MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II November 21, 2012 Prof. Alan Guth QUIZ 2 SOLUTIONS QUIZ DATE: NOVEMBER 15, 2012
PROBLEM 1: THE MAGNETIC FIELD OF A SPINNING, UNIFORMLY CHARGED SPHERE (25 points) This problem is based on Problem 1 of Problem Set 8.
A uniformly charged solid sphere of radius R carries a total charge Q,andisset spinning with angular velocity ω about the z axis. (a) (10 points) What is the magnetic dipole moment of the sphere? (b) (5 points) Using the dipole approximation, what is the vector potential A(r)atlarge distances? (Remember that A is a vector, so it is not enough to merely specify its magnitude.) (c) (10 points) Find the exact vector potential INSIDE the sphere. You may, if you wish, make use of the result of Example 5.11 from Griffiths’ book. There he considered a spherical shell, of radius R, carrying a uniform surface charge σ, spinning at angular velocity ω directed along the z axis. He found the vector potential µ Rωσ 0 r sin θ φ,ˆ (if r ≤ R) 3 A(r, θ, φ)= (1.1) µ R4ωσ sin θ 0 φ,ˆ (if r ≥ R). 3 r2
PROBLEM 1 SOLUTION:
(a) A uniformly charged solid sphere of radius R carries a total charge Q, hence it has 4 3 charge density ρ = Q/( 3 πR ). To find the magnetic moment of sphere we can divide the sphere into infinitesimal charges. Using spherical polar coordinates, we can take dq = ρ dτ = ρr2 dr sin θ dθ dφ, with the contribution to the dipole moment given by 1 × dm = 2r J dτ. One method would be to write down the volume integral directly, using J = ρv = ρω × r. We can, however, integrate over φ before we start, so we are breaking the sphere into rings, where a given ring is indicated by its coordinates r and θ, and its size dr and dθ. The volume of each ring is dτ =2πr2 dr sin θ dθ.The current dI in the ring is given by dq/T ,whereT =2π/ω is the period, so
dq ωρdτ dI = = = ωρr2 dr sin θ dθ. (1.2) T 2π 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 2
The magnetic dipole moment of each ring is then given by 1 1 2 2 dm ring = r × J dτ = dI r × d =dI(πr sin θ)ˆz. (1.3) 2 ring 2 ring
The total magnetic dipole moment is then m = ωρr2 sin θ (πr2 sin2 θ)dr dθ zˆ
R π = πωρ r4 dr (1 − cos2 θ)sinθ dθ zˆ 0 0
QR5 4 1 = πω = QωR2 z.ˆ (1.4) 4 3 3 πR 5 3 5
(b) The vector potential in dipole approximation is,
µ m × r µ |m | sin θ µ QωR2 sin θ A =0 = 0 φˆ = 0 φ.ˆ (1.5) 4π r3 4π r2 4π 5 r2
(c) To calculate the exact vector potential inside the sphere, we split the sphere into shells. Let r be the integration variable and the radius of a shell, moreover let dr denote the thickness of the shell. Then we can use the results of Example 5.11 (pp. 236-37) in Griffiths, if we replace σ by its value for this case. The value of σ is found equating charges Q σ(4πr 2)= (4πr 2)dr (1.6) 4 3 3 πR and therefore we must replace
Q σ → dr . 4 3 3 πR Making this replacement in Griffiths’ Eq. (5.67), quoted above as Eq. (1.1), we now have r r if r
