Physics 3323, Fall 2014 Problem Set 12 due Nov 21, 2014

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1 Physics 333, Fall 014 Problem Set 1 due Nov 1, 014 Reading: Griffiths Ch Square loops Griffiths 7.3 (formerly 7.1). A square loop of wire, of side a lies midway between two long wires, 3a apart, and in the same plane. (Actually, the long wires are sides of a large rectangular loop, but the short ends are so far away that they can be neglected.) A clockwise current I in the square loop is gradually increasing: di/dt = k (a constant). Find the emf induced in the big loop. Which way will the induced current flow? SOLUTION: The EMF induced in the big loop will have will be counter-clockwise. Lets label the big loop and the small loop 1. E = dφ di 1 = M 1 (1.1) dt dt Now M 1 is hard to calculate, since the magnetic field from the small loop is hard to calculate. But the flux through the small loop from the large loop is easier, since the large loop is just two long straight currents. The contribution to the flux from each current is the same. We can use the relation M 1 = M = M 1 to calculate the mutual inductance. Now di 1 dt M = Φ a 1 µ 0 = I a πs a ds = µ 0a ln (1.) π = k, so E = µ 0ak ln π (1.3) Figure 1: Square Loop in the Big Loop magnitude E = dφ dt. Its direction is given by Lenz s law. Since the clockwise current in the small loop is increasing, the flux into the page is increasing. This means that the induced current in the big loop will create flux out of the page and. Conductive capacitor A thin metal plate capacitor of plate separation d is filled with a medium of conductivity σ and dielectric constant ε. The plates of the capacitor are circular. A variable voltage V = V 0 sin ωt is applied to the capacitor a shown in the figure. Assuming that the electric field between the plates is homogeneous, find H in the capacitor. SOLUTION: Since the electric field is homogeneous, we can simply write it down as follows. 1

2 ε, σ Figure : Conductive Capacitor Arrangement 3. Cable Inductance Consider a coaxial cable consisting of two long concentric hollow conducting cylinders with radii a and b. A current I travels up the inner cylinder, and returns down the outer cylinder. Determine the self-inductance per unit length, both from using our identity L = Φ/I, and from comparing the magnetic energy with the standard circuit form 1 LI. E = V d ẑ = V 0 d Now we can use Ampere s law H = J f + D sin ωt ẑ (.1) (.) along with the constitutive relation D = εe and ohms law J = σe to get H. The cylindrical symmetry requires H = H ˆφ. Consider an Amperian loop centered on the axis of symmetry with radius s. H d l = [ Jf + D d a = [ σe + ε E d a (.3) H(πs) = V 0 d (σ sin ωt + εω cos ωt) πs (.4) H = V 0s d (σ sin ωt + εω cos ωt) ˆφ (.5) We see that the answer has a resistive component in phase with the voltage, and a capacitive component out of phase with the voltage by π/. SOLUTION: The magnetic field is circumferential, so the relevant flux is that which passes between radius a and b. The magnetic field inside the cable is B = µ 0I πs. First we ll find the inductance per unit length by using the identity L = Φ/I. Φ = B d a = µ 0I π b a l ds s L l = Φ Il = µ 0 π ln b a = µ 0Il π ln b a (3.1) (3.) Now we will derive the same formula by comparing the magnetic energy with the standard circuit form U = 1 LI. U = 1 µ 0 B dv = 1 b µ 0 I µ 0 a 4π πsl ds (3.3) s U = µ 0I l 4π b a ds s = µ 0I l 4π ln b a L l = U I l = µ 0 π ln b a 4. From standing waves to travelling (3.4) (3.5) a) Write the traveling wave Ψ(z, t) = A cos(ωt kz) as a

