Electromagnetism II - Theory and Exercises, Esercizi di Fisica

Scarica Electromagnetism II - Theory and Exercises e più Esercizi in PDF di Fisica solo su Docsity! Chapter 14 Electromagnetism II Abstract Chapters 13 and 14 are devoted to electromagnetism concerned with motion of charged particles in electric and magnetic fields, Lorentz force, cyclotron and betatron, magnetic induction, magnetic energy and torque, magnetic dipole moment, Faraday’s law, Hall effect, RLC circuits, resonance frequency, Maxwell’s equations, electromagnetic waves, Poynting vector, phase velocity and group veloc- ity, dispersion relations, waveguides and cut-off frequency. 14.1 Basic Concepts and Formulae Self-Inductance (L) L = Nϕ i = Total flux linkage Current linked (14.1) ξ = −L di dt (14.2) For a long solenoid or toroid L = μ0 N 2 A l (14.3) where N is the number of turns, A is the area of cross-section of each turn and l is the length of the coil. The energy stored in an inductor is W = 1/2Li2 (14.4) L–R Circuit Differential equation. L di dt + i R = ξ (for charging process) (14.5) 631 632 14 Electromagnetism II Solution i = ξ R ( 1 − e− Rt L ) (14.6) Inductive time constant τL = L/R (14.7) Differential equation L di dt + iR = 0 (discharging process) (14.8) Solution i = ξ R e−Rt/L (14.9) C–R Circuit Ri + q/c = ξ (charging) (14.10) q = q0(1 − e−t/RC ) (14.11) Capacitive Time Constant τ = RC (14.12) Ri + q/C = 0 (discharging) (14.13) q = q0e−t/RC (14.14) Inductors in Series Leq = L1 + L2 + 2M (coil currents in the same sense) (14.15) Leq = L1 + L2 − 2M (coil currents in the opposite sense) (14.16) where M is the mutual inductance Leq = L1 + L2 (inductors well separated) (14.17) Inductors in Parallel Leq = L1L2/(L1 + L2) (14.18) M = (L1L2) 1/2 (inductors are closely placed) (14.19) The Alternating Current (AC) The effective or root-mean-square value: Ie = I0/ √ 2 (14.20) where I0 is the peak current 14.1 Basic Concepts and Formulae 635 Now the flux of current from a closed surface is the decrease of the charge inside the surface. In general this cannot be zero because charges can be moved from one place to another. This difficulty is avoided by adding the term ∂ D/∂t , where D = ε0 E , on the right-hand side of (32). Integral Form ϕE = ∮ B.ds = 0 (Gauss’ law for magnetostatics) (14.36) ϕE = ∮ E.ds = qenc ε0 (Gauss’ law for electrostatics) (14.37) ∮ E.ds = −dϕB dt (Faraday’s law of induction) (14.38) ∮ B.ds = μ0ε0 dϕE dt + μ0ienc (Ampere-Maxwell law) (14.39) Electromagnetic Waves The E- and H -waves are transverse to the direction of propagation and are perpen- dicular to each other. ∇ × (∇ × E) = −∇2E + ∇(∇.E) (14.40) Wave Equation ∇2E = −ω2μ0ε0E (14.41) The Intrinsic Impedance η = √ μ ε (14.42a) η0 = √ μ0 ε0 = 377  (free space) (14.42b) The Poynting vector (S) represents the power in the electromagnetic wave and is given by S = E × H (14.43) and points in the direction of propagation of the wave. The skin thickness (δ) represents the depth to which an electromagnetic wave of frequency f = ω/2π can penetrate a medium and is given by 636 14 Electromagnetism II δ = √ 2 μσω (14.44) where μ is the permeability and σ is the conductivity. Phase Velocity (vph) and Group Velocity (vg) Phase velocity is just due to the nodes of the wave that are moving and not energy or information. In fact the phase velocity can be greater than c, the velocity of light vph = ω/k (14.45) In order to know how fast signals will travel one must calculate the speed of pulses or modulation caused by the interference of a wave of one frequency with one or more