Solutions: Homework 5

Transcription

1 Ex. 5.1: Capacitor Solutions: Homework 5 (a) Consider a parallel plate capacitor with large circular plates, radius a, a distance d apart, with a d. Choose cylindrical coordinates (r,φ,z) and let the z axis be aligned to the capacitor symmetry axis. Clearly, the charge on the top plate is q(t) = it, (1) for a fixed current i, assuming q() =. Ignoring edge effects, applying the integral form of Gauss Law as usual, we have πr E(t) = 4πitr /a so E(t) = 4it a e z. () Note the charge on the top plate is chosen to be positive, so E points downward. Between the plates, Ampère s law is B = + E/c t. The integral form of this, applied to a horizontal Ampèran loop of radius r centred on the z-axis is Bπr = 1 E c t da = 1 4it c a πr, (3) where da = e z da and we ignore edge effects. Hence which is constant in time, as expected. (b) The energy density B(t) = ir ca e φ, (4) E(t) = 1 [ E (t)+b (t) ] 8π = 4i 8πc a 4 ( 4c t +r ) (5) Since ct a r, then E B, so the electric contribution to the energy density is much greater than that of the magnetic field. (c) The poynting vector S = c 4π E B = c 4π = i π Now, the divergence of S in cylindrical coordinates However, the rate of change of energy density between the plates 8i rt ca 4 e z e φ rt a 4e r. (6) S = 4i t πa 4. (7) E t = 4i t πa 4. (8) 1

2 Hence we have E t + S =, (9) as expected from energy conservation, since there are no free currents between the plates. (d) The components of the stress tensor in Cartesian coordinates are T ij = 1 [ (Ei E j +B i B j ) (1/)(E +B ] )δ ij. (1) 4π The electric field points in the z direction, so there are no off-diagonal electric terms. Since E B as above, we may neglect the magnetic field terms in the diagonal components T ii. So T = 1 E B sinφ B sinφ E, 8π +E = 4i 8πa 4 4t (r /c )sinφ (r /c )sinφ 4t, (11) +4t noting B x = Bsinφ and B y = Bcosφ. Obviously, the off-diagonal terms are much smaller than the diagonal ones. (e) Consider the top plate. The force on this plate df = TdA = Te z da, (1) where the area vector points into the region between the plates. Hence the total force F = i t a e z. (13) Note that the charge density on a plate ρ s = it/(πa ), so we can rewrite the pressure as πρ s exactly as found in HW 1. Ex. 5.: Current Loop (a) Consider a circular loop radius a centred at the origin inthe xy plane, carrying a current I. This loop has a magnetic dipole moment m = Iπa e z. (14) c Far away from the loop, r a, the magnetic field is asymptotically the dipole field B(r) = 1 r 5 [ 3(m r)r mr ]. (15) In spherical coordinates(r,θ,φ), we let r = re r and note e z e r = cosθ, e z e θ = sinθ and e z e φ =, so that e z = e r cosθ e θ sinθ. (16) It follows that B(r) = 1 r 5 [ 3mr cosθe r mr (cosθe r sinθe θ ) ] = m r 3 ( er cosθ+e θ sinθ ), (17) where m = Iπa /c.

3 (b) Consider a second loop Γ with radius a, fixed in the xy plane. At some time t let Γ be centred at y = r a,a, and moving at velocity ve y. First, in the xy plane θ = π/ and e θ = e z so B(r) = m/r 3 e z. The flux through the loop Γ is da Φ B (r ) = B(r) da = m r 3. (18) S Here r is the distance of da from the origin, which depends on r and the polar coordinates r,φ whose origin is the centre of Γ. Explicitly r = r +(r ) +r r cosφ by the cosine rule. However, since r a r, then to zeroth order (in a ) r r, and so the flux A Φ B (r ) = Iπ a (a ) cr 3 Φ () B. (19) Aside: Higher order corrections can be found by expanding 1/r 3 in powers of r, although zeroth order will suffice for this problem. E.g. at first order we have ( Φ B (r ) = Φ () B 1 3 (a ) ) (r ). () By Faraday s law in integral form, the induced emf around the loop E = Φ B (r ) c t = v a (a ) 3Iπ c, (1) by the chain rule, and since r / t = v. (c) The smaller loop has resistance R, so the power dissipated in the loop, by Ohm s Law, is P = E R = 9I π 4 ( ) ( ) v aa 4 R c. () (d) Since magnetic fields do no work, the power dissipated in the small loop must produce a change in the loop kinetic energy E k. Now, by energy conservation de k = F dx = Fdy, where F is the force acting on the loop due to the power dissipation. Since we assume the current flows uniformly around the loop, the only Lorentz force acts in the y direction. Hence r r 4 P = de k /dt = Fdy/dt = Fv (3) since v = ve y and we assume that the loop s velocity is kept constant by another external force - clearly the net force is zero on the loop since its velocity is constant. It follows, after application of Lenz s law, that the force on the loop due to the Lorentz interaction is F = 9I π 4 R v c 4 ( ) aa 4 e y. (4) r Ex. 5.3: Rotating conducting sphere (a) Consider a wire element dl moving with velocity u in a magnetic field F. The Lorentz force on a charge q in this wire is clearly just F = q u B. (5) c Now, by definition, the motional EMF E is the work done per charge by the force F, so qde = dw = F dl. Hence we have de = 1 u B dl. (6) c 3

