EE 6340 Intermediate EM Waves Fall 2016 Prof. David R. Jackson Dept. of EE Notes 1 1
Maxwell s Equations E D rt 2, V/m, rt, Wb/m T ( ) [ ] ( ) ( ) 2 rt, /m, H ( rt, ) [ A/m] B E = t (Faraday's Law) D H = J + t (Ampere's Law) D = ρ v B (Gauss's Law) B =0 (Magnetic Gauss's Law) 2
urrent Density J ρ v v 2 [A/m ] The current density vector points in the direction that positive charges are moving. J v J v Positive charges ρ ( ) v r ρ v Note: The charge density ρ v is the "free charge density" (the charge density that is free to move in the material usually electrons or ions). 3
urrent Density (cont.) The charge density ρ v is the "free charge density" (the charge density that is free to move in the material). J v ρ v A copper wire: ρ v is the charge density of the electrons in the conduction band that are free to move (the wire itself is neutral, however). altwater: ρ v is the charge density of the Na (+) or l (-) ions that are free to move. (There will be two currents densities, one from the Na ions and one from the l ions. The two current densities will be in the same direction.) A electron beam: ρ v is the charge density of the electrons in the beam (a negative number). 4
urrent Density (cont.) A small surface d is introduced perpendicular to the direction of current flow. di = current flowing through d dv J = J ˆv ++++++ ++++++ ++++++ dl ˆv d di = dq / dt J harge calculation: dq = = = = Hence ρv ( dv ) ρv ( ) ρv ( ) ( ρ ) v dl d v dt d v dt d = J dt d di = J d 5
urrent Density (cont.) A more general case of an arbitrary orientation of the surface is now considered. ( ˆ) ˆ ( ) n t J = J + J = J n n+ J tˆ tˆ ˆt di ( J ˆ) = n d urface d J t ˆn J n J Integrating over the surface, we have I ( ˆ) = J n d ( J ˆn ) = component of current density crossing d 6
ontinuity Equation From Gauss s Law, D = ρ v D ρ ( ) v D = = t t t (The two derivative operators commute.) From Ampere s Law, so D H = J + t D ( H ) = J + t 7
ontinuity Equation (cont.) D ( H ) = J + t D = J t From the last slide: ρ t v = D t Hence J ρv = t ontinuity Equation 8
Integral Form of ontinuity Equation V ρv J dv = t V dv V (closed) ˆn V Apply divergence theorem: J dv = J nˆ d Flux per unit volume Net flux out of 9
Integral Form of ontinuity Equation (cont.) Hence we have J ρv nˆ d = t V dv Assume V is stationary (not changing with time): V ρv d dq dv = ρvdv = t dt dt V encl Hence J nˆ d = dq dt encl charge conservation 10
Integral Form of ontinuity Equation (cont.) dq dt encl = nˆ d J + + (stationary) + Q encl + J + + ˆn harge that enters the region from the current flow must stay there, and this results in an increase of total charge inside the region. 11
Generalized ontinuity Equation Assume is moving (e.g., expanding) v s (r) Q encl J Here v s denotes the velocity of the surface. dq dt encl nˆd ρ v nˆd = J + v s Note : dq dt encl v ( ) ˆ = ρ v v n d s Difference in charge and surface velocities! 12