waves of slightly different frequencies. The speed of the envelope of such a group of waves is called the group velocity and is given by vg = dω/dk (14.46) Waveguides are hollow metallic structures in which the electromagnetic waves are guided to travel from one place to another without much attenuation. Here, we shall be concerned only with rectangular guide of cross-section of x = a and y = b, with the wave propagated in the z-direction. k2 = ω2 c2 − (mπ a )2 − (nπ b )2 (14.47) where c is the velocity of light in free space, m and n are integers. Equation (14.47) is valid both for TEmn and THmn waves. For the simplest case m = 1 and n = 0. For TE10 wave. vph = c√ 1 − ( λ0 2a )2 (14.48) where λ0 is the free space wavelength. vg = c √ 1 − ( λ 2a )2 (14.49) vph.vg = c2 (14.50) λg = λ0√ 1 − ( λ0 2a )2 (14.51) 14.2 Problems 637 14.2 Problems 14.2.1 The RLC Circuits 14.1 A 2.5 μF capacitor is connected in series with a non-inductive resistor of 300  across a source of PD of rms value 50 V, alternating at 1000/2π Hz. Calculate (a) thermal values of the current in the circuit and the PD across the capacitor. (b) the mean rate at which the energy is supplied by the source. [Joint Matriculation Board of UK] 14.2 A 3  resistor is joined in series with a 10 mH inductor of negligible resis- tance, and a potential difference (rms) of 5.0 V alternating at 200/π Hz is applied across the combination. (a) Calculate the PD VR across the resistor and VL across the inductor. (b) Determine the phase difference between the applied PD and the current. [Joint Matriculation Board of UK] 14.3 An inductance stores 10 J of energy when the current is 5 A. Find its value. 14.4 A tuning circuit in a radio transmitter has a 4 × 10−6 H inductance in series with a 5 × 10−11 F capacitance. Find (a) the frequency of the waves transmitted. (b) their wavelength. 14.5 A 6  resistor, a 12  inductive reactance and a 20  capacitive reactance are connected in series to a 250 V rms AC generator. (a) Find the impedance. (b) Estimate the power dissipated in the resistor. 14.6 At 600 Hz an inductor and a capacitor have equal reactances. Calculate the ratio of the capacitive reactance to the inductive reactance at 60 Hz. 14.7 A capacitance has a reactance of 4  at 250 Hz. (a) Find the capacitance. (b) Calculate the reactance at 100 Hz. (c) What is the rms current, if it is connected to a 220 V 50 Hz line? 14.8 When an impedance, consisting of an inductance L and a resistance R in series, is connected across a 12 V 50 Hz supply, a current of 0.05 A flows which differs in phase from that of the applied potential difference by 60◦. Find the value of R and L . Find the capacitance of the capacitor which when connected in series in the above circuit has the effect of bringing the current into phase with the applied potential difference. [University of London] 14.9 When a 0.6 H inductor is connected to a 220 V 50 Hz AC line, what is (a) the rms current and (b) peak current? 640 14 Electromagnetism II Fig. 14.3 (iv) Find the phase angle between supply voltage and current through the circuit. (v) Find the voltages across R, C and L and show these on a phasor dia- gram. (vi) What is the condition for resonance to occur in this type of circuit and at what frequency would this occur? [University of Aberystwyth, Wales 2001] 14.22 A 40  resistor and a 50 μF capacitor are connected in series, and an AC source of 5 V at 300 Hz is applied. What is the magnitude of the current flowing through the circuit? [University of Manchester 2007] 14.23 For the circuits shown in Fig. 