some radius r Doing the integrals one finds 3 2 µ0 Qω 3r rR Aφ(r, θ, φ)= sin θ − + . (1.9) 4π R3 10 2 PROBLEM 2: SPHERE WITH VARIABLE DIELECTRIC CONSTANT (35 points) A dielectric sphere of radius R has variable permittivity, so the permittivity throughout space is described by (R/r)2 if r dln ∂V ∇2V + =0. (2.2) dr ∂r (b) (4 points) Explain why the solution can be written as ∞ {} V (r, θ)= V(r) zˆi1 ...zˆi rˆi1 ...rˆi , (2.3a) =0 or equivalently (your choice) ∞ V (r, θ)= V(r)P(cos θ) , (2.3b) =0 where { ...} denotes the traceless symmetric part of ... ,andP(cos θ)istheLeg- endre polynomial. (Your answer here should depend only on general mathematical principles, and should not rely on the explicit solution that you will find in parts (c) and (d).) 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 4 (c) (9 points) Derive the ordinary differential equation obeyed by V(r)(separatelyfor r d dP(cos θ) sin θ = −( +1)sinθP(cos θ) . (2.4) dθ dθ The relevant formulas for the traceless symmetric tensor formalism are in the formula sheets. (d) (9 points) Using appropriate boundary conditions on V (r, θ)atr =0,r = R,and r →∞, determine V (r, θ)forr PROBLEM 2 SOLUTION: (a) Since we don’t have free charges anywhere, ∇·D = ∇· ( E ), = E · (∇ )+ ∇· E =0. (2.5) d The permittivity only depends on r,sowecanwrite∇ = eˆr. Then putting this dr result into Eq. (2.5) with E = −∇ V , we find d 2 0=(∇ V ) · eˆr + ∇ V dr ∂V d 1 = + ∇2V ∂r dr ∂V dln =⇒ 0= + ∇2V. (2.6) ∂r dr (b) With an external field along the z-axis, the problem has azimuthal symmetry, imply- ing ∂V/∂φ =0,soV = V (r, θ). The Legendre polynomials P(cos θ)areacomplete set of functions of the polar angle θ for 0 ≤ θ ≤ π, implying that at each value of r, V (r, θ) can be expanded in a Legendre series. In general, the coefficients may be functions of r,sowecanwrite ∞ V (r, θ)= V(r)P(cos θ) . (2.7) =0 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 5 { } The same argument holds for an expansion in zˆi1 ...zˆi rˆi1 ...rˆi , since these are in fact the same functions, up to a multiplicative constant. Note that if depended on θ as well as r, then the completeness argument would still be valid, and it would still be possible to write V (r, θ) as in Eqs. (2.3). In that case, however, the equations for the functions V(r) would become coupled to each other, making them much more difficult to solve. dln 2 (c) For r 2 1 ∂ 2 ∂V dV 2 ( +1) d V ( +1) r + − − V = − V =0. (2.9) r2 ∂r ∂r dr r r2 dr2 r2 The general solution to Eq. (2.9) is +1 B V(r)=A r + . (2.10) r (This can be verified by inspection, but it can also be found by assuming a trial p function in the form of a power, V ∝ r . Inserting the trial function into the differential equation, one finds p(p − 1) = ( +1) . One might see by inspection that this is solved by p = +1orp = −, or one can solve it as a quadratic equation, finding 1 ± (2 +1) p = = +1or − .) 