Generalized ontinuity Equation (cont.) Define the current density in the moving coordinate system: Define: J = ρ ( ) v v v s Q encl J v s (r) Then we have: J nˆ d = dq dt encl 13
Relaxation Equation tart from the continuity equation: ρ t v = J Assume a homogeneous conducting medium that obeys Ohm s law: J = σ E We then have ρv σ σ = σ E = D= ρv t ε ε Hence ρ t v σ = ρ v ε 14
Relaxation Equation (cont.) ρ t v σ = ρ v ε For an initial charge density ρ v (r,0) at t = 0 we have: ρ v ( ) ( ) ( / ) rt, = ρ r,0 e σ ε v t In a conducting medium the charge density dissipates very quickly and is zero in steady state! Example: σ = 4 /m [ ] ( sea water) 0 12 ( ) [ ] ε = 81ε = 81 8.854 10 F/m (use low-frequency permittivity) σ ε = [ ] 9 5.58 10 1/s 15
Time-Harmonic teady tate harge Density An interesting fact: In the time-harmonic (sinusoidal) steady state, there is never any charge density inside of a homogenous, source-free* region. Time - harmonic, homogeneous, source - free ρ v = 0 (You will prove this in HW 1.) * The term source-free means that there are no current sources that have been placed inside the body. Hence, the only current that exists inside the body is conduction current, which is given by Ohm s law. 16
Integral Forms of Maxwell s Eqs. D H = J + t tokes s Theorem: (open) Ampere's law ˆn Note: right-hand rule for. ( H ) ˆ n d = H dr irculation per unit area of irculation on boundary of 17
Integral Forms of Maxwell s Eqs. (cont.) Integrate Ampere s law over the open surface : n d n d n d t ( H ) ˆ = J ˆ + ˆ D Apply tokes s theorem: H D dr = J nˆd + nˆd t or H D dr = i ˆ s + n d t (The current i s is the current though the surface.) 18
Integral Forms of Maxwell s Eqs. (cont.) H D dr = i ˆ s + n d t Ampere s law Note: In statics, i s is independent of the shape of, since the last term is zero and the LH is independent of the shape of. imilarly: E B dr = t nˆ d Faraday s law Note: In statics, the voltage drop around a closed path is always zero (the voltage drop is unique). 19
Integral Forms of Maxwell s Eqs. (cont.) D = ρ v Electric Gauss law (closed) Divergence theorem: V ˆn V ˆ D dv = D n d Flux per unit volume Net flux out of Integrate the electric Gauss law throughout V and apply the divergence theorem: D nˆ d = Q D encl nˆ d = ρv dv V Hence we have 20
Integral Forms of Maxwell s Eqs. (cont.) B =0 Magnetic Gauss law Apply divergence theorem B nˆ d = 0 21
Boundary Form of Maxwell s Eqs. δ ( σ, ε, µ ) τˆ 1 1 1 + τ J s ˆn ( σ, ε, µ ) 2 2 2 Note: The surface current does not have to be perpendicular to the surface. We allow for a surface current density. The narrow rectangular path is chosen arbitrarily oriented in the tangential direction as shown. δ 0 The unit normal points towards region 1. H D dr = i ˆ s + n d t τˆ Evaluate the LH of Ampere s law on the path shown: lim H dr = lim dr + dr + dr H H H + δ δ 0 δ 0 so lim dr ˆ τ ( ) δ 0 H H 1 H 2 τ 22