14.4a, b consisting of resistors, capacitors and inductors (i) Derive the expression to represent the complex impedances for each of the networks. (ii) Work out the magnitude of the impedance for each of the networks given that the frequency of the supply voltage is 150 Hz. [University of Aberystwyth, Wales 2006] Fig. 14.4a Fig. 14.4b 14.2 Problems 641 14.24 (a) Define what is meant by electric current and current density. (b) When we refer to a quantity of charge we say that the value is quantized. Explain what is meant by quantized. (c) A thin copper bar of rectangular cross-section of width 5.6 mm and height 50 μm has an electron density of n = 8.5 × 1028/m3. If a uniform current of i = 2.4 × 10−4 A flows through the strip (i) Find the magnitude of the current density in the strip. (ii) Find the magnitude of the drift speed of the charge carriers. (iii) Briefly explain why the current is relatively high for such a small drift speed. [University of Aberystwyth, Wales 2008] 14.25 Two series resonant circuits with component values L1C1 and L2C2, respec- tively have the same resonant frequency. They are then connected in series; show that the combination has the same resonant frequency. [University of Manchester 1972] 14.26 An inductance and condenser in series have a capacitative impedance of 500  at 1 kHz and an inductive impedance of 100  at 5 kHz. Find the values of inductance and capacitance. [University of Manchester 1972] 14.27 A condenser of 0.01 μF is charged to 100 V. Calculate the peak current that flows when the charged condenser is connected across an inductance of 10 mH [University of Manchester 1972] 14.28 An inductance of 1 mH has a resistance of 5 . What resistance and con- denser must be put in series with the inductance to form a resonant circuit with a resonant frequency of 500 kHz and a Q of 150? [University of Manchester 1972] 14.29 A parallel resonant circuit consists of a coil of inductance 1 mH and resis- tance 10  in parallel with a capacitance of 0.0005 μF. Calculate the reso- nant frequency and the Q of the circuit. [University of Manchester 1972] 14.30 The voltage on a capacitor in a certain circuit is given by V (t) = V0e−t/RC . Find the fractional error in the voltage at t = 50 μs if R = 50 k ± 5% and C = 0.01 μF ± 10%. [University of Manchester 1972] 14.31 A condenser of 10 μF capacitance is charged to 3000 V and then discharged through a resistor of 10, 000 . If the resistor has a temperature coefficient of 0.004/◦C and a thermal capacity of 0.9 cal/◦C, find (a) the time taken for the 642 14 Electromagnetism II voltage on the condenser to fall to 1/e of its initial value; (b) the percentage error which would have been introduced if thermal effects had been ignored. [University of Manchester 1958] 14.32 Show that the fractional half-width of the resonance curve of an RLC circuit is given by ω ω = √ 3 Q where Q is the quality factor given by Q = ω L/R. 14.2.2 Maxwell’s Equations, Electromagnetic Waves, Poynting Vector 14.33 A plane em wave E = 100 cos (6×108 t + 4x) V/m propagates in a medium. What is the dielectric constant of the medium? [Indian Administrative Services] 14.34 An infinite wire with charge density λ and current I is at rest in the Lorentz frame S. Show that the speed of reference frame S′ where the electric field is zero, i.e. that frame in which one observes pure magnetic field, is given by v = λc2 I . 14.35 Show that for a magnetic field B the wave equation has the form ∇2B = μ0ε0 ∂2B ∂t2 14.36 Use Maxwell’s equation to show that ∇. ( j + 1 ε0 ∂E ∂t ) = 0. 