Solutions: Homework 2

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1 Solutions: Homework 2 Ex 2.1: Particle moving in Magnetic Field (a) If the initial velocity is perpendicular to the magnetic field B, the magnitude of the Lorentz Force (Eq 1.52) is F = 1 c q qu B ub (1) c The direction of this force is perpendicular to the plane containing u and B. Thus the magnetic force does no work on the particle, and its speed u = u and kinetic energy 1/2mu 2 remain constant. Since the force is always perpendicular to the velocity, the particle moves in a circle, and its acceleration has the centripetal form u 2 /R. Newton s equation of motion is The radius of the circular orbit is thus q ub = mu2 c R R = mcu q B (2) (3) (b) The angular velocity has the magnitude ω = u R = q B mc (4) The sense of gyration is such that for a positive particle the vector angular velocity ω is antiparallel to B; hence, ω = q mc B (5) (c) Any component of initial velocity parallel to B produces no net force and therefor remains constant. This general motion has the form of a helix. Source: Heald & Marion Solutions Manual Ex. 2.2: Higher-Order Mulitpole Moments (a) Consider a system of charges defined by a density ρ(r). (This may include delta functions for lowerdimensional or discrete distributions.) The dipole moment of this distribution is by definition p = d 3 rρ(r)r, (6) where we have implicity defined a coordinate system for r with origin of course r =. A change of coordinates may be implemented by a simple shift r r = r a, so that the new coordinate system has its origin at r = a in the old coordinates: see the figure. The transformed density is ρ (r ) = ρ(r), so that explicitly e.g. ρ () = ρ(a). Note also d 3 r = d 3 r. 1

2 In this new coordinate system, the transformed dipole moment is p = d 3 r ρ (r )r = d 3 rρ(r)(r a) = p a d 3 rρ(r). (7) But, clearly d 3 rρ(r) = q, the total charge. For zero total charge, it follows that p = p, so dipole moment is independent of choice of origin. (b) For Q, we have Clearly, if we choose a = (1/q)p, then p =. p = p qa. (8) (c) The quadrupole tensor has components Q ij = ρ(r) ( 3r i r j r 2 δ ij ) d 3 r, (9) where we write the coordinates of r as (r 1,r 2,r 3 ). Under the same coordinate shift r r = r a as above, we have Q ij = ρ (r ) [ 3r i r j (r ) 2 ] δ ij d 3 r = ρ(r) [ 3(r i a i )(r j a j ) (r a) 2 ] δ ij d 3 r = Q ij + ρ(r) [ 3( r i a j r j a i +a i a j )+(2r a+a 2 )δ ij d 3 r = Q ij 3(p i a j +p j a i )+2p aδ ij +(3a i a j a 2 δ ij )q. (1) For p = and q =, then Q ij = Q ij as required. (d) Suppose the net charge q =, but p. Then we have the relation Q ij = Q ij 3(p i a j +p j a i )+2p aδ ij. (11) The tensor Q is symmetric and traceless, so in general it has five independent components. Since a has only three independent components a i, from (11) we see that we can only set three of the five independent components of Q to zero. Hence we cannot set Q = under choice of origin alone. Ex. 2.3: Multipole Expansion 2

3 (a) From simple application of superposition, we have (b) (i) The monopole moment is the net charge Φ(r 1 ) = 2e r 1 e r 1 1, Φ(r 2 ) = 2e r 2 e (r2 2. (12) +1)1/2 q = α q α = e. (13) (ii) For a discrete distribution, the dipole moment is p = α q α r α = ee x. (14) (iii) Finally, the quadrupole tensor has components Q ij = α q α (3x α,i x α,j r 2 α δ ij). (15) Since the charge distribution has symmetry about the x-axis, we expect that Q should be diagonal, and that Q 22 = Q 33 (Note we use subscripts 1, 2 and 3 to denote the x, y and z coordinates respectively). We can check this by directly calculating all the components of Q, which is simple to do for this system. We have in matrix notation as expected. (c) In this case, the first three terms of the multipole expansion are Φ(r) = q r + p e r r Q = 2e e (16) e where r = (x,y,z). Hence, for r 1 = r 1 e x and r 2 = r 2 e y, ij ( 3xi x j r 2 ) δ ij Q ij r 5 = q r + p e r r Q 11 2 r 5 (x2 y 2 /2 z 2 /2), (17) Φ(r 1 ) = e r 1 e r 1 2 e r 1 3, Φ(r 2 ) = e r 2 + e 2 r 2 3. (18) Clearly these are different from the potentials in (12): these potentials are expansions of the latter in powers of (1/r). Another way to say this is that the potentials (18) converge asymptotically to those in (12). A brief check of this is to note that in the limits r 1,2, we have from (12) Φ(r 1 ) e/ r 1, and Φ(r 2 ) e/ r 2, which are the monopole terms. More carefully, note that for r 1 > 1 we have a geometric series Φ(r 1 ) = 2e e ( ) 1 r 1 r 1 1 1/r 1 = 2e e ( ) r 1 r 1 r 1 r , (19) 3