Boundary Form of Maxwell s Eqs. (cont.) δ ( σ, ε, µ ) τˆ 1 1 1 + τ J s ˆn ( σ, ε, µ ) 2 2 2 Evaluate the RH of Ampere s law for path: i = lim J nˆ d s 0 0 s δ s τ ( nˆ ˆ) ˆ ( J ˆ ) s n J τ τ = τ τ nˆ = nˆ ˆ τ where we have used the following vector identity: A ( B ) = ( A B) = B ( A) Next, for the displacement current term in ampere's law, we have lim D nˆ d = 0 t δ 0 ( since 0) 23
Boundary Form of Maxwell s Eqs. (cont.) δ ( σ, ε, µ ) τˆ 1 1 1 + τ J s ˆn ( σ, ε, µ ) 2 2 2 Hence, we have ( H ) ( ) 1 H 2 τ Js n ˆ τ = ˆ ˆ ince τˆ is an arbitrary unit tangent vector, This may be written as: ( H ) ( ˆ ) 1 H 2 = Js n t ( ) ( ) 1 2 s nˆ H H = nˆ J nˆ = J t s Hence ( ) n ˆ H H = J 1 2 s 24
Boundary Form of Maxwell s Eqs. (cont.) δ ( σ, ε, µ ) τˆ 1 1 1 + τ J s ˆn ( σ, ε, µ ) 2 2 2 imilarly, from Faraday's law, we have ( E E ) n ˆ = 0 1 2 Note: The existence of an electric surface current does not affect this result. 25
Boundary Form of Maxwell s Eqs. (cont.) nˆ + + nˆ ( σ, ε, µ ) δ + + + 1 1 1 + ρ s ( σ, ε, µ ) 2 2 2 A surface charge density is now allowed. There could also be a surface current, but it does not affect the result. A pillbox surface is chosen. δ 0 The unit normal points towards region 1. Evaluate LH of Gauss s law over the surface shown: ˆ D n d = Q encl lim D nˆ d = lim nˆ d + ( nˆ) d + nˆ d D D D + δ δ 0 δ 0 so δ 0 ( ) lim D nˆ ˆ d n D 1 D 2 26
Boundary Form of Maxwell s Eqs. (cont.) + + nˆ δ + + + ( σ, ε, µ ) 1 1 1 + ρ s ( σ, ε, µ ) 2 2 2 Hence we have ( D D ) nˆ = ρ 1 2 s RH of Gauss s law: limq encl δ 0 δ 0 limq δ 0 encl Hence Gauss s law gives us ( D D ) 1 2 = lim ρs d ρ nˆ = ρ s s imilarly, from the magnetic Gauss law: ( B B ) nˆ = 0 1 2 27
Boundary Form of Maxwell s Eqs. (cont.) We also have a boundary form of the continuity equation. D = ρ v From B.. derivation with Gauss s law: ( D D ) nˆ = 1 2 ρ s ρ = t = ρs t v J nˆ ( J ) 1 J2 Note: If a surface current exists, this may give rise to another surface charge density term: Noting the similarity, we can write: s J s ρs = t (2-D continuity equation) In general: ( ) s nˆ ρ J 1 J + 2 = s Js t 28
ummary of Maxwell Equations Point Form Integral Form Boundary Form D D H = J + dr = i ˆ ˆ s + n d n ( 1 2) = t H H H J t B B E = dr = nˆd nˆ ( 1 2) = 0 t E E E t ( ) D = ρ D nˆd = Q nˆ D D = ρ v encl 1 2 s = 0 nˆ d = 0 nˆ B1 B2 = 0 B B ( ) ontinuity equation : ρ dq v ˆ encl ˆ ( ) ρ J = n d = n s 1 2 + s s = t J J J J dt t s 29
Faraday s Law E B dr = t nˆ d If is stationary: B d dψ nˆd = B nˆd = t dt dt ψ = magnetic flux through Hence: E dr = dψ dt (open) ˆn 30
Faraday s Law (cont.) If is moving (e.g., expanding): v s Previous form is still valid: E B dr = t nˆ d ( ) ( ) = t = t However, B! d dψ nˆd B nˆd = t dt dt 31
Faraday s Law (cont.) Vector identity for a moving path: d dt B B nˆd = nˆd ( v ) s B dr t v s Helmholtz identity (ee appendix for derivation.) 32
Faraday s Law (cont.) tart with E B dr = nˆ d t Then use Hence d dt B B nˆd = nˆd ( vs B) dr t dψ E dr = ( v ) s B dr dt Helmholtz identity or + v B dr = ( ) E s dψ dt 33