14.37 The free-space wave equation for a medium without absorption is ∇2E − μ0ε0 ∂2E ∂t2 = 0 Show that this equation predicts that electromagnetic waves are propagated with velocity of light given by c = 1/ √ μ0ε0. 14.38 An electromagnetic wave of wavelength 530 nm is incident onto a sheet of aluminium with resistivity ρ = 26.5 × 10−9m. Estimate the depth that the wave penetrates into the aluminium. The expression for the skin depth, δ, is δ = √ 2/μ0σω. [University of Manchester 2008] 14.39 Consider an electromagnetic wave with its E-field in the y-direction. Apply the relation ∂ Ey ∂x = − ∂ Bz ∂t to the harmonic wave E = E0 cos (kx − ωt), B = B0 cos (kx − ωt) to show that E0 = cB0. 14.2 Problems 645 14.60 At the orbit of the earth, the power of sunlight is 1,300 Wm−2. Estimate the amplitude of the electric field if we assume that all the power arrives on the earth in a monochromatic wave. [University of Aberystwyth, Wales 2004] 14.61 A typical value for the amplitude of the E-field for sunlight at the surface of Mars is 300 V/m Calculate the amplitude of the corresponding B-field and estimate the flux of radiation at the surface of Mars. [University of Manchester 2006] 14.62 Show that |E | |H | = 377  [University of Aberystwyth, Wales] 14.63 Calculate the skin depth in copper (conductivity 6×107 −1/m) of radiation of frequency 20 kilocycles/s. Take μ, the relative permeability of copper, as unity. [University of Newcastle upon Tyne 1964] 14.64 Copper has an electrical conductivity σ = 5.6 × 107 −1/m and a magnetic permeability μ = 1. On this basis estimate the order of magnitude of the depth to which radiation at a frequency of 3000 Mc/s can penetrate a large copper screen. [University of Bristol 1959] 14.65 Show that the skin depth in a good conductor is [ 1 2 ωσμμ0 ]−1/2 where the symbols have their usual meaning. [University of Newcastle upon Tyne 1964] 14.66 If the maximum electric field in a light wave is 10−3 V/m, find how much energy is transported by a beam of 1 cm2 cross-sectional area. [University of Durham 1962] 14.67 Prove Poynting’s theorem, namely div (E × H) + E. ∂D ∂t + H. ∂E ∂t + E. j = 0 What is the interpretation of this equation? [University of Durham 1962][University of New Castle upon Tyne 1965] 14.68 From Maxwell’s equations, one can derive a wave equation for a dielectric of the form ∇2 E − μ0ε0εr ∂2 E ∂t2 − μ0σN ∂ E ∂t = 0 where E is the electric field, t is the time, εr is the relative permittivity and σN is the electrical conductivity. Hence by substituting a travelling wave solution into the wave equation derive a dispersion relation of the form 646 14 Electromagnetism II k2 = μ0ε0εrω 2 + iμ0σN ω where k is the wave vector and ω is the angular frequency. [University of Durham 2006] 14.69 Show that the general identity ∇ × (∇ × E) = −∇2E + ∇(∇.E) is true for the specific vector field F = x2z3 î . 14.70 Use the Poynting vector to determine the power flow in a coaxial cable by a DC current I when voltage V is applied. Neglect the resistance of the conductors. How are the results affected if this assumption is not made? 14.71 A wire with radius a and of conductivity σE carries a constant, uniformly distributed current I in the z-direction. Apply Poynting’s theorem to show that power dissipated in the wire is given by the familiar expression I 2 R for Joule’s heat. 14.72 A super-conductor is a material which offers no DC resistance and satisfies the equation B = − me ne2 ∇ × J where n is the number of conduction electrons per unit volume, B is the magnetic field and J is the current