Electromagnetism II, Final Formula Sheet

3y ago
15 Views
4 Downloads
306.01 KB
19 Pages
Last View : 14d ago
Last Download : 2m ago
Upload by : Fiona Harless
Transcription

MASSACHUSETTS INSTITUTE OF TECHNOLOGYPhysics DepartmentPhysics 8.07: Electromagnetism IIDecember 18, 2012Prof. Alan GuthFORMULA SHEET FOR FINAL EXAMExam Date: December 19, 2012 Some sections below are marked with asterisks, as this section is. The asterisksindicate that you won’t need this material for the quiz, and need not understand it. It isincluded, however, for completeness, and because some people might want to make useof it to solve problems by methods other than the intended ones.Index Notation: ·B Ai Bi ,A B i "ijk Aj Bk ,A"ijk "pqk δip δjq δiq δjpdet A "i1 i2 ···in A1,i1 A2,i2 · · · An,inRotation of a Vector:A i Rij Aj ,Orthogonality: Rij Rik δjkj 1Rotation about z-axis by φ: Rz (φ)ijRotation about axis n̂ by φ: j 2 i 1 cos φ sin φ i 2 sin φ cos φi 300(RT T I)j 3001 R(n̂, φ)ij δij cos φ n̂i n̂j (1 cos φ) "ijk nˆ k sin φ .Vector Calculus: i Gradient: ϕ)i i ϕ ,( Divergence: ·A i A i Curl: A) i "ijk j Ak( Laplacian: · ( ϕ) 2 ϕ xi 2ϕ xi xiFundamental Theorems of Vector Calculus: bGradient: ϕ · d , ϕ( b) ϕ( a) a Divergence:V · d aA ·A d3 x Swhere S is the boundary of V · d ,A A) · d a ( Curl:SPwhere P is the boundary of S

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 2Delta Functions: ϕ( r )δ 3 ( r r ) d3 x ϕ( r )ϕ(x)δ(x x ) dx ϕ(x ) , ddϕ ϕ(x) δ(x x ) dx dxdx x x δ(x xi ), g(xi ) 0δ(g(x)) g (xi ) i 1 r r · 4πδ 3 ( r r ) 2 3 r r r r x r̂j1δij 3r̂i r̂j4πjδij δ 3 ( r) i 3 i j i23rr3rr · 3(d · r̂)r̂ d 8π (d · )δ 3 ( r ) 3r3 δ 3 ( r ) 3(d · r̂)r̂ d 4π d 33rElectrostatics: , whereF qE 11( r r ) qi( r r ) E( r ) r ) d3 x 3 ρ( 4π"0 i r r 34π"0 r r "0 permittivity of free space 8.854 10 12 C2 /(N·m2 )1 8.988 109 N·m2 /C24π"0 r1ρ( r ) 3 E( r ) · d, d xV ( r ) V ( r 0 ) 4π"0 r r r0 ·E ρ , E 0, V E "0ρ(Poisson’s Eq.) ,ρ 0 2 V 0 (Laplace’s Eq.) 2 V "0Laplacian Mean Value Theorem (no generally accepted name): If 2 V 0, thenthe average value of V on a spherical surface equals its value at the center.Energy:1 1W 2 4π"01W 2 ijqi qj1 1 rij2 4π"0i j 1d xρ( r )V ( r ) "023 d3 x d3 x 2 3 E d xρ( r )ρ( r ) r r