Faraday s Law (cont.) + v B dr = ( ) E s dψ dt Define EMF + v B dr ( ) E s EMF around a closed path Then EMF = dψ dt 34
Two Forms of Faraday s Law B = dr = The direct or voltage V E nˆ d Form of Faraday s law t EMF = + v B dr = ( ) E s dψ dt The alternative or EMF form of Faraday s law Note: The path means (t), a fixed path that is evaluated at a given time t. That is, = (t) is a snapshot of the moving path. The same comment for = (t). Practical note: The voltage form of Faraday s law is usually easier to work with if you want to calculate the voltage drop around a closed path. The EMF form of Faraday s law is usually easier to work with if you wish to calculate the EMF around a closed path. Either equation can be used to calculate either voltage drop or EMF, however. 35
Two Forms of Ampere s Law H D dr = i ˆ s + n d t Ampere s law d H dr = i ˆ ( ) s + D n d + vs D dr dt Alternative Ampere s law 36
Example y B t z ωt ( ) = ˆcos ( t) ρ = t ρ x Note: t is in [s], ρ is in [m]. 1) Find the voltage drop V(t) around the closed path. 2) Find the electric field on the path 3) Find the EMF drop around the closed path. Practical note: At a nonzero frequency the magnetic field must have some radial variation, but this is ignored here. 37
Example (cont.) (1) Voltage drop (from the voltage form of Faraday s law) y Unit normal: V B = E dr = nˆ d t = Bz d t Bz = d t Bz ( 2 = πρ ) t = ω sinωt πρ ( 2 ) B z is independent of (x,y). ρ B ( ) = ˆcos t z ωt ( t) ρ = t ˆn x V ( t) = E dr =ωπt 2 sin ωt [V] Would this result change if the path were not moving? No! 38
Example (cont.) (2) Electric field (from the voltage drop) 2 ( ) = t Eφ 2πρ ωπ sinωt y E dr ( 2 ) dr = E φ πρ ˆ( d ) = φ ρ φ E E φ φ ωπt 2 sinωt = 2π ρ 2 ωt = sinωt 2ρ ρ x E so E φ ωt = sinωt 2 ˆ ωt =φ sin ωt [V/m] 2 Note: There is no ρ or z component of E (seen from the curl of the magnetic field): H = zf ˆ ( ρ) cos( ωt) 39
Example (cont.) (3) EMF (from the EMF form of Faraday s law) + v B dr = ( ) E s ψ = B nˆ d = = B B z z d ( 2 πρ ) = cosωt ( 2 πt ) dψ dt B y ( t) = zˆ cos( ωt) ( t) ρ ρ = t Unit normal: ˆn x 40
Example (cont.) ( 2 t ) ψ = π cosωt y dψ = dt 2πtcosωt ωπt 2 Path moving sinωt ρ x Field changing EMF = dψ dt EMF = 2πtcosωt + ωπ t 2 sin ωt [V] 41
Example (cont.) Electric field and voltage drop (Alternative method: from the EMF form of Faraday s law) EMF = dψ dt dψ ( 2πρ )( E + vs B) = φ dt ( )( dψ 2πρ E v ) φ sρ Bz = dt y ρ B ( t) = zˆ cos( ωt) ( t) ρ = t x ince v sρ = 1 [m/s]: dψ 2πρ φ z = dt 1 dψ Eφ = Bz 2πρ dt ( )( E B ) so E φ ( ωt) = cos 1 dψ 2π t dt 42
Example (cont.) E φ 1 dψ = cos( ωt) 2π t dt 1 2 = cos( ωt) 2πtcos ωt πt ωsinωt 2π t ωt = cosωt cosωt sinωt 2 E φ = ωt 2 sin ωt [V/m] V = ωt 2 sin ωt [V] 2 ( πρ ) E = ˆ ωt φ sin ωt [V/m] 2 V = πω t 2 sin ωt [V] 43
Example (cont.) EMF (Alternative method: from the voltage form of Faraday s law) V ( t) = ωπt 2 sinωt (from Faraday s law) B y ( t) = zˆ cos( ωt) ( t) ρ = t EMF ( E v B) = + dr s ( ) ( B) = V t + v dr s ρ x 2π ( ) ( ) ( ) ( ( )) ˆ v 1 ˆ s B dr = ρ zˆ cos ωt φ( ρdφ) 0 2π = cos 0 ( ωt) ( ωt)( πρ ) = cos 2 ρdφ v sρ = 1 [m/s] 44