2 For r>R, 1 ∂ 2 ∂V ( +1) r − V =0. (2.11) r2 ∂r ∂r r2 The general solution to Eq. (2.11) is, D V(r)= Cr + . (2.12) r+1 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 6 (d) The coefficients B are zero, B = 0, to avoid a singularity at r =0. Thepotential goes as V (r)=−E0r cos θ for r R;thisgivesC = 0except for C1 = −E0.The potential V (r, θ) is continuous at r = R, implying that +1 D A R = +1 for =1 R (2.13) D1 A R2 = −E R + for =1. 1 0 R2 In addition, the normal component of the displacement vector is continuous on the boundary of the sphere. Since is continuous at r = R, this means that Er = −∂V/∂r is continuous, which one could also have deduced from Eq. (2.2), since any discontinuity in ∂V/∂r would produce a δ-function in ∂2V/∂r2. Setting ∂V/∂r at r = R− equal to its value at r = R+, we find D − ( +1)AR = ( +1) +2 for =1 R (2.14) D1 2A R = −2 − E for =1 . 1 R3 0 Solving Eq. (2.13) and Eq. (2.14) as two equations (for each ) for the two unknowns A and D,weseethatA = D =0for =1,andthat 3E E R3 A = − 0 ,C= −E, and D = 0 . (2.15) 1 4R 1 0 1 4 Then we find the potential as 3E r2 − 0 cos θ for r (e) Eq. (2.16) tells us that for r>R, the potential is equal to that of the applied external field, Vext = −E0r cos θ, plus a term that we attribute to the sphere: E R3 V (r, θ)= 0 cos θ. (2.17) sphere 4r2 This has exactly the form of an electric dipole, 1 p· rˆ Vdip = 2 , (2.18) 4π 0 r if we identify 3 p = π 0R E0 z.ˆ (2.19) 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 7 PROBLEM 3: PAIR OF MAGNETIC DIPOLES (20 points) Suppose there are two magnetic dipoles. One has dipole moment m 1 = m0zˆ and 1 − is located at r 1 =+2 a zˆ; the other has dipole moment m 2 = m0zˆ, and is located at − 1 r 2 = 2 a zˆ. (a) (10 points) For a point on the z axis at large z, find the leading (in powers of 1/z) behavior for the vector potential A(0, 0,z) and the magnetic field B (0, 0,z). (b) (3 points) In the language of monopole ( =0),dipole( =1),quadrupole( =2), octupole ( = 3), etc., what type of field is produced at large distances by this current configuration? In future parts, the answer to this question will be called a whatapole. (c) (3 points) We can construct an ideal whatapole — a whatapole of zero size — by n taking the limit as a → 0, keeping m0a fixed, for some power n. What is the correct value of n? (d) (4 points) Given the formula for the current density of a dipole, 3 Jdip(r)=−m × ∇ r δ (r − r d) , (3.1) where r d is the position of the dipole, find an expression for the current density of the whatapole constructed in part (c). Like the above equation, it should be expressed in terms of δ-functions and/or derivatives of δ-functions, and maybe even higher derivatives of δ-functions. PROBLEM 3 SOLUTION: (a) For the vector potential, we have from the formula sheet that µ m × rˆ A(r)= 0 , (3.2) 4π r2 which vanishes on axis, since m = m0zˆ,andrˆ=ˆz on axis. Thus, A(0, 