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 3Conductors: σ n̂Just outside, E"0Pressure on surface:1 2 σ E outsideTwo-conductor system with charges Q and Q: Q CV , W 12 CV 2N isolated conductors:Vi Pij Qj ,Pij elastance matrix, or reciprocal capacitance matrixCij Vj ,Cij capacitance matrixjQi jaa2Image charge in sphere of radius a: Image of Q at R is q Q, r RRSeparation of Variables for Laplace’s Equation in Cartesian Coordinates: V cos αxsin αx cos βysin βy cosh γzsinh γz where γ 2 α2 β 2Separation of Variables for Laplace’s Equation in Spherical Coordinates:Traceless Symmetric Tensor expansion: 1 12 ϕr 2 2θ ϕ 0 , ϕ(r, θ, φ) 2r rr rwhere the angular part is given by 1 ϕ1 2ϕ2sin θ θ ϕ sin θ θ θsin2 θ φ22( )( ) 2θ Ci1 i2 .i nˆ i1 nˆ i2 . . . n̂i ,(, 1)Ci1 i2 .i n̂i1 n̂i2 . . . n̂i ,( )where Ci1 i2 .i is a symmetric traceless tensor andn̂ sin θ cos φ ê1 sin θ sin φ ê2 cos θ ê3 .General solution to Laplace’s equation: ( ) C( )i2 .i rˆi1 r̂i2 . . . r̂i ,V ( r ) Ci1 i2 .i r i1 1r 0where r rr̂

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 4Azimuthal Symmetry: B A r 1 { ẑi1 . . . ẑi } r̂i1 . . . r̂i V ( r ) r 0where { . . . } denotes the traceless symmetric part of . . . .Special cases:{1} 1{ ẑi } ẑi{ ẑi ẑj } ẑi ẑj 13 δij ẑi δjk ẑj δik ẑk δij { ẑi ẑj ẑk ẑm } zˆi ẑj ẑk ẑm 71 ẑi ẑj δkm ẑi ẑk δmj ẑi ẑm δjk ẑj zˆk δim 1δij δkm δik δjm δim δjk ẑj ẑm δik ẑk ẑm δij 35{ ẑi ẑj ẑk } ẑi ẑj ẑk 15Legendre Polynomial / Spherical Harmonic expansion:General solution to Laplace’s equation: B m V ( r ) A m r 1 Y m (θ, φ)r 0 m 2πOrthonormality:πdφ00sin θ dθ Y m (θ, φ) Y m (θ, φ) δ δm mAzimuthal Symmetry: B V ( r ) A r 1 P (cos θ)r 0Electric Multipole Expansion:First several terms: 1 Q p · rˆ 1 r̂i r̂jQ ···, whereV ( r ) 2 ijr2 r34π"0 r 33Q d x ρ( r ) , pi d x ρ( r ) xi Qij d3 x ρ( r )(3xi xj δij r 2 ) , dip ( r ) 1 E4π"0 E dip ( r ) 0 , p · r̂r2 1 3(p · r̂)r̂ p 1 pi δ 3 ( r )34π"0r3"0 ·E dip ( r ) 1 ρdip ( r ) 1 p · δ 3 ( r ) "0"0

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 5Traceless Symmetric Tensor version:V ( r ) 14π"0 1( )r 1 0Ci1 .i rˆi1 . . . r̂i ,where( )Ci1 .i (2, 1)!! ,!1 r r ρ( r ) { xi1 . . . xi } d3 x( r rr̂ xi eˆi )(2, 1)!! r { r̂i1 . . . r̂i } r̂i 1 . . . r̂i ,,!r 1 0(2, 1)!! (2, 1)(2, 3)(2, 5) . . . 1 for r r(2,)!, with ( 1)!! 1 .2 ,!Reminder: { . . . } denotes the traceless symmetric part of . . . .Griffiths version:1V ( r ) 4π"0 1 0r 1r ρ( r )P (cos θ ) d3 x where θ angle between r and r .1 r r r P (cos θ ) , 1 r 01P (x) 2 ,! ddx 1 1 2λx λ2λ P (x) 0 (x2 1) ,(Rodrigues’ formula) P (1) 1 1 P ( x) ( 1) P (x) 1dx P (x)P (x) 2δ 2, 1Spherical Harmonic version: 1V ( r ) 4π"0 0 m where q m 1 r r 4π q mY m (θ, φ)2, 1 r 1 Y mr ρ( r ) d3 x 0 m 4π r Y (θ , φ )Y m (θ, φ) ,2, 1 r 1 mfor r r