Example (cont.) V ( t) = E dr =ωπt 2 sin ωt [V] Recall: y B ( t) = zˆ cos( ωt) ( t) ρ = t Hence ( )( ) = ωπ ω ω πρ 2 EMF t sin t cos t 2 ρ x so ( )( ) 2 EMF = sin cos 2 ωπ t ωt ωt π t v sρ = 1 [m/s] 45
Example ummary Voltage drop V(t) around the closed path: V ( t) = ωπt 2 sin ωt [V] Electric field on the path: ˆ ωt E = φ sin ωt [V/m] 2 EMF drop around the closed path: 2 EMF = 2π tcosωt+ ωπ t sin ωt [V] Note: These results can be obtained using either the voltage or the EMF forms of Faraday s law. 46
Example Find the voltage readout on the voltmeter. + - V m y a x Perfectly conducting leads V 0 + - Voltmeter B Applied magnetic field: B ( t) = zˆ cos( ωt) Note: The voltmeter is assumed to have a very high internal resistance, so that negligible current flows in the circuit. (We can neglect any magnetic field coming from the current flowing in the loop.) 47
Example (cont.) V B = E dr = nˆ d t Bz = d t Bz = d t Bz = t ( 2 π a ) = ωsinωt π ( 2 a ) 2 0 sin ( ) Vm V =ωπ a ωt + V m nˆ = zˆ - Voltmeter B B y a x ( t) = zˆ cos( ωt) Bz = ω t sin ( ωt) V 0 + - 2 0 sin ( ) Vm = V +ωπ a ωt Practical note: In such a measurement, it is good to keep the leads close together (or even better, twist them.) 48
Generalized Ohm's Law A resistor is moving in a magnetic field. A R σ i + V - v B B EMF = Ri ( v B ) EMF = EMFAB E + s dr B A You will be proving this in the homework. B ( E + vs B ) dr = Ri V ( s B) A Note that there is no EMF change across a PE wire (R = 0). B + v dr = Ri A 49
Perfect Electric onductor (PE) A PE body is moving in the presence of a magnetic field. B v Inside the PE body: E = v B You will be proving this in the homework. 50
Example Find the voltage V m on the voltmeter. y liding bar (perfect conductor) Velocity v 0 V m + W Voltmeter V 0 - + - dw = v0 dt L Applied magnetic field: B ( t) = zˆ x We neglect the magnetic field coming from the current in the loop itself (we have a high-impedance voltmeter). 51
Example (cont.) EMF ( v B ) = + dr = E s dψ dt (EMF Faraday s law) ψ = B z ( LW ) y dψ = dt = Hence = dw Bz L dt B Lv z Lv 0 0 V m + Voltmeter - Velocity v 0 x V 0 + - x EMF = Lv 0 L B ( t) = zˆ 52
Example (cont.) EMF Hence ( v B ) = + dr E = V + 0 + ( V ) + 0 m s 0 PE wires y Note: The voltage drop across the sliding PE bar is not zero, but the EMF drop is. PE bar EMF = Ri ( R = 0) Vm V = Lv 0 0 Velocity v 0 so Vm = V Lv 0 0 B V m Voltmeter ( t) = zˆ + - L V 0 + - x 53
Example (cont.) Now let's solve the same problem using the voltage form of Faraday's law. E B dr = nˆ d t y Velocity v 0 = = 0 0 d V m + Voltmeter - V 0 + - x so L E V m 0 ( 0) top dr = + + V + dr = 0 B ( t) = zˆ 54
Example (cont.) For the top wire we have ( ) ( B) ( ) E ˆ ˆ ˆ x = x v = x yv zˆ = v 0 0 Hence 0 0 ( ) top L L ˆ x x E dr = E xdx = E dx = LE = Lv y 0 B ( t) = zˆ V m + + V0 + E dr = top 0 ( ) 0 Velocity v 0 Vm V + Lv = 0 0 0 V m + Voltmeter - x V 0 + - x L 55