0,z)=0. (3.3) This does not mean that B = 0, however, since B depends on derivatives of A with respect to x and y. From the formula sheet we have µ 3(m · rˆ)ˆr− m B (r)= 0 , (3.4) dip 4π r3 wherewehavedroppedtheδ-function because we are interested only in r =0. Evaluating this expression on the positive z axis, where rˆ =ˆz, we find µ 2m zˆ µ m zˆ B (0, 0,z)= 0 0 = 0 0 . (3.5) dip 4π r3 2π r3 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 8 For 2 dipoles, we have µ m 1 1 0 0 − B2dip(0, 0,z)= 3 3 zˆ 2π z − 1 a z + 1 a 2 2 µ m 1 1 = 0 0 − zˆ 2πz3 1 a 3 1 a 3 1 − z 1+ z 2 2 µ m 1 1 ≈ 0 0 − zˆ 3 − 3 a 3 a 2πz 1 2 z 1+ 2 z µ m 3 a 3 a ≈ 0 0 1+ − 1 − zˆ 2πz3 2 z 2 z µ m a ≈ 0 0 3 zˆ 2πz3 z 3µ m a = 0 0 z.ˆ (3.6) 4πz4 (b) Since it falls off as 1/z4, it is undoubtedly a quadrupole ( =2). For either the E or B fields, the monopole falls off as 1/r2, the dipole as 1/r3, and the quadrupole as 1/r4. (c) We wish to take the limit as a → 0in such a way that the field at large z approaches a constant, without blowing up or going to zero. From Eq. (3.6), we see that this goal will be accomplished by keeping m0a fixed, which means n =1. (d) For the two-dipole system we add together the two contributions to the current density, using the appropriate values of r d and m : − × ∇ 3 − a × ∇ 3 − a J2dip(r )= m0zˆ r δ r 2 zˆ + m0 zˆ r δ r 2 zˆ . (3.7) Rewriting, 3 a −3 − a δ (r + 2 zˆ) δ (r 2 zˆ) J (r )=m azˆ × ∇ r . (3.8) 2dip 0 a Now we can define Q ≡ m0a, and if we take the limit a → 0with Q fixed, the above expression becomes ∂ 3 J (r )=Qzˆ × ∇ r δ (r) . (3.9) 2dip ∂z 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 9 Since partial derivatives commute, this could alternatively be written as ∂ 3 J (r )=Qzˆ × ∇ r δ (r) . (3.10) 2dip ∂z PROBLEM 4: UNIFORMLY MAGNETIZED INFINITE CYLINDER (10 points) Consider a uniformly magnetized infinite circular cylinder, of radius R, with its axis coinciding with the z axis. The magnetization inside the cylinder is M = M0zˆ. (a) (5 points) Find H (r ) everywhere in space. (b) (5 points) Find B (r) everywhere in space. PROBLEM 4 SOLUTION: (a) The magnetization inside the cylinder is M = M0zˆ.ThecurloftheH (r)fieldis ∇×H (r)=Jfree =0, (4.1) and the divergence is B (r) 1 ∇·H ( r)=∇· − M ( r) = ∇·B − ∇· M =0. (4.2) µ0 µ0 Note that for a finite length cylinder, the divergence would be nonzero because of the abrupt change in M at the boundaries. Since H (r) is divergenceless and curl-free, we can say H (r) = 0everywhere in space. (4.3) (b) Having H (r) = 0everywhere in space, we can find magnetic field as B (r ) µ M zˆ for r In this question we could alternatively find the bound currents as Jb = ∇×M =0and Kb = M × nˆ = M0φˆ. Then, using Amp`ere’s law as we did for a solenoid, we could find the magnetic field and then also H , obtaining the same answers as above. 