4 which matches the asymptotic expression above. Similarly, finding the Taylor expansion for Φ(r 2 ) in powers of 1/r 2 directly yields Φ(r 2 ) = 2e e ( ) 1 r 2 r 2 1+(1/r2 ) 2 = 2e e ( 1 1 ) r 2 r 2 2r , (2) as above. Ex. 2.4: Charged Ring a) It s just a step function: b) By definition of the density ρ l = ρ l (θ), we have dq = ρ l (θ)adθ, and hence Q ij = 2π dθ ρ l (θ)a ( 3x i x j r 2 δ ij ). (21) Note that x 3 = for all the contributions, and (x 1,x 2 ) = (acosθ,asinθ). c) Since x 3 =, it follows that Q 32,Q 31,Q 23,Q 13 =. Further, Q 33 is zero since Q 33 = 2π dθρ l (θ)a 2 =. (22) The 11 and 22 components are both zero because the integrand contains the step function multiplied by an always-positive function of twice the periodicity. That is, Q 11 = 2π dθρ l (θ)a 3( 3cos 2 θ 1 ) π/2 = 2a 3 dθρ l (θ) ( 3cos(2θ)+2 ) π +2a 3 dθρ l (θ) ( 3cos(2θ)+2 ) π/2 = 2a 3 dθρ l (θ) [ 3cos(2θ)+2 ] +2a 3 λ π/2 π/2 π/2 = 6a 3 dθ [ ρ l (θ) ρ l (θ) ][ 3cos(2θ)+2 ] ρ l (θ +π/2)dθ [ 3cos(2θ+π)+2 ] =. (23) Since Q is traceless, then clearly Q 22 =, too. 4

5 Only the 12 elements are nonzero: Q 12 = Q 21 = 2π = 3a3 2 dθ ρ l (θ)a 3 (3cosθsinθ) (24) 2π = 3a3 2 4λ π/2 dθ ρ l (θ)(sin2θ) (25) dθ (sin 2θ) (26) = 6a 3 λ (27) Hence the quadrupole moment is Q αβ = 6a3 λ 6a 3 λ. (28) Ex. 2.5: Force on a Magnetic Dipole (a) Consider a magnetic dipole m in a uniform magnetic field B. The dipole moment by definition is m = 1 dv [r J(r )], (29) c and we showed in class that dv J(r ) =. (3) 5

6 An alternative way: The Lorentz force acting on such an ensemble J(r ) is simply F = 1 dv J(r ) B = 1 ( ) dv J(r ) B = (31) c c since B is uniform. (b) A magnetic dipole moment produces a magnetic field which is not uniform: it behaves as 1/r 3 at large r. A second dipole moment will therefore experience a net force due to this field, which is the interaction force between the dipoles. (a) ConsideramagneticdipolemandamagneticfieldB. Thetorqueactingonthisdipoleis(bydefinition) τ = m B = mbsin(θ)e m B, (32) where θ is defined as the angle from B to m, whence the minus sign. Noting θ is the canonical variable, the potential energy of this configuration is found by rotating m such that U(θ) U() = θ τdθ = mb(cosθ 1) = m B(θ)+m B() (33) so we choose the gauge U = m B. It follows that the net force on the dipole This a general result for the force on a dipole in any magnetic field. F = (m B). (34) 6