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 6Electric Fields in Matter:Electric Dipoles: p d3 x ρ( r ) r r δ 3 ( r r d ) , where r d position of dipoleρdip ( r ) p · (p · )E (p · E) F(force on a dipole) p E(torque on a dipole) U p · EElectrically Polarizable Materials: ( r ) polarization electric dipole moment per unit volumeP · n̂ρbound · P ,σbound P "0 E P ,D ·D ρfree , E 0 (for statics) Boundary conditions: Eabove Ebelow σ"0 Eabove Ebelow 0 Dabove Dbelow σfree Dabove Dbelow Pabove PbelowLinear Dielectrics: "0 χe E, Pχe electric susceptibility "E " "0 (1 χe ) permittivity,D" 1 χe relative permittivity, or dielectric constant"r "0N α/"0, where N number density of atoms1 Nα3 0 or (nonpolar) molecules, α atomic/molecular polarizability (P αE) 1 ·E d3 x(linear materials only)Energy: W D2 W (Even if one or more potential differences areForce on a dielectric: F held fixed, the force can be found by computing the gradient with the totalcharge on each conductor fixed.)Clausius-Mossotti equation: χe Magnetostatics:Magnetic Force: q (E v B) dp ,Fdtwhere p γm0 v ,1γ 1 v2c2

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 Fp. 7 Id, B d3 xJ BCurrent Density: J · d aCurrent through a surface S: IS SCharge conservation: ρ · J tMoving density of charge: J ρ vBiot-Savart Law: d, ( r r )K( r ) ( r r ) µ0µ0 B ( r ) I da4π r r 34π r r 3 µ0J( r ) ( r r ) 3 d x4π r r 3where µ0 permeability of free space 4π 10 7 N/A2Examples: µ0 I φ̂Infinitely long straight wire: B2πr µ0 nI0 ẑ , where n turns perInfinitely long tightly wound solenoid: Bunit length 0, z) Loop of current on axis: B(0,µ0 IR2ẑ2(z 2 R2 )3/2 r ) 1 µ0 K n̂ , n̂ unit normal toward rInfinite current sheet: B( 2Vector Potential: )coul µ0A(r4π J ( r ) 3 d x , r r A ,B ·A coul 0 ·B 0 (Subject to modification if magnetic monopoles are discovered) ( r ) A( r ) Λ( r ) for any Λ( r ). B A isGauge Transformations: Aunchanged.Ampère’s Law: · d , µ0 IencB B µ0 J , or equivalently P

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 8Magnetic Multipole Expansion:Traceless Symmetric Tensor version: { r̂i1 . . . r̂i }r 1 0 (2, 1)!!( )d3 xJj ( r ){ xi1 . . . xi }where Mj ;i1 i2 .i ,! Current conservation restriction:d3 x Sym(xi1 . . . xi 1 Ji ) 0Aj ( r ) µ04π( )Mj;i1 i2 .i i1 .i whereSymmeans to symmetrize — i.e. average over alli1 .i orderings — in the indices i1 . . . i Special cases: , 1: d3 x Ji 0d3 x (Ji xj Jj xi ) 0, 2: r̂ r ) µ0 mLeading term (dipole): A( ,4π r 2where1(1)mi "ijk Mj;k2 11 m I r d, d3 x r J I a ,22 P where a d a for any surface S spanning PS r̂µ0 3(m · r̂)r̂ m 2µ0 dip ( r ) µ0 m m δ 3 ( r )B 234πr34πr ·B dip ( r ) 0 , B dip ( r ) µ0 J dip ( r ) µ0 m δ 3 ( r ) Griffiths version: r ) µ0 IA( 4π 01r 1(r ) P (cos θ )d , Magnetic Fields in Matter:Magnetic Dipoles: 11 d3 x r J I am I r d, 2 P2