Example (cont.) Vm V + Lv = 0 0 0 B ( t) = zˆ Hence, we have y Vm = V Lv 0 0 Velocity v 0 V m + Voltmeter - x V 0 + - x L 56
Time-Harmonic Representation Assume f (r,t) is sinusoidal: ( ω φ ) f rt, = Ar cos t+ r ( ) ( ) ( ) = Re { ( ) jφ ( r) jωt Are e } Define F( r) ( ) j Are φ (phasor form of f ) ( r) From Euler s identity: jz e = cos z+ jsin z jz ( e ) cos z = Re, if z real arg ( ) = Ar ( ) F( r) = φ ( r) Fr Then f( rt, ) = Re{ F( r) e jωt } 57
Time-Harmonic Representation (cont.) f( rt, ) = Ar ( ) cos( ωt+ φ( r) ) F r j Are φ ( ) ( ) ( r) f( rt, ) = Re{ F( r) e jωt } Notation: (, ) ( ) ( rt, ) F( r) f rt F r F for scalars for vectors 58
Time-Harmonic Representation (cont.) Derivative Property: f t { jωt = Re F( r) e } t Re ( jωt F( r) e ) = t = Re ( ) { jωt jω F r e } (sinusoidal) Hence: phasor for the derivative (, ) f rt t F ( rt, ) t ( ) jω F r ( ) jω F r for scalars for vectors 59
Time-Harmonic Representation (cont.) onsider a differential equation such as Faraday s Law: E rt, = ( ) B rt, ( ) t Assume sinusoidal fields. This can be written as { ( ) j ω t } ω ( ) jωt Re E r e = Re{ j B r e } or {( ) j t E jω B e ω } Re + = 0 60
Time-Harmonic Representation (cont.) Let c = c + jc = E + jω B ( ),, r i x y z Then { j t} Re ce ω = 0 Does this imply that c = 0? Phasor interpretation: hoose c = 0 ω t = 0 r j t ce ω Im hoose c = 0 ωt = π /2 Hence: c = 0 i ωt j t Re ce ω Re 61
Time-Harmonic Representation (cont.) Therefore so or ( E+ jω B),, = 0 x y z ( E+ jω B) = 0 E = jω B Hence B ( rt, ) E ( rt, ) E = jω B = t Time-harmonic (sinusoidal) steady state 62
Time-Harmonic Representation (cont.) We can also reverse the process: E = jω B {( ) j t E jω B e ω } Re + = 0 Re{ ( ) j ω t } Re{ ( ) jωt E r e jω B r e } = E rt, = ( ) B rt, ( ) t 63
Time-Harmonic Representation (cont.) Hence, in the sinusoidal steady-state, these two equations are equivalent: B E = t E = jω B 64
Maxwell s Equations in Time-Harmonic Form E = jω B Η = J + jω D D = ρv B = 0 65
ontinuity Equation Time-Harmonic Form J ρv = t J = jωρ v urface (2D) di d Wire (1D) s Js = jωρs = j l ωρ I is the current in the direction l. 66
Frequency-Domain url Equations (cont.) At a non-zero frequency, the frequency domain curl equations imply the divergence equations: tart with Faraday's law: E = jω B ( ) E = jω B Hence B = 0 67
Frequency-Domain url Equations (cont.) tart with Ampere's law: Η = J + jω D ( Η) J jω ( D) = jω( ρ D) = + v J = jωρ v (We also assume the continuity equation here.) Hence D = ρ v 68
Frequency-Domain url Equations (cont.) E = jω B Η = J + jω D ω 0 (with the continuity equation assumed) B = 0 D = ρ v Hence, we often consider only the curl equations in the frequency domain, and not the divergence equations. 69