density. Using Maxwell’s equations, show that the equation for a superconductor leads to the relation ∇2B = μ0ne2 me B [University of Durham 2004] 14.73 An oscillating voltage of high frequency is applied to a load by means of copper wire of radius 1 mm. Given that the skin depth is 6.6 × 10−5 m for this frequency, what is the high-frequency resistance per unit length of the wire in terms of its direct current resistance per unit length? [University of Durham 1966] 14.74 Use Stokes’ theorem to derive the expression Curl E = −∂B ∂t [University of New Castle upon Tyne 1964] 14.2 Problems 647 14.75 Explain the difference between the vectors B and H in the theory of mag- netism. Derive the expression B = μ0(H + M). Indicate briefly how B depends on it in the case of (a) paramagnetics and (b) ferromagnetics. [University of Durham 1962] 14.76 Show that Gauss’ and Ampere’s laws in free space, subject to the Lorentz condition, can be expressed in the usual notation as −∇2φ + 1 c2 ∂2φ ∂t2 = ρ ε0 and −∇2A + 1 c2 ∂2A ∂t2 = μ0J respectively. [The University of Aberystwyth, Wales 2005] 14.77 Using Faraday’s law, ∇ × E = −∂ B/∂t , for the propagation of electromag- netic waves travelling along the z-axis, show that Ey = Z0 Hx where Z0 = √ μ0/ε0 = 376.6 , is the wave impedance of free space. 14.78 (a) The electric field of an electromagnetic wave propagating in free space is described by the equation E(z, t) = E0[x̂sin(kz − ωt) + ŷcos(kz − ωt)] where x̂ and ŷ are unit vectors in the x- and y-direction, respectively. What is this wave’s direction of propagation? What is the polarization of the wave? (b) State and prove the boundary conditions satisfied by the magnetic inten- sity H and the magnetic field B at the boundary between two media with different magnetic properties. (c) Show that tan θ1 tan θ2 = μ1 μ2 where θ1 and θ2 are incident and refraction angles. 14.79 A plane wave is normally incident on a dielectric discontinuity. Use appro- priate boundary conditions to calculate R, the reflectance, and T , the trans- mittance, and show that T + R = 1. 14.80 An electromagnetic wave, propagating and linearly polarized in the xz-plane, is incident onto an interface between two non-conducting media as shown in Fig. 14.10. The electric fields and propagation vectors of the incident, 650 14 Electromagnetism II 14.91 Show that the group velocity can be expressed as vg = c n + ω ( dn dω ) 14.92 For a rectangular guide of width 2.5 cm what free-space wavelength of radi- ation is required for energy to traverse 50 m of length of the guide in 1 μs. What would be the phase velocity under these conditions? 14.93 Show that for light waves of angular frequency ω in a medium of refractive index n, the group velocity vg and the phase velocity vp are related by the expression 1 vg = 1 vp + ω c dn dω where c is the velocity of light in free space. [University of Manchester 1972] 14.94 Prove that the usual expression for the group velocity of a light wave in a medium can be rearranged as vg = c dv d(nv) , where c is the phase velocity of the waves in free space, v is the frequency and n is the refractive index of the medium. [University of Durham 1961] 14.95 Show that the group velocity associated with a free non-relativistic particle is the classical velocity of the particle. [University of Manchester 1972] 14.96 Calculate the group velocity of light of wavelength 500 nm in glass for which the refractive index μ at wavelength λ (meters) is μ = 1.420 + 3.60 × 10−14 λ2 [University of Manchester 1972] 14.2.4 Waveguides 14.97 For a rectangular guide of width 2.5 cm, calculate (a) the phase velocity; (b) the group velocity; (c) guide wavelength for the free-space wavelength of 4 cm. Assume the dominant mode. 14.98 A rectangular guide has a width a = 3 cm. What should be the free-space wavelength if the guide wavelength is to be thrice the free-space wavelength? 