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 9 r δ 3 ( r r d ), where r d position of dipole J dip ( r ) m (m F · B)(force on a dipole) m B U m ·B(torque on a dipole)Magnetically Polarizable Materials: ( r ) magnetization magnetic dipole moment per unit volumeM M , bound M n̂KJ bound J free , , H ·B 0 1B M Hµ0Boundary conditions: HabBabove Bbelow 0ove Hbelow (Mabove Mbelow ) HBabove Bbelow µ0 (K n̂)above Hbelow Kfree n̂Linear Magnetic Materials: χm H, χm magnetic susceptibilityM µH Bµ µ0 (1 χm ) permeability,Magnetic Monopoles: r ) µ0 qm r̂ ; B(Force on a static monopole: F qm B4π r 2 µ0 qe qm r̂ , where r̂ pointsAngular momentum of monopole/charge system: L4πfrom qe to qmµ0 qe qm1 h̄ integerDirac quantization condition:4π2Connection Between Traceless Symmetric Tensors and Legendre Polynomialsor Spherical Harmonics:(2,)!{ ẑi1 . . . ẑi } n̂i1 . . . n̂i P (cos θ) 2 (,!)2For m 0,( ,m)Y m (θ, φ) Ci1 .i n̂i1 . . . n̂i ,( ,m)ˆ ˆim 1 . . . ẑi } ,where Ci1 i2 .i d m { û i1 . . . uim z 2m (2, 1)( 1)m (2,)!,with d m 2 ,!4π (, m)! (, m)!1and û (êx iêy )2 Form m 0, Y , m (θ, φ) ( 1)m Y m(θ, φ)

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 10More Information about Spherical Harmonics: 2 1 ( m)! mP (cos θ)eimφY m (θ, φ) 4π ( m)! where P m (cos θ) is the associated Legendre function, which can be defined byP m (x) m( 1)m2 m/2 d(1 x)(x2 1) m2 !dxLegendre Polynomials:SPHERICAL HARMONICS Ylm(θ , φ)l 01Y00 4π3sin θeiφ8πY11 l 13cos θ4πY10 Y22 l 21415sin2 θe2iφ2π15sin θ cosθeiφ8πY21 -Y20 5( 32 cos2θ4π1)235sin3 θe3iφ4πY33 -14Y32 14105sin2 θ cos θe2iφ2πY31 -1421sinθ (5cos2θ -1)eiφ4πl 3Y30 7( 5 cos3θ4π 232cos θ)Image by MIT OpenCourseWare.

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 11Maxwell’s Equations: E B ,(iii) t ·E 1ρ(i) "0 B µ0 J 1 E(iv) c2 t ·B 0(ii) 1c2 v B) Lorentz force law: F q(Ewhere µ0 "0 ρ · J tMaxwell’s Equations in Matter:Charge conservation: :Polarization P and magnetization M · P ,ρb M ,J b ρ ρf ρb ,J J f J bAuxiliary Fields: B M ,Hµ0 "0 E P DMaxwell’s Equations: ·D ρf(i) ·B 0(ii) E B ,(iii) t H J f D(iv) tFor linear media: "E ,D 1B Hµwhere " dielectric constant, µ relative permeability D displacement currentJ d tMaxwell’s Equations with Magnetic Charge: µ0 J m B , E(iii) t µ0 J e 1 E ·B µ0 ρm B(iv) (ii) c2 t 1 Magnetic Lorentz force law: F qm B 2 v Ec ·E 1 ρe(i) "0

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 12Current, Resistance, and Ohm’s Law: v B) , where σ conductivity. ρ 1/σ resistivityJ σ(EResistors: V IR ,P IV I 2 R V 2 /R,Resistance in a wire: R ρ , where , length, A cross-sectional area, and ρ Aresistivity V0 t/RCCharging an RC circuit: I ,Q CV0 1 e t/RCeREMF (Electromotive force): E v B) · d , , where v is either the velocity(Eof the wire or the velocity of the charge carriers (the difference points along thewire, and gives no contribution)Inductance:Universal flux rule: Whenever the flux through a loop changes, whether due to a or motion of the loop, E dΦB , where ΦB is the magnetic fluxchanging Bdtthrough the loopMutual inductance: Φ2 M21 I1 , M21 mutual inductance(Franz) Neumann’s formula: M21 M12 Self inductance: Φ LI ,E LdI;dtµ04πP1P2d ,1 · d ,2 r 1 r 2 L inductanceSelf inductance of a solenoid: L n2 µ0 V , where n number of turns per length,V volume V0 RtL1 eRising current in an RL circuit: I RBoundary Conditions:D1 D2 σf1E1 E2 σ"0 E 0E12 D P P D1212B1 B2 0H1 H2 M2 M1 H n̂ K fH12 µ0 n̂ K BB12