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 10 PROBLEM 5: ELECTRIC AND MAGNETIC UNIFORMLY POLARIZED SPHERES (10 points) Compare the electric field of a uniformly polarized sphere with the magnetic field of a uniformly magnetized sphere; in each case the dipole moment per unit volume points along zˆ. Multiple choice: which of the following is true? (a) The E and B field lines point in the same direction both inside and outside the spheres. (b) The E and B field lines point in the same direction inside the spheres but in opposite directions outside. (c) The E and B field lines point in opposite directions inside the spheres but in the same direction outside. (d) The E and B field lines point in opposite directions both inside and outside the spheres. PROBLEM 5 SOLUTION: E field of a uniformly B field of a uniformly polarized sphere magnetized sphere The answer is (c), E and B field lines point in opposite directions inside the spheres but in the same direction outside, as shown in the diagrams, which were scanned from the first edition of Jackson. Note that the diagram on the left shows clearly that ∇·E =0 at the boundary of the sphere, so it could not possibly be a picture of B . Itisatleast visually consistent with ∇×E =0,orequivalently E · d = 0for any closed loop, as it must be to describe an electrostatic field. The diagram on the right, on the other hand, shows clearly that ∇×B =0,orequivalently B · d =0,soitcouldnotpossiblybea picture of an electrostatic field. It is at least qualitatively consistent with ∇·B =0,as it must be. 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 11 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II November 13, 2012 Prof. Alan Guth FORMULA SHEET FOR QUIZ 2, V. 2 Exam Date: November 15, 2012 ∗∗∗ Some sections below are marked with asterisks, as this section is. The asterisks indicate that you won’t need this material for the quiz, and need not understand it. It is included, however, for completeness, and because some people might want to make use of it to solve problems by methods other than the intended ones. Index Notation: A· B = AiBi ,A× B i = ijkAjBk , ijk pqk = δipδjq − δiqδjp ··· det A = i1i2···in A1,i1 A2,i2 An,in Rotation of a Vector: T Ai = RijAj , Orthogonality: Rij Rik = δjk (RT = I) j j j =1 =2 =3 i − =1 cos φ sin φ 0 Rotation about z-axis by φ: Rz(φ)ij = i=2 sin φ cos φ 0 i=3 001 Rotation about axis nˆ by φ:∗∗∗ R(ˆn, φ)ij = δij cos φ +ˆninˆj(1 − cos φ) − ijknˆk sin φ. Vector Calculus: ∂ Gradient: (∇ ϕ)i = ∂iϕ, ∂i ≡ ∂xi Divergence: ∇·A ≡ ∂iAi Curl: (∇×A )i = ijk∂jAk ∂2ϕ Laplacian: ∇2ϕ = ∇· (∇ ϕ)= ∂xi∂xi Fundamental Theorems of Vector Calculus: b Gradient: ∇ ϕ· d = ϕ( b) − ϕ(a) a Divergence: ∇·A d3x = A · da V S where S is the boundary of V Curl: (∇× A) · da = A · d S P where P is the boundary of S 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 12 Delta Functions: ϕ(x)δ(x − x )dx = ϕ(x ) , ϕ(r)δ3(r − r )d3x = ϕ(r ) d dϕ ϕ(x) δ(x − x )dx = − dx dx x=x