimentally that a current was induced in a conducting loop when the magnetic flux linking the loop changed. d electromotive force (emf): V V dt Faraday’s law • Fundamental postulate for electromagnetic induction is dB B V V EE dl dl d sE dtCC S t t • The electric field intensity in a region of time-varying magnetic flux density is therefore non conservative and cannot be expressed as the negative gradient of a scalar potential. • The negative sign is an assertion that the induced emf will cause a current to flow in the closed loop in such a direction as to oppose the change in the linking magnetic flux Lentz’s law d B V E dt t 7-2.3. A moving conductor in a static magnetic field • Charge separation by magnetic force FuBm q • To an observer moving with the conductor, there is no apparent motion and the magnetic force can be interpreted as an inducted electric field acting along the conductor and producing a voltage. 2 VdluB 21 1 • Around a circuit, motional emf or flux cutting emf VdluB 21 C A moving conductor in a static magnetic field •Example 7-2 (a) Open voltage V0 ? (b) Electric power in R (c) Mechanical power required to move the sliding bar 1 ()a V V VuB dl a u a B a dl uB h 012C 2 xz 0 y 0 2 VuBh uB0 h ()bI 00 P IR2 W RRe R ()cP MMFuF , Mmechanical force to counteract the magnetic force Fm 1 FBaFFIdl IBh N mag2 x0 M mag 222 uB h uBh00 u Bh IP0 Fa W RRRMx M PPeM A moving conductor in a static magnetic field • Example 7-3. Faraday disk generator 4 VdlrBdruB = a a a 00C 3 zr 2 0 Bb Brdr 0 V 0 b 2 Magnetic force & electric force FEuB q When a charge q0 moves parallel to the current on a wire, the magnetic force on q0 is equivalent to the electric force on q0. At the rest frame on wire At the moving frame on charge ' FvBB q0 FEE q0 I 22 L L001/ vc /1/ vc E' I To observer moving with q0 under E and B fields, there is no apparent motion. But, the force on q0 can be interpreted as caused by an electric field, E’. 7-2.4. A moving circuit in a time-varying magnetic field To observer moving with q0 under E and B fields, there is no apparent motion. But, the force on q0 can be interpreted as caused by an electric field, E’. EEuB Now, consider a conducting circuit with contour C and surface S u moves with a velocity u under static E and B fields. Changing in magnetic flux due to the circuit movement produces an emf, V: d V dt B On the other hand, the moving circuit experiences an emf, V’, due to E’: E'dl V C E ' Is it true that VVBE ' VVBE ' ?? EEuB B From the Faraday's law of Esdl d , CS t by replacing EEEuB with = - , u B EsuBdl d dl V CSCt time variation motional emf at rest dd Note that Vdl E VdB Bs E ' C dt dt S d B Therefore, we need to prove that Bsdd s uB dl dtSS t C VVBE ' ?? • Time-rate of change of magnetic flux through the contour dd 1 Bsdttdtdlim B s21 B s dt dtSSt 0 t 21 S Bt BBtt t t H.O.T., t d Bt 1 u Bsdd slim Bs dd21 Bs H.O.T. dtSS tt 0 t S21 S • In going from C1 to C2, the circuit covers a region bounded by S1, S2, and S3 BBsBsBs dddd 0 VSSS213 213 emf = V E dl ddlts u Bs d Bs d t uB dl C 321SS C 21 d Bt Bsdd s uB dl dtSS t C • Therefore, the emf induced in the moving circuit C is equivalent to the emf induced by the change in magnetic flux d VV V : same form as not in motion. EB' dt A moving circuit in a time-varying magnetic field •Example 7.3 – Determine the open-circuit voltage of the Faraday disk generator 2 bt b Bsd B rdrd B t S 0000 2 d Bb2 V 0 B dt 2 Compare! B VdldEsuB dl E ' CSCt 4 VdlrBdruB = a a a Ezr'0C 3 2 0 Bb Brdr 0 0 b 2 VVB E ' B dd VdldE ' EsuB dlVd Bs CSCt B dt dt S (Example 7-4) Find the induced emf in the rotating loop under Ba()tBt yo sin (a) when the loop is at rest with an angle . BsdBthwBhwt asin a sin cos yo n 0 d VBStShw cos cos , : the area of the loop a dt 0 (b) When the loop rotates with an angular velocity ω about the x-axis 13ww VdlaaBtadxaaBtadx uB [( ) ( sin )] [( ) ( sin )] anyxnyxC 242200 w 2(BtthBSt sin sin ) sin sin ( = t) 2 00 22 VVVBSEaa'0 (cos t sin t ) BS 0 cos 2 t Compare! 