4 (b) Consider a conducting spherical shell, radius a, rotating with angular frequency ω = ωe z (i.e. in a right handed sense about the z-axis), in a magnetic field B = B e z. In the spherical coordinates (r,θ,φ), the velocity u of a point r = ae r on the shell is dependent only on the declination angle θ: The perpendicular distance of this point from the axis of rotation - the z axis - is r = asinθ, so that u(θ) = rω( sinφ,cosφ,) = aωe φ. (7) Note e φ is the spherical coordinate azimuthal angle vector, not the cylindrical one. Now, consider the arc of a great circle from the north pole of the shell to its equator. We may write such a curve as l(θ) = ae r (θ), for θ π/ and the tangent to this curve is dl(θ)/dθ = ae θ, so that dl = ae θ dθ. (8) Note that this is just a more formal way of writing the elemental arc length along a circle dl = adθ, in terms of our spherical coordinates. Applying (6) we have Integrating from the pole to the equator we have de(θ) = a ω c B (e φ e z e θ )dθ = a ω c B sinθcosθdθ. (9) so that E E(π/) E() = π/ a ω c B sinθcosθdθ, (3) E = a ωb c. (31) Ex. 5.4: Energy of Uniformly Charged Sphere Consider a sphere V, having radius a with uniform charge density ρ. (a) By Gauss Law, with respect to the spherical coordinate r, the electric field for this system is clearly E int = 4π 3 ρ r, r a The total energy is then E ext = 4π 3 ρ a 3 r, r a. (3) E int E = V 8π dv + Eext R 3 /V 8π dv = π [ a ] 9 ρ 4πr dr + 4π a6 r 4r dr = 16π ρ a5 15 a. (33) (b) The energy of the system may also be written as E = 1 4 V ρ ΦdV (34)

5 provided that we choose the gauge for Φ such that Φ as r. From (3), we must then have Φ ext (r) = (4π/3)ρ a 3 /r, so the potential on the surface of the sphere is and it follows that Φ(a) = 4πρ a 3, (35) Hence the energy is Φ int (r) = 4π 6 ρ (3a r ). (36) E = ρ which is the same as above, as expected. a = 4π ρ 3 4πr Φ int (r)dr a = 16π ρ a 5 15 (3a r )r dr, (37) Ex. 5.5: Coulomb Gauge The vector and scalar potentials are defined via E = Φ 1 c ta B = A. (38) Substituting these into the Gauss law and applying the Coulomb gauge condition A =, we have ( 4πρ = E = Φ 1 ) c ta which is the Poisson equation. Similarly from Ampère s law we find 4π c J 1 ( c t Φ 1 ) c ta = ( A) = Φ, (39) = ( A) A = A 1 c ta = 1 c tφ 4π c J. (4) Now, let J = J 1 +J such that J = and J 1 =. Taking the divergence of eq. (4) and noting that A = A = in the Coulomb gauge, then it follows that ( ) 4πJ 1 + t Φ =. (41) By eq. (39) we can rewrite this as J 1 = t ρ, (4) which is actually just the continuity equation in disguise. Since J 1 is curl-less, then by analogy to electrostatics, the solution is J 1 (r,t) = dv tρ(r,t)/4π r r = 1 4π t dv ρ(r,t) r r. (43) 5

6 But from eq. (39) so then 4πJ 1 = t Φ. Φ(r,t) = dv ρ(r,t) r r, (44) 6