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(4πr2)= (4πr2)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 rr if r<r Q µ0ω dAφ(r, θ, φ)= dr sin θ r 4 (1.7) 4 πR3 3 if r>r . 3 r2 Note that the R of Griffiths has been replaced by r, which is the radius of the integration shell. Now we can calculate the vector potential inside the sphere at 8.07 QUIZ 2 SOLUTIONS, FALL 2012 p. 3 some radius r<R. The integration will require two pieces, a piece where 0 <r <r and the other where r<r <R, thus using the two options in Eq. (1.7): r 4 R µ0 Qω r φ A (r, θ, φ)= 3 sin θ dr 2 + dr rr . (1.8) 4π R 0 r 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<R (r)= 0 (2.1) 0 , if r>R. There are no free charges anywhere in this problem. The sphere is embedded in a constant external electric field E = E0zˆ, which means that V (r ) ≡−E0r cos θ for r R. (a) (9 points) Show that V (r) obeys the differential equation 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<Rand r>R) and give its two independent solutions in each region. Hint: they are powers of r. You may want to know that 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<Rand r>R. (e) (4 points) What is the net dipole moment of the polarized sphere? 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<Rwe have = − . Using the hint, Eq. (2.4) in the problem statement, dr r we write ∞ 2 ∂V dln 1 ∂ 2 ∂V dV 2 ( +1) ∇ V + = P(cos θ) r + − − V =0. ∂r dr r2 ∂r ∂r dr r r2 =0 (2.8) For this equation to hold for all r<Rand for all θ, the term inside the square brackets should be zero. (To show this, one would multiply by P (cos θ)sinθ and then integrate from θ =0toθ =2π. By the orthonormality of the Legendre polynomials, only the = term would survive, so it would have to vanish for every .) Thus, 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.