Physics 3 Midterm Nov 30, 015 name: Box your final answer. 1 (15 pt) (50 pt) 3 (0 pt) 4 (15 pt) total (100 pt) 1
1. (15 pt) An infinitely long cylinder of radius R whose axis is parallel to the ẑ axis has a uniform magnetization M = Mẑ. Compute H everywhere. ans Method 1: (5 pt H = 0; 3 pt H = M; 5 pt z [MΘ(s R)] = 0; pt concluding H = 0) Since J f = 0, we know H = 0 and H = M = [MẑΘ(s R)] = [MΘ(s R)] = 0. z Hence H = 0 everywhere. Method : (5 pt K b = M ˆn = M ˆφ; 5 pt writing down Ampere s law; 3 pt computing B z = µ 0 M; pt computing H = 0) Since M = 0, we conclude J b = 0. Since K b = M ˆn, we conclude K b = Mẑ ŝ = M ˆφ. Using Ampere s law, we find B z l = µ 0 K b ˆφl B z = µ 0 M H = B µ 0 M = Mẑ + Mẑ = 0 inside. Outside, there is no B for an infinite solenoid: H = 0.
. Suppose for t < 0, the capacitor is charged to V 0 in the circuit below. The switch is closed at t = 0. a) (10 pt) The current behaves as a function of time as I(t) = (envelope function of time) cos( ft + δ) where f > 0 for (R/L) < 4/(LC). What is f? ans (5 pt IR V L di dt = 0; 3 pt iωr + 1 C Lω = 0; pt taking the real part of quadratic equation solution correctly) Following the current and the potential drops (where we denote V as the capacitor potential), we have IR V L di dt = 0 Putting in we find Hence, we conclude f = Rω = IR Q C LdI dt = 0 di dt R + I C + Ld I dt = 0 I = I 0 e iωt, iωr + 1 C Lω = 0 ω = ir ± 4L C R L 4L C R L = 1 LC R 4L b) (15 pt) Suppose the inductance L is due to the self-inductance of a toroidal coil with rectangular cross section (innder radius a, outer radius b, and height h as shown in the figure below) carrying a total of N turns. Express L in terms of N, h, a, and b. ans (4 pt Ampere s law; 5 pt Φ = LI; 5 pt Φ computation; 1 pt answer) 3
Use Ampere s law: we have d l B = µ 0 NI, B φ = µ 0NI πs Using Φ = LI, we conclude Φ = Nh ˆ b a dsb φ = µ 0N Ih ln b π a L = µ 0N h π ln b a c) (10 pt) Suppose the toroidal coil described in part b) has a magnetic field B(s,φ,z) = f (s) ˆφ (where s is a cylindrical coordinate variable: e.g. the inner radius is described as s = a). In this magnetic field, suppose a pointlike magnetic dipole with m = mẑ is placed at rest at s = u (a,b) inside the torus. What are the forces and torques on the dipole due to the magnetic field? ans (3 pt N = m B; pt N = ŝm f (u); 3 pt F = ( m B); pt F = 0) We have the torque being The magnetic force is N = m B = ŝm f (u) F = ( m B) = 0 d) (15 pt) Suppose the capacitor is made of parallel circular plates each with radius R with vacuum in between. Suppose the charge density on one of the plates for t > 0 is σ(t) and the uniform electric field region between the capaciator plates is then characterized by E σ(t) ẑ. Neglecting the edge effects as usual and assuming rotational symmetry about the axis passing through the center of the plates, find the resulting B φ and the Poynting vector at s = R, somewhere between the parallel plates in the uniform field region. ans (3 pt B = µ 0 E ; pt B φ = µ 0 s σ Since the sign) R s=r = µ 0 σ ; 5 pt S = 1 µ 0 E B s=r ; 5 pt S = ŝ σ(t) B = µ 0 E R σ where one point is for B φ πs = µ 0 πs E z s σ B φ = µ 0 R s=r = µ 0 σ 4
Hence, we find As a check, note S = 1 µ 0 E B s=r = ŝ σ(t) R σ which is the energy leaving the uniform field region. πrhs = πrh σ(t) R σ = πrh 1 R σ R E = πrh = = πr h u 3. Suppose you are given that the Maxwell equations were replaced by Ẽ = 0 Ẽ = B L Ẽ B = Ñ Ẽ L Ẽ B = 0 where L and Ñ are complex constants and the tilde indicates that these fields are complex. a) (10 pt) What is the wave equation governing Ẽ? ans (3 pt ( Ẽ) = ( Ẽ) Ẽ; 3 pt Ẽ = 0; 3 pt for substituting B = Ñ Ẽ L Ẽ;1 pt Ẽ = Ñ Ẽ ) Use ( Ẽ) = ( Ẽ) Ẽ Ẽ = ( B + L Ẽ) = (Ñ Ẽ L E + L Ẽ) = Ñ Ẽ b) (10 pt) Find a plane wave solution Ẽ to this Maxwell equation system proportional to e iωt, traveling in the y direction, having a single wavelength, and the boundary condition Ẽ(t = 0, x = 0) = C. Be sure to specify the condition on C coming from the Maxwell equation system. ans (4 pt plane wave form; 3 pt transverse; 3 pt dispersion) Use Ẽ = i( ky ωt) Ce ŷ C = 0 k = ω Ñ k = ω Ñ 5
4. (15 pt) A plane wave is incident normal to a dielectric interface as shown: where the index of refraction of medium i {1,} is n i and µ 1 = µ. The incident wave is and the electric field in the rest of the regions can be written as ẼI (t,z) = Ẽ 0I exp[i(k 1 z ωt)] ˆx ẼR (t,z) = Ẽ 0R exp[i( k 1 z ωt)] ˆx ẼT (t,z) = Ẽ 0T exp[i(k z ωt)] ˆx. Recall that one can derive from one of the Maxwell equations the magnetic field for the reflected wave being B(t,z) R = ωµ 1ε 1 Ẽ 0R e i( k 1z ωt)ŷ k 1 where the minus sign in the amplitude comes from the minus sign in k = k 1 ẑ for the reflected wave. Use E continuity and H continuity to find Ẽ 0T just in terms of { Ẽ 0I, n 1, n }. ans (5 pt Ẽ 0I + Ẽ 0R = Ẽ 0T ; 5 pt for ωµ 1ε 1 k 1 Ẽ 0I ωµ 1ε 1 The E boundary condition says while the H boundary condition says k 1 Ẽ 0R = ωµ ε k Ẽ 0R ; 3 pt ω/k i = v i ; pt Ẽ 0T = Ẽ 0I 1+ n n 1 ) Ẽ 0I + Ẽ 0R = Ẽ 0T H (+) = H ( ) Using ω/k i = v i, we find Hence, ωµ 1 ε 1 Ẽ 0I ωµ 1ε 1 Ẽ 0R = ωµ ε k 1 k 1 k 1 v 1 (Ẽ 0I Ẽ 0R ) = 1 v Ẽ 0T Ẽ 0I = (1 + v 1 v )Ẽ 0T Ẽ 0R Ẽ 0T = Ẽ 0I 1 + n n 1 6
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