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 13Conservation Laws: 11 22 B Energy density: uEM "0 E 2µ0 1 E B Poynting vector (flow of energy): Sµ0Conservation of energy: d · d aIntegral form:[UEM Umech ] Sdt u , where u uEM umech ·SDifferential form: t1 1; 2 Si is the density of momentum in the i’thMomentum density: EM 2 Sccdirection 11122 Maxwell stress tensor: Tij "0 Ei Ej δij E Bi Bj δij B 2µ02where Tij Tji flow in j’th direction of momentum in the i’th directionConservation of momentum: d13Si d x Tij daj , for a volume VIntegral form:Pmech,i 2dtc VSbounded by a surface S Differential form:( mech,i EM,i ) j Tji tAngular momentum: B)] Angular momentum density (about the origin): ,EM r EM "0 [ r (EWave Equation in 1 Dimension:1 2f 2f 0 , where v is the wave velocityv 2 t2 z 2Sinusoidal waves:f (z, t) A cos [k(z vt) δ] A cos [kz ωt δ]whereω angular frequency 2πνν frequencyωv phase velocityδ phase (or phase constant)kk wave numberλ 2π/k wavelengthT 2π/ω periodA amplitudeEuler identity: eiθ cos θ i sin θ i(kz ωt) ] , where à Aeiδ ; “Re” is usuallyComplex notation: f (z, t) Re[Aedropped.ωdω group velocityWave velocities: v phase velocity; vgroup kdk

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 14Electromagnetic Waves: 1 2E 0,c2 t2Linearly Polarized Plane Waves: Wave Equations: 2 E 2 B 1 2B 0c2 t2 r ωt) ( r , t) E 0 ei(k· nˆ , where Ẽ0 is a complex amplitude, n̂ is a unit vector,Eand ω/ k vphase c.n̂ · k 0(transverse wave) 1 k̂ E BcEnergy and Momentum:u "0 E02 cos2 (kz ωt δ) , ( k k ẑ)averages to 1/2! " 1 1E B uc zˆ , "0 E02SI (intensity) S µ021 u EM 2 S ẑccElectromagnetic Waves in Matter:#µ"n index of refractionµ0 "0cv phase velocity n 1 2 1 B 2u " E2µ n k̂ E Bc 1E B uc ẑSµnReflection and Transmission at Normal Incidence:Boundary conditions:"1 E1 "2 E2 B1 B2 i(k1 z ωt)ETElV2V1BlIncident wave (z 0): I (z, t) Ẽ0,I eEX , EE121 1 B1 B .µ1µ2 2BTZêx I (z, t) 1 Ẽ0,I ei(k1 z ωt) êy .Bv1ERBRV1InterfaceYImage by MIT OpenCourseWare.

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 15Transmitted wave (z 0): T (z, t) Ẽ0,T ei(k2 z ωt) êxE T (z, t) 1 Ẽ0,T ei(k2 z ωt) êy .Bv2Reflected wave (z 0): R (z, t) Ẽ0,R ei( k1 z ωt) êxE R (z, t) 1 E 0,R ei( k1 z ωt) êy .Bv1ω must be the same on both sides, soωcωc v1 , v2 n1k2n2k1Applying boundary conditions and solving, approximating µ1 µ2 µ0 , n1 n2 2n1 0,IE0,R E0,IEE0,T n1 n2n1 n2Electromagnetic Potentials: A ,The fields: B V AE t Λ , AGauge transformations: A ·A 0Coulomb gauge: ·A 1 VLorentz gauge: c2 t22V 1ρ,"02V V 2 V 1ρ"0 Λ t is complicated)(but A µ0 J ,Awhere2 2 1 2c2 t2 D’AlembertianRetarded time solutions (Lorentz gauge): r , tr )r , tr )113 ρ( 3 J( V ( r , t) ,A( r,t) dxd x4π"0 r r 4π"0 r r wheretr t r r c