14.99 (a) Calculate the guide wavelength for a rectangular waveguide of width a = 5 cm if the free-space wavelength is 8 cm. (b) What is the cut-off wavelength for the guide? 14.3 Solutions 651 14.100 Calculate the number of states of electromagnetic radiation between 5000 and 6000 Å in wavelength using periodic boundary conditions in a cubical region 0.5 cm on a side. [University of Manchester 1972] 14.101 Consider a car entering a tunnel of dimensions 15 m wide and 4 m high. Assuming the walls are good conductors, can AM radio waves (530– 1600 kHz) propagate in the tunnel? [The University of Aberystwyth, Wales 2006] 14.102 Calculate the least cut-off frequency for TEmn waves for a rectangular waveguide of dimensions 5 cm × 4 cm. 14.103 Calculate how the wave and group velocities of the TE01 wave in a rectan- gular waveguide with a = 1 cm and b = 2 cm vary with frequency. [The University of Wales, Aberystwyth 2004] 14.104 Consider a rectangular waveguide of dimensions x = a and y = b, the TMmn wave travelling in the z-direction which is the axis of the guide. Given that the z-component Ez satisfies the equation ( ∂ z ∂xz + ∂ z ∂yz ) Ez = (k2 − ω2με)Ez obtain (a) the solution for Ez and (b) the cut-off frequency. 14.105 Consider a rectangular waveguide of dimensions x = a and y = b, the wave travelling along the z-direction, the axis of the guide. Given that the z-component Hz satisfies the equation ( ∂2 ∂x2 + ∂2 ∂y2 ) Hz = (k2 − ω2με)Hz (a) obtain the solution for Hz . (b) Obtain the cut-off frequency. (c) What are the similarities and differences between TMmn mode and TEmn mode? 14.3 Solutions 14.3.1 The RLC Circuits 14.1 (a) Reactance of capacitor Xc = 1 2π fc = 1 2π × (1000/2π) × 2.5 × 10−6 = 400  Impedance of the circuit Z = √ R2 + X2 c = √ (300)2 + (400)2 = 500  652 14 Electromagnetism II The rms current, Ie = Ve Z = 50 500 = 0.1 A PD across the capacitor, Vc = Ie Xc = 0.1 × 400 = 40 V (b) Power, P = Ve Ie cos α = Ve Ie R Z = 50 × 0.1 × 300 500 = 3.0 W 14.2 (a) XL = 2π fL = 2π × 200 π × 10 × 10−3 = 4  Z = √ R2 + X2 = √ 32 + 42 = 5  Ie = Ve/Z = 5/5 = 1.0 A VR = Ie R = 1.0 × 3 = 3.0 V VL = Ie XL = 1.0 × 4 = 4.0 V (b) Phase angle between Ve and Ie is given by tan α = XL R = 4 3 ⇒ α = 53◦ Ve leads Ie by 53◦ 14.3 U = 1 2 L I 2 L = 2U I 2 = 2 × 10 52 = 0.8 H 14.4 (a) f = 1 2π √ LC = 1 2π √ 4 × 10−6 × 5 × 10−11 = 1.125 × 10−7 Hz (b) λ = c f = 3 × 108 1.125 × 107 = 26.67 m 14.5 (a) Z = √ (XL − XC)2 + R2 = √ (12 − 20)2 + 62 = 10  (b) P = IeVe R Z = V 2 e R Z2 = (250)2 × 6 102 = 3750 W 14.6 At ω0 = 600 rad/s, XL = XC ∴ ω0L = 1 ω0C ⇒ 1 LC = ω2 0 At ω = 60 rad/s, XC XL = 1/ωc ωL = 1 ω2LC = ω2 0 ω2 = (600)2 602 = 100 14.3 Solutions 655 Energy of an inductance, E = 1 2 Li2 ∴ [L] = [Ml2T −2 A−2] (3) Using (1), (2) and (3), it is observed that the given combinations have the dimension of time and therefore are expressed in seconds [RC] = [ L/R ] = [√ LC ] = [T ] 14.15 ω = 1√ LC (1) ω′ = √ 1 LC − ( R 2L )2 (2) ω − ω′ ω = 1 − ω′ ω = 1 − √ 1 − R2C 4L  1 − ( 1 − R2C 8L ) = R2C 8L (3) where we have used (1) and (2) and expanded the radical binomially. q = qme−Rt/2L , q/qm = 1 2 whence t = 2L R ln ( qm q ) = 2L R ln 2 (4) If n is the number of cycles and ν the oscillating frequency t = n ν = 2πn √ LC = 2L R ln 2 or CR2 L = (ln 2)2 (πn)2 (5) Combining (3) and (5) ω − ω′ ω = ω ω = (ln 2)2 8π2n2 = 0.006085 n2 = 0.00038 where we have put n = 4. 