1 tsinsinsBa ttSBSt cos BStcos t BStin 2 n 002 0 dd 1 VBStBStB 00sin 2 cos 2 VV dt dt 2 BE' 7-3. How Maxwell fixed Ampere’s law? • We now have the following collection of two curl eqns. and two divergence eqns. B EHJ , = t DB , =0 • Charge conservation law the equation of continuity J t • The set of four equations is now consistent with the equation of continuity? Taking the divergence of HJ= , HJ 0 H 0 from the vector null identit y J does not vanish in a time-varying situation and this equation is, in general, not true. Maxwell’s equations • A term ∂ρv /∂t must be added to the equation. D HJ 0 J D tt D HJ t E B J 000t • The additional term ∂D/∂t means that a time-varying electric field will give rise to a magnetic field, even in the absence of a free current flow (J=0). • ∂D/∂t is called displacement current (density). Maxwell’s equations • Maxwell’s equations Continuity equation B E t J D t HJ= t Lorentz’s force D B=0 F qE() B • These four equations, together with the equation of continuity and Lorentz’s force equation form the foundation of electromagnetic theory. These equations can be used to explain and predict all macroscopic electromagnetic phenomena. • The four Maxwell’s equations are not all independent – The two divergence equations can be derived from the two curl equations by making use of the equation of continuity Maxwell’s equations • Integral form & differential form of Maxwell’s equations Differential form Integral form Significance B Bd E Edl dS Faraday’s law t CStdt D D HJ = Hsdl I d Ampere’s law t CSt D DsdQ Gauss’s law C B=0 Bsd 0 No isolated S magnetic charge Maxwell’s equations •Example 7-5 – Verify that the displacement current = conduction current in the wire. iC : conduction current in the connecting wire d iCC CVcos t C 110dt A C 1 d E C : uniform between the plates d C Vt0 sin V DE 0 sin t d D V0 idD s cos tACVti10 cos C A td Maxwell’s equations •Example 7-5 – Determine the magnetic field intensity at a distance r from the wire Two typical open surfaces with rim C may be chosen: (a) a planar disk surface SS12,b a curved surface For the surface S1, D = 0 CV HJsdl2coscos rH d i C V t H10 t C 10 CS1 2 r Since the surface S2 passes through the dielectric medium, no conducting current flows through S2 displacement current D 2 rH idi s DCA t CV Ht10cos 2 r 7-5. Electromagnetic Boundary Conditions B Edl dS 0 when h 0 CSt EE12tt DD Hssdl J d; d 0 when h 0 CStt S aHHJHHJnsttsn212 12 DaDDdS d whenh 0 sV ns2 12 DD12nns B dS 0 s BB12nn0 1122H n H n Electromagnetic Boundary Conditions Both static and time-varying electromagnetic fields satisfy the same boundary conditions: EE12tt The tangential component of an E field is continuous across an interface. H12ttsnHJ The tangential component of an H field is discontinuous across an interface where a surface current exists. B12nn B The normal component of an B field is continuous across an interface. DD12nns The normal component of an D field is discontinuous across an interface where a surface charge exists. Boundary conditions at an interface between two lossless linear media Between two lossless media () with = 0, and S = 0, Js = 0 D1t 1 EE12tt D22t B B1t 1 DEH , HH12tt B22t D12nnDEE 1122 n n B12nnBHH 1122 n n Boundary conditions at an interface between dielectric and perfect conductor In a perfect conductor ( infinite, for example, supercondiuctors), EDHB22220, 0 Medium2 Medium 1(dielectric) (perfect metal) EE12tt 0 0 HJ12tts H 0 DD12nns 0 BB12nn 0 0 Boundary conditions • Table 7-4. Potential functions • Vector potential BA T B 0 • Electric field for the time-varying case. A EAE 0 tt A E V t A E V V/m t Due to charge distribution Due to time-varying current J Wave equation for vector potential A A From HB AD, E V , t D HJ t A AJ V tt 2 2 V A AA J 2 or tt 2A V 2 (Lorenz condition, or gauge) A 2 JA 0 tt 2A (# Show that the Lorentz condition is consistent 2A J with the equation of continuity.
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