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 16Liénard-Wiechert Potentials (potentials of a point charge):V ( r , t) 1q 4π"0 r r p 1 q vp ( r , t) µ0A 4π r r p 1 vpc vpc·ˆ·ˆ vpV ( r , t)c2where r p and vp are the position and velocity of the particle at the retardedtime tr , and r r p , r r p ,ˆ r r p r r p Fields of a point charge (from the Liénard-Wiechert potentials): r , t) E( q r r p (c2 vp2 ) u ( r r p ) ( u ap )4π"0 ( u · ( r r p ))3 r , t) 1 ˆ E( r , t)B(cwhere u c ˆ vpRadiation:Radiation from an oscillating electric dipole along the z axis:p(t) p0 cos(ωt) ,p0 q0 dApproximations: d λ r, cos θp0 ωV (r, θ, t) sin[ω(t r/c)]4π"0 cr r , t) µ0 p0 ω sin[ω(t r/c)] ẑA( 4πr 2µpωsinθ00 r , t) 1 r̂ E( r , t) Ecos[ω (t r/c)] θˆ ,B( c4πr 2 1 µ0 p0 ω 2 sin θ Poynting vector: S (E B ) r̂cos[ω(t r/c)]µ0c4πr! " µ p2 ω 4 sin2 θ 2% 10 0 cos Intensity: I Srˆ,usingr2232π 2 c ! "2 4 · d a µ0 p0 ωTotal power: P S12πc

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 17Magnetic Dipole Radiation:Dipole moment: m (t) m0 cos(ωt) ẑ , at the origin µ0 m0 ω 2 sin θ r , t) 1 r̂ E( r , t)E cos[ω(t r/c)] φˆ ,B( 4πcrcm0,θ̂ φ̂Compared to the electric dipole radiation, p0 cGeneral Electric Dipole Radiation: r , t) 1 r̂ E( r , t) µ0 [r̂ p] ( r , t) µ0 [(ˆEr · p )ˆr p ] ,B( 4πrc4πrcMultipole Expansion for Radiation:The electric dipole radiation formula is really the first term in a doubly infiniteseries. There is electric dipole, quadrupole, . . . radiation, and also magneticdipole, quadrupole, . . . radiation.Radiation from a Point Particle:When the particle is at rest at the retarded time,q rad E[ ˆ ( ˆ ap )]24π"0 c r r 2 2 rad 1 E rad 2 ˆ µ0 q aPoynting vector: S16π 2 cµ0 cwhere θ is the angle between ap and ˆ .Total power (Larmor formula): P sin2 θ2 ˆµ0 q 2 a 26πc(valid for vp 0 or vp c)Liénard’s Generalization if vp 0: 2 µ0 q 2 γ 6 v a P a2 6πcc µ0 q 2 dpµ dpµ6πm20 c dτ dτFor relativists onlyRadiation Reaction:Abraham-Lorentz formula:2 rad µ0 q ȧF6πcThe Abraham-Lorentz formula is guaranteed to give the correct average energyloss for periodic or nearly periodic motion, but one would like a formulathat works under general circumstances. The Abraham-Lorentz formulaleads to runaway solutions which are clearly unphysical. The problem ofradiation reaction for point particles in classical electrodynamics apparentlyremains unsolved.