δ(x− xi) δ(g(x)) = ,g(xi)=0 |g (xi)| i r − r 1 ∇· = −∇2 =4πδ3(r− r ) |r − r |3 |r − r | − rˆj xj 1 δij 3ˆri rˆ j 4π 3 ∂i ≡ ∂i = −∂i∂j = + δij δ (r) r2 r3 r r3 3 3(d · rˆ)ˆr − d 8π ∇· =− (d · ∇ )δ3(r) r3 3 3(d · rˆ)ˆr − d 4π ∇× = − d × ∇ δ3(r) r3 3 Electrostatics: F = qE,where − − 1 (r r ) qi 1 (r r ) 3 E(r)= 3 = 3 ρ( r )d x 4π 0 i |r − r | 4π 0 |r − r | −12 2 2 0 =permittivity of free space = 8.854 × 10 C /(N·m ) 1 =8.988 × 109 N·m2/C2 4π 0 r − · 1 ρ(r ) 3 V (r)=V (r 0) E(r ) d = | − |d x r 0 4π 0 r r ρ ∇·E = , ∇× E =0,E = −∇ V 0 ρ ∇2V = − (Poisson’s Eq.) ,ρ=0= ⇒∇2V = 0(Laplace’s Eq.) 0 Laplacian Mean Value Theorem (no generally accepted name): If ∇2V =0,then the average value of V on a spherical surface equals its value at the center. Energy: 1 1 qiqj 1 1 3 3 ρ(r)ρ(r ) W = = d x d x ij | − | 2 4π 0 ij r 2 4π 0 r r i= j 1 1 2 W = d3xρ(r)V (r )= E d3x 2 2 0 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 13 Conductors: σ Just outside, E = nˆ 0 1 | | Pressure on surface: 2 σ E outside − 1 2 Two-conductor system with charges Q and Q: Q = CV , W = 2 CV N isolated conductors: Vi = Pij Qj ,Pij = elastance matrix, or reciprocal capacitance matrix j Qi = Cij Vj ,Cij = capacitance matrix j a a2 Image charge in sphere of radius a: Image of Q at R is q = − Q, r = RR Separation of Variables for Laplace’s Equation in Cartesian Coordinates: cos αx cos βy cosh γz V = where γ2 = α2 + β2 sin αx sin βy sinh γz Separation of Variables for Laplace’s Equation in Spherical Coordinates: Traceless Symmetric Tensor expansion: 1 ∂ ∂ϕ 1 ∇2ϕ(r, θ, φ)= r2 + ∇2 ϕ =0, r2 ∂r ∂r r2 θ where the angular part is given by 2 2 1 ∂ ∂ϕ 1 ∂ ϕ ∇θ ϕ ≡ sin θ + sin θ ∂θ ∂θ sin2 θ ∂φ2 ∇2 () − () θ Ci1 i2...i nˆi1 n ˆi2 ...n ˆi = ( +1) Ci1 i 2...i nˆi1 nˆi2 ...nˆ i , () where Ci1i2...i is a symmetric traceless tensor and nˆ =sinθ cos φ eˆ1 +sinθ sin φ eˆ2 +cosθ eˆ3 . General solution to Laplace’s equation: ∞ () () Ci1i2...i V( r)= C r + rˆi rˆi ...rˆi , where r = rrˆ i1 i 2...i r+1 1 2 =0 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 14 Azimuthal Symmetry: ∞ B V (r)= A r + { zˆi ...zˆi } rˆi ...rˆi r+1 1 1 =0 where { ...} denotes the traceless symmetric part of ... . Special cases: { 1 } =1 { zˆi } =ˆzi 1 { zˆizˆj } =ˆzizˆj − δij 3 1 { zˆizˆjzˆk } =ˆzizˆjzˆk − zˆiδjk +ˆzj δik +ˆzkδij 5 1 { zˆizˆjzˆkzˆm } =ˆzizˆjzˆkzˆm − zˆizˆjδkm +ˆzizˆkδmj +ˆzizˆmδjk +ˆzj zˆkδim 7 1 +ˆzjzˆmδik +ˆzkzˆmδij + 35 δij δkm + δikδjm + δimδjk Legendre Polynomial / Spherical Harmonic expansion: General solution to Laplace’s equation: ∞ Bm V (r)= Am r + Ym(θ, φ) r+1 =0 m=− 2π π ∗ Orthonormality: dφ sin θ dθY m (θ, φ) Ym(θ, φ)=δ δm m 0 0 Azimuthal Symmetry: ∞ B V (r)= A r + P(cos θ) r+1 =0 Electric Multipole Expansion: First several terms: 1 Q p · rˆ 1 rˆirˆj V (r)= + + Qij + ··· ,where 4π r r2 2 r3 0 3 3 3 2 Q = d xρ(r) ,pi = d xρ(r) xi Qij = d xρ(r)(3xixj −δij |r | ) , · · − 1 p rˆ 1 3(p rˆ)ˆr p 1 3 − ∇ − i Edip(r)= 2 = 3 p δ (r) 4π 0 r 4π 0 r 3 0 1 1 3 ∇×E dip(r)=0, ∇·Edip(r )= ρdip(r )=− p · ∇δ (r) 0 0 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 15 Traceless Symmetric Tensor version: ∞ 1 1 () V (r)= C rˆi ...rˆi , +1 i1...i 1 4π 0 r =0 where − () (2 1)!! 