3 superposition of two standing waves. b) Write the standing wave Ψ(z, t) = A cos ωt cos kz as a superposition of two traveling waves that travel in opposite directions. c) Rewrite the following superposition of two traveling waves Ψ(z, t) = A cos(ωt kz) + RA cos(ωt + kz), where R is also a constant, as a superposition of standing waves. SOLUTION: These problems are applications of the following identities: a) b) c) cos (α + β) = cos α cos β sin α sin β (4.1) cos α cos β = 1 [cos (α + β) + cos (α β) (4.) Ψ(z, t) = A cos (ωt kz) (4.3) = A cos ωt cos kz + A sin ωt sin kz (4.4) Ψ(z, t) = A cos ωt cos kz (4.5) = A [cos (ωt + kz) + cos (ωt kz) (4.6) Ψ(z, t) = A cos (ωt kz) + RA cos (ωt + kz) (4.7) = A[cos ωt cos kz+sin ωt sin kz+ra[cos ωt cos kz sin ωt sin kz (4.8) = (1 + R)A cos ωt cos kz + (1 R)A sin ωt sin kz (4.9) 5. Am I a wave? Griffiths 9.1. By explicit differentiation, check that the functions f 1 (z, t) = Ae b(z vt), and f 3 (z, t) = satisfy the wave equation and f (z, t) = A sin[b(z vt), A b(z vt) + 1 f 4 (z, t) = Ae b(bz +vt), and f 5 (z, t) = A sin(bz) cos(bvt) 3 do not. SOLUTION: 1.. f [ 1 z = 4b (z vt) b Ae b(z vt) (5.1) f 1 = [ 4b (z vt) b v Ae b(z vt) = v f 1 z (5.) f z = b A sin [b(z vt) (5.3) f = (bv) A sin [b(z vt) = v f z (5.4) 3

4 f 3 z = Ab (b(z vt) + 1) [ 4b(z vt) 1 (5.5) f 3 = Abv [ 4b(z vt) (b(z vt) + 1) 1 = v f 3 f 4 z = b [ b z 1 Ae b(bz +vt) (5.6) (5.7) f 4 = b v Ae b(bz +vt) v f 4 z (5.8) f 5 z = b A sin (bz) cos 3 (bvt) (5.9) f 5 = [ 3A(bv) cos (bvt) sin (bvt) cos 3 (bvt) v f 5 z (5.10) would you construct a wave circling the other way? (In optics, the clockwise case is called right circular polarization, and the conterclockwise, left circular polarization.) (b) Sketch the string at time t = 0. (c) How would you shake the string in order to produce a circularly polarized wave? SOLUTION: We can write this wave as a superposition of two perpendicular linearly polarized waves with the same amplitude, but out of phase by π/. ) f(z, t) = (Ãx ˆx + Ãy ŷ e i(kz ωt) (6.1) ( ) = A ˆx + e iπ/ ŷ e i(kz ωt) (6.) f(z, t) = A (ˆx + i ŷ) e i(kz ωt) (6.3) 6. Circular polarization on a string Griffiths 9.8. Equation 9.36 describes the most general linearly polarized wave on a string. Linear (or plane ) polarization (so called because the displacement is parallel to a fixed vector ˆn) results from the combination of horizontally and vertically polarized waves of the same phase (Eq. 9.39). If the two components are of equal amplitude, but out of phase by 90 (say, δ v = 0, δ h = 90 ), the result is a circularly polarized wave. In that case: (a) At a fixed point z, show that the string moves in a circle about the z axis. Does it go clockwise or counterclockwise, as you look down the axis toward the origin? How a) Let s pick the point z = 0 to see which way the vector rotates. f(z, t) = A (ˆx + i ŷ) e iωt = A (ˆx + i ŷ) (cos (ωt) i sin (ωt)) (6.4) Remember that the actual vector is the real part of Eq. (??). This gives the following parametrization. f x = A cos (ωt) (6.5) f y = A sin (ωt) (6.6) This is a parametric equation for a circle of radius A rotating counter-clockwise as t increases. To get a wave that circles the other way, we could start with Ãy = Ae iπ/ instead of the positive phase difference. 4

5 Figure 3: Plot of circular wave at t = 0 b) See Figure?? c) To create this wave, you could just grab one end of the string and swing your arm in a circle. 7. Hey, don t push me around! Griffiths The intensity of sunlight hitting the earth is about 1300 W/m. If sunlight strikes a perfect absorber what pressure does it exert? How about a perfect reflector? What fraction of atmospheric pressure does this amount to? SOLUTION: For a perfect absorber P = I/c. With I = 1300 W/m we have: P = I/c = Pa (7.1) A perfect reflector has twice the pressure P = Pa (7.) These correspond to atm and atm respectively. 5

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