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012p. 18Vector Identities:Triple ProductsA . (B x C) B . (C x A) C . (A x B)A x (B x C) B(A . C) - C(A . B)Products Rules (f g) f ( g) g ( f) (A . B) A x ( x B) B x ( x A) (A . )B (B . )A (f A) f ( . A) A . ( f) (A x B) B . ( x A) - A . ( x B) x (f A) f ( x A) - A x ( f) (A x B) (B . )A - (A . )B A ( . B) - B( . A) Second Derivatives . ( x A) 0 x ( f) 0 x ( x A) ( . A) - 2A Image by MIT OpenCourseWare.

MIT OpenCourseWarehttp://ocw.mit.edu8.07 Electromagnetism II)DOO For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

FORMULA SHEET FOR FINAL EXAM Exam Date: December 19, 2012 Some sections below are marked with asterisks, as this section is. The asterisks ndneednotunderstandit. Itis included, however, forcompleteness, and because some people might want to makeuse

Related Documents:

ELTR 105 (DC 2), section 2 Recommended schedule Day 1 Topics: Magnetism, electromagnetism, and electromagnetic induction Questions: 1 through 20 Lab Exercises: Electromagnetism (question 71) Day 2 Topics: Applications of electromagnetism and induction, Lenz’s Law Questions: 21 through 40 Lab Exercise: Electromagnetic induction (question 72) Day 3 Topics: Introduction to Th evenin’s and .

0 Time Reversal for Electromagnetism: Applications in Electromagnetic Compatibility Ibrahim El Baba 1,2, Sébastien Lalléchère 1,2 and Pierre Bonnet 1,2 1 Clermont University, Blaise Pascal University, BP 10448,F-63000, Clermont-Ferrand 2 CNRS, UMR 6602, LASMEA, F-63177,Aubière France 1.Introduction ElectroMagnetic Compatibility (EMC) is the branch of electromagnetism that studies

A Guide to Electromagnetism Teaching Approach Learners should know about magnetism from grade 10. Electromagnetism describes the interaction between current and electric and magnetic fields. An electric current creates a magnetic fiel

Table 68: Shirt Laundry Formula 04: White (No Starch) Table 69: Shirt Laundry Formula 05: Colored (No Starch) Table 70: Shirt Laundry Formula 06: Delicates Table 71: Shirt Laundry Formula 07: Stain Treatment Table 72: Shirt Laundry Formula 08: Oxygen Bleach Table 73: Shirt Laundry Formula 09: Stain Soak Table 74: Shirt Laundry Formula 10 .

the empirical formula of a compound. Classic chemistry: finding the empirical formula The simplest type of formula – called the empirical formula – shows just the ratio of different atoms. For example, while the molecular formula for glucose is C 6 H 12 O 6, its empirical formula

A Note about Array formulas (not for Excel 365 / Excel 2021) Sometimes, you will need to enter a formula as array formula. In Excel 365/Excel 2021, all formulas are treated as Array formula, hence you need not enter any formula as Array formula. Only for older versions of Excel, you might need to enter a formula as Array formula.

The F1 FORMULA 1 logo, F1 logo, FORMULA 1, FORMULA ONE, F1, FIA FORMULA ONE WORLD CHAMPIONSHIP, GRAND PRIX and related marks are trade marks of Formula One Licensing BV, a . FORMULA 1 HEINEKEN DUTCH GRAND PRIX 2022 - Zandvoort Race History Chart. LAP 6 GAP TIME 1 1:16.350 16 1.051 1:16.213 . Race History Chart. LAP 11 GAP TIME 1 1:16.671 16 .

The F1 FORMULA 1 logo, F1 logo, FORMULA 1, FORMULA ONE, F1, FIA FORMULA ONE WORLD CHAMPIONSHIP, GRAND PRIX and related marks are trade marks of Formula One Licensing BV, a . FORMULA 1 HEINEKEN AUSTRALIAN GRAND PRIX 2022 - Melbourne Race History Chart. LAP 6 GAP TIME 16 2:24.953 . Race History Chart. LAP 11 GAP TIME 16 1:23.356 1 3.085 1:24 .