3 C = ρ(r) { xi ...xi } d x (r ≡ rrˆ ≡ xieˆi) i1...i ! 1 ∞ 1 (2 − 1)!! r = { rˆi ...rˆi } rˆ ...rˆ ,forr Griffiths version: ∞ 1 1 3 V (r)= +1 r ρ(r )P (cos θ )d x 4π 0 r =0 where θ = angle between r and r . ∞ ∞ 1 r< 1 = P(cos θ ) , √ = λ P( x) |r − r | +1 − 2 =0 r> 1 2λx + λ =0 1 d 2 P(x)= (x − 1) , (Rodrigues’ formula) 2! dx 1 2 P(1) = 1 P(−x)=(−1) P(x) dxP (x)P(x)= δ −1 2 +1 Spherical Harmonic version:∗∗∗ ∞ 1 4π qm m V (r)= +1 Y (θ, φ) 4π 0 2 +1r =0 m=− ∗ 3 where qm = Ymr ρ(r )d x ∞ 1 4π r ∗ = Y (θ ,φ)Ym(θ, φ), forr Electric Fields in Matter: Electric Dipoles: p = d3xρ(r) r 3 ρdip(r)=−p · ∇ r δ (r − r d),wherer d = position of dipole F =(p · ∇ )E = ∇ (p · E )(forceonadipole) = p × E (torque on a dipole) U = −p · E Electrically Polarizable Materials: P (r ) = polarization = electric dipole moment per unit volume ρbound = −∇ · P , σbound = P · nˆ D ≡ 0E + P , ∇·D = ρfree , ∇×E = 0(for statics) Boundary conditions: ⊥ − ⊥ σ ⊥ − ⊥ Eabove Ebelow = Dabove Dbelow = σfree 0 − − − Eabove Ebelow =0 Dabove Dbelow = Pabove Pbelow Linear Dielectrics: P = 0χeE , χe = electric susceptibility ≡ 0(1 + χe) = permittivity, D = E r = =1+χe = relative permittivity, or dielectric constant 0 Nα/ 0 Clausius-Mossotti equation: χe = , where N = number density of atoms 1 − Nα 3 0 or (nonpolar) molecules, α = atomic/molecular polarizability (P = αE ) 1 Energy: W = DE· d3x (linear materials only) 2 Force on a dielectric: F = −∇ W (Even if one or more potential differences are held fixed, the force can be found by computing the gradient with the total charge on each conductor fixed.) Magnetostatics: Magnetic Force: dp 1 F = q ( E + v× B )= , where p = γm0v, γ= dt v2 1 − c2 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 17 F = Id × B = J × B d3x Current Density: Current through a surface S: IS = J · da S ∂ρ Charge conservation: = −∇·J ∂t Moving density of charge: J = ρv Biot-Savart Law: µ d × ( r − r ) µ K (r )(×r − r ) B (r)= 0 I = 0 da 4π |r − r |3 4π |r − r |3 µ J(r ) × (r − r ) = 0 d3x 4π |r − r |3 −7 2 where µ0 = permeability of free space ≡ 4π × 10 N/A Examples: µ I Infinitely long straight wire: B = 0 φˆ 2πr Infintely long tightly wound solenoid: B = µ0nI0 zˆ ,wheren = turns per unit length µ IR2 Loop of current on axis: B (0, 0,z)= 0 zˆ 2(z2 + R2)3/2 1 Infinite current sheet: B (r)= µ K × nˆ ,ˆn = unit normal toward r 2 0 Vector Potential: µ J(r ) A( r ) = 0 d3x , B = ∇×A , ∇· A =0 coul 4π |r − r | coul ∇·B = 0(Subject to modification if magnetic monopoles are discovered) Gauge Transformations: A (r)=A(r)+∇ Λ(r ) for any Λ(r). B= ∇× A is unchanged. Amp`ere’s Law: ∇×B = µ0J, or equivalently B · d = µ0Ienc P 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 18 Magnetic Multipole Expansion: Traceless Symmetric Tensor version: ∞ { } µ0 () rˆi1 ...rˆi Aj(r )= M 4π j;i1i2...i r+1 =0 − () (2 1)!! 