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Physics 70006 Answers to Final Exam May 8, 2006 1. A charge Q is distributed uniformly along the z axis from z = −a to z = a. Write out the ﬁrst few terms in the Legendre polynomials expansion of the potential for the case r > a. Answer: The potential at a point z > a on the axis may be written z+a 1 Q a dz 1 Q Φ(z) = ln = . 4π0 2a −a z − z 4π0 2a z−a Expanding in powers of a/z, we ﬁnd a 1 a 2 1 a 3 1 Q − + + ··· Φ(x) = 4π0 2a z 2 z 3 z 2 a 1 a 1 a 3 − − − − + ··· z 2 z 3 z 1 Q a 1 a 3 1 a 5 = + + + ··· 4π0 a z 3 z 5 z Replacing 1/z (l+1) by 1/r(l+1) Pl (cos θ), we obtain 1 1 a2 Q 1 a4 Φ(x) = + P2 (cos θ) + P4 (cos θ) + · · · 4π0 r 3 r3 5 r5 Verify that your answer is correct for r a. Answer: For r a, the above potential reduces to the potential of a point charge at the origin. Q 1 Φ(x) → . 4π0 r 2. A point dipole p is imbedded at the center of a dielectric sphere (radius R and dielectric constant r ). Find the potential inside and outside of the sphere. Answer: Set up a boundary value problem: Assume that the potential has the form 1 p P1 (cos θ) + ArP1 (cos θ) Φin = 4π0 r2 B Φout = 2 P1 (cos θ), r where A and B are expansion constants. Boundary conditions at r = R lead to the two equations: B p 1 = AR + 2 R 4π0 R2 2B 2p 1 = r −A + . R3 4π0 R3 1 Solving, we ﬁnd 2(r − 1) 1 p (r + 2) R3 4π0 p 3r B= (r + 2) 4π0 A= By what factor is the dipole moment p enhanced by the presence of the dielectric sphere? Answer: pout = 3r p (r + 2) 3. A long cylinder of radius R has magnetization vector M = kρ3 ẑ, where k is a constant and ρ is the radius in cylindrical coordinates. (a) Ignoring end eﬀects, determine the bound current densities Jb = ∇ × M and Kb = M × n. Answer: Jb = ∇ × M = − dM φ̂ = −3kρ2 φ̂ dρ Kb = M × n = kR3 φ̂ (b) From the bound currents determine B inside and outside the cylinder using Ampère’s law. Answer: For ρ < R choose a rectangular loop; one side of the loop of length l coincides with the axis, an opposite side is parallel to the axis at a distance ρ away. These two sides are connected by sides perpendicular to the axis. The total bound current through this loop is ρ Ib = l 3kρ2 dρ = lkρ3 0 This current ﬂows in the −φ̂ direction. From Ampère’s law, taking into account the sense of the current, one ﬁnds that B is in the +z direction and that Bz (ρ) = μ0 kρ3 = μ0 M (ρ), ρ < a. If the above loop encloses the the surface of the cylinder, then the total enclosed current will be R 3kρ2 dρ − lkR3 = 0. Ib = l 0 and we will have Bz (ρ) = 0, 2 ρ < a. (c) Determine H inside and outside the cylinder. Answer: Outside H = B/μ0 = 0. Inside H = B/μ0 − M = 0. Therefore, H = 0, everywhere. 4. An inﬁnite straight wire carries a current 0 t≤0 I(t) = I0 t > 0 (a) Show that the (retarded) vector potential A is in the z direction and that Az (ρ, t) = 0 = μ0 I0 ln 2π ct + c2 t2 − ρ2 ρ ct < ρ ct ≥ ρ. Answer: For ct < ρ no signal reaches ρ. For ct > ρ, only that segment of the wire with |z| ≤ c2 t2 − ρ2 contributes to the potential at ρ. For ct > ρ, we have √c2 t2 −ρ2 dz μ0 I0 Az = 4π −√c2 t2 −ρ2 z 2 + ρ2 c2 t2 − ρ2 + ct μ0 I0 ln = 4π − c2 t2 − ρ2 + ct μ0 I0 ct + c2 t2 − ρ2 = ln 2π ρ (b) Find B(ρ, t) and E(ρ, t). Show that your ﬁelds have the correct static limit as t → ∞. Answer: E has only a z component since E=− We ﬁnd μ0 I0 Ez = 2π 1 ∂A . ∂t ct + c2 t2 − ρ2 μ0 I0 c = . 2 2π c t2 − ρ2 As t → ∞, Ez → 0, as expected. 3 c+ c2 t c2 t2 − ρ2 B has only a φ component and ∂Az ∂φ 1 ρ μ0 I0 1 + = 2π ρ ct + c2 t2 − ρ2 c2 t2 − ρ2 Bφ = − μ0 I0 1 ct 2π ρ c2 t2 − ρ2 μ0 I0 1 as t → ∞. → 2π ρ = 5. A transmission line consists of two long conducting ribbons of width w, parallel to one another and separated by distance h. A uniformly distributed current I runs up one conductor and back down the other as illustrated in the ﬁgure. (a) Determine the direction and magnitude of the B ﬁeld between the ribbons. Answer: First note that the magnitude of the surface current density is K = I/w. Using the coordinates shown in the ﬁgure, the B ﬁeld between the ribbons is in the −z direction: B = −μ0 K ẑ = −μ0 I ẑ. w (b) Determine the magnetic energy stored in a section of length l. Answer: 1 μ0 hI 2 Wm = l d3 rB · H = 2 2w (c) Find the self inductance per unit length of the transmission line. Answer: Use Wm = (1/2)LI 2 to ﬁnd L/l = μ0 h . w (d) Assuming that one end of the transmission line is connected to a power supply with terminal voltage V and the other end is terminated by a resistor R, determine the direction and magnitude of Poynting vector. Express the magnitude in terms of V and R. Answer: Assuming the upper ribbon is at potential V , the electric ﬁeld is downward and has value E=− V ŷ h The Poynting vector is S =E×H = 4 V I V2 x̂ = x̂, hw RA where A = wh is the area between the ribbons normal to the direction of energy ﬂow (x̂). (e) Determine the direction and magnitude force/length exerted on the upper ribbon by the E ﬁeld and by the B ﬁeld. Answer: The contribution to the force from the E ﬁeld is 1 2 0 V 2 2 da, dFy = 0 Ey − E (−1) da = − 2 2 h2 thus 0 V 2 w 2h2 The contribution from the B ﬁeld is 1 1 μ0 I 2 da, dFy = 0 − B 2 (−1) da = + μ0 2 2 w2 Fye /l = − and Fym /l = 5 μ0 I 2 . 2w