Time Averaging of Periodic Quantities 1 T Define: f ( t) = ( ) T = ( ) = cos( ω + α) ( ) = cos( ω + β) f t A t g t B t T 0 f t dt [ ] period s Assume a product of sinusoidal waveforms: F = Ae G = Be jα jβ ( ) ( ) = cos( ω + α) cos( ω + β) f t g t AB t t 1 = AB cos + cos 2 + + 2 ( α β) ( ωt α β) { } 70
Time Average (cont.) 1 f t g t AB cos cos 2 t 2 ( ) ( ) = ( α β) + ( ω + α + β) { } The time-average of a constant is simply the constant. The time-average of a sinusoidal wave is zero. Hence: f ( t) g( t) = ABcos( α β) 1 2 71
Time Average (cont.) The phasors are denoted as: F = Ae jα G = Be jβ onsider the following: so Re FG * = ABe ( α β) ( * FG ) = A B cos( α β) j Hence f t g t 1 Re 2 ( ) ( ) ( * = FG ) The same formula extends to vectors as well. 72
Example: tored Energy Density 1 1 U = D E = DE + DE + DE 2 2 ( ) E x x y y z z D = Re j t xyz,, Dxyz,, e ω etc. D E = DE + DE + DE x x y y z z 1 1 1 = Re + Re + Re 2 2 2 1 Re ( * * * = DE ) x x + DE y y + DE z z 2 ( *) ( *) ( * DE ) x x DE y y DE z z or 1 D E = Re 2 ( * D E ) 73
Example: tored Energy Density (cont.) Hence, we have U E = 1 2 D E U E 1 1 = D E = Re 2 4 ( * D E ) imilarly, 1 1 UH = B H = Re 2 4 ( * B H ) 74
Example: tored Energy Density (cont.) U U E H 1 = Re 4 1 = Re 4 ( * D E ) ( * B H ) 75
Example: Power Flow = E H (instantaneous Poynting vector) 1 = E H = Re 2 ( * E H ) Define 1 ( * E H ) (complex Poynting vector [VA/m 2 ]) 2 Then = ( ) 2 Re [W/m ] This formula gives the time-average power flow. 76
Appendix: Proof of Moving urface (Helmholtz) Identity (t) ˆn (t+ t) Note: (t) is fist assumed to be planar, for simplicity. d dt 1 B nˆd = lim B( t + t) nˆd B( t) nˆd t ( + ) ( ) t 0 t t t ( ) ( ) ( ) ( ) ( ) B t + t nˆd = B t + t nˆd + B t + t nˆd t+ t t 77
Proof of Moving urface Identity (cont.) Hence: d dt 1 B nˆd = lim ( B( t + t) B ( t) ) nˆd + B( t + t) nˆd t ( ) t 0 t so d 1 nˆd nˆd lim ( t t) nˆd dt = B B + + t t B t 0 t ( ) For the last term: B t + t nˆd B t nˆd ( ) ( ) 78
Proof of Moving urface Identity (cont.) Examine term: B ( t) ˆ n d d dr vs t t ˆn t+dt Hence B ( ( )) ˆ d = dr v t n ( ) ( ) s ( B ) ( ) t nˆ d t nˆ dr v nˆ t s 79
Proof of Moving urface Identity (cont.) ( ) dr v s ˆn ince only has an component, we can write ( B( ) ˆ) ( ) ( ˆ) B( ) ( ) t n dr v n = t dr v s s Hence B t nˆ d B t dr v t ( ) ( ) ( ) s Therefore, summing all the d contributions: lim t 0 1 t B s t nˆ d = B dr v ( ) ( ) s 80
Proof of Moving urface Identity (cont.) Therefore d dt B B nˆd = nˆd B dr vs t ( ) t ( ) Vector identity: B dr v = v B dr ( ) ( ) s s Hence: d dt B B nˆd = nˆd ( v ) s B dr t ( ) t 81
Proof of Moving urface Identity (cont.) Notes: If the surface is non-planar, the result still holds, since the magnetic flux through any two surfaces that end on must be the same, since the divergence of the magnetic field is zero (from the magnetic Gauss law). Hence, the surface can always be chosen as planar for the purposes of calculating the magnetic flux through the surface. The identity can also be extended to the case where the contour is nonplanar, but the proof is omitted here. 82