3 where M = d xJj(r){ xi ...xi } j;i1i2...i ! 1 3 Current conservation restriction: d x Sym(xi1 ...xi −1 Ji )=0 i1...i where Sym means to symmetrize — i.e. average over all i1...i orderings — in the indices i1 ...i Special cases: 3 =1: d xJi =0 3 =2: d x (Jixj + Jjxi)=0 µ m × rˆ Leading term (dipole): A( r)= 0 , 4π r2 where −1 M(1) mi = ijk j;k 2 1 1 m = I r × d = d3xr × J = Ia, 2 P 2 where a = da for any surface S spanning P S µ m × rˆ µ 3(m · rˆ)ˆr − m 2µ B (r)= 0 ∇× = 0 + 0 mδ 3(r ) dip 4π r2 4π r3 3 3 ∇·B dip(r)=0, ∇×B dip(r )=µ0Jdip(r)=−µ0m × ∇ δ (r) Griffiths version: ∞ µ0I 1 A( r )= (r) P(cos θ)d 4π r+1 =0 Magnetic Fields in Matter: Magnetic Dipoles: 1 1 m = I r × d = d3xr × J = Ia 2 P 2 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 19 3 Jdip(r )=−m × ∇ r δ (r − r d), where r d = position of dipole F =∇ (m · B )(forceonadipole) = m × B (torque on a dipole) U = −m · B Magnetically Polarizable Materials: M (r) = magnetization = magnetic dipole moment per unit volume Jbound = ∇×M , Kbound = M × nˆ 1 H ≡ B − M , ∇×H = Jfree , ∇·B =0 µ0 Boundary conditions: ⊥ − ⊥ ⊥ − ⊥ − ⊥ − ⊥ Babove Bbelow =0 Habove Hbelow = (Mabove Mbelow) − × − × Babove Bbelow = µ0(K nˆ) Habove Hbelow = Kfree nˆ Linear Magnetic Materials: M = χmH , χm = magnetic susceptibility µ = µ0(1 + χm) = permeability, B = µH Magnetic Monopoles: µ0 qm B ( r)= rˆ ; Force on a static monopole: F = qmB 4π r2 µ qe q m Angular momentum of monopole/charge system: L = 0 rˆ ,whererˆpoints 4π from qe to qm µ qe q m 1 Dirac quantization condition: 0 = ¯h × integer 4π 2 Connection Between Traceless Symmetric Tensors and Legendre Polynomials or Spherical Harmonics: (2)! P(cos θ)= { zˆi ...zˆi } nˆi ...nˆi 2(!)2 1 1 For m ≥ 0, (,m) Ym(θ, φ)=Ci1...i nˆi1 ...nˆi , (,m) { + + } where Ci i ...i = dm uˆi ...uˆi zˆim+1 ...zˆi , 1 2 1 m (−1)m(2)! 2m (2 +1) with dm = , 2! 4π ( + m)! ( − m)! + 1 and uˆ = √ (ˆex + ieˆy) 2 m ∗ Form m<0, Y,−m(θ, φ)=(−1) Ym(θ, φ) 8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012 p. 20 More Information about Spherical Harmonics:∗∗∗ 2 +1( − m)! m imφ Ym(θ, φ)= P (cos θ)e 4π ( + m)! m where P (cos θ) is the associated Legendre function, which can be defined by (−1)m d+m P m(x)= (1 − x2)m/2 (x2 − 1) 2! dx+m Legendre Polynomials: SPHERICAL HARMONICS Ylm(θ , φ) 1 l = 0 Y00 = 4π 3 iφ Y11 = - sin θe 8π l = 1 3 Y10 = cos θ 4π 151 2 2iφ Y22 = sin θe 4 2π 15 iφ l = 2 Y21 = - sin θ cosθe 8π 5 3 2θ 1 Y20 = ( 2 cos 2 ) 4π 1 35 3 3iφ Y33 = - sin θe 4 4π 1 105 2 2iφ Y32 = sin θ cos θe 4 2π l = 3 1 21 2 eiφ Y31 = - sinθ (5cos θ -1) 4 4π 7 5 3 3 Y30 = ( cos θ cos θ) 4π 2 2 Image by MIT OpenCourseWare. MIT OpenCourseWare http://ocw.mit.edu 8.07 Electromagnetism II Fall 2012 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.