14.16 q = qme−Rt/2L cos ω′t (charge oscillation of damped oscillator) Differentiating with respect to time i = dq dt = −qmω′e−Rt/2L ( R 2Lω′ cos ω′t + sin ω′t ) = −qmω′e−Rt/2L (tan φ cos ω′t + sin ω′t) 656 14 Electromagnetism II where we have set R/2Lω′ = tan φ. This gives i = −qmω′e−Rt/2L cos φ (sin φ cos ω′t + cos φ sin ω′t) = −qmω′e−Rt/2L sin(ω′t + φ) cos φ But for low damping φ → 0 as R/2Lω′ → 0. ∴ cos φ → 1, so that i = −qmω′e−Rt/2L sin(ω′t + φ) 14.17 Equation of the circuit is L d2q dt2 + 1 C q = ξ (1) Multiply (1) by i = dq/dt Li di dt + 1 C q dq dt = ξ t or d dt ( 1 2 Li2 ) + d dt ( q2 2C ) = Pinput (2) Thus, the input power is the sum of the powers delivered to the inductor and the capacitor. 14.18 The circuit equation is L d2q dt2 + R dq dt + 1 C q = 0 (1) Multiply (1) by i = dq dt Li di dt + Ri2 + 1 C q dq dt = 0 or d dt ( 1 2 Li2 + q2 2C ) = −Ri2 or dE dt = −i2 R (2) where E = 1 2 Li2 + q2 2C = total energy 14.3 Solutions 657 14.19 If U is the total field energy then U = UB + UE = 1 2 Li2 + 1 2 Q2 C (1) which shows that at any time the energy is stored partly in the magnetic field in the conductor and partly in the electric field in the capacitor. In the presence of the resistance R the energy is transferred to Joule heat, being given by dU dt = −i2 R (2) the minus sign signifying that the stored energy U decreases with time. Dif- ferentiating (1) with respect to time and equating the result with (2) gives Li di dt + Q C dQ dt = −i2 R (3) Substituting i = dQ/dt and di/dt = d2 Q/dt2, (3) becomes d2 Q dt2 + R L dQ dt + Q LC = 0 (4) Writing R/L = 2γ and 1/LC = ω2 0, (4) takes the required form d2 Q dt2 + 2γ dQ dt + ω2 0 Q = 0 (5) (a) f0 = 1 2π √ LC = 1 2π √ 80 × 10−3 × 700 × 10−12 = 2.128 × 104 Hz (b) τ = L R = 80 × 10−3 100 = 8 × 10−4 s 14.20 d2 Q dt2 + 2γ dQ dt + ω2 0 Q = 0 (1) Let Q = eλt so that dQ/dt = λeλt and d2 Q/dt2 = λ2eλt . The characteristic equation is then λ2 + 2γ λ + ω2 0 = 0 whose roots are λ = −γ ± √ γ 2 − ω2 0 (2) Calling α = √ γ 2 − ω2 0 λ1 = −γ + α, λ2 = −γ − α 660 14 Electromagnetism II f0 = 1 2π √ LC = 1 2π √ 0.2 × 10 × 10−6 = 112.6 Hz 14.22 XC = 1 2π f C = 1 (2π)(300)(50 × 10−6) = 10.6  Z = √ R2 + X2 C = √ (300)2 + (10.6)2 = 300.19  I = V Z = 5 300.19 = 0.0166 A 14.23 Impedance of a capacitor i. Let an AC emf be applied across a capacitor. The potential difference across the capacitor will be VC = V0 sin ωt (1) where V0 is the amplitude of the AC voltage of angular frequency ωt = 2π f , across the capacitor. qC = CVC = CV0 sin ωt (2) The current ic = dqC dt = ωCV0 cos ωt = ωCV0 sin(ωt + 90◦) (3) or ic = V0 Xc sin (ωt + 90◦) (4) where XC = 1/ωC (5) Comparison of (4) with (1) shows that ic leads VC by 90◦ or quarter of a cycle. Further, the current amplitude I0 = V0 XC (6) By Ohm’s law XC is to be regarded as impedance offered by the capaci- tor. In complex plane Zc = −j ωC (7) where j is imaginary. Impedance of an inductance On applying an AC across an inductance the potential difference will be VL = V0 sin ωt (8) where V0 is the amplitude of VL. By Faraday’s law of induction (ξ = −L di/dt) we can write 14.3 Solutions 661 VL = L diL dt (9) Combining (8) and (9) diL dt = V0 L sin ωt (10) ∴ iL = ∫ diL = V0 L ∫ sin ωt dt = − ( V0 ωL ) cos ωt ∴ iL = ( V0 XL ) sin(ωt − 90◦) (11) where X L = ωL (12) is known as the inductive impedance. In complex plane X L = jωL , where j is imaginary. Comparison of (11) with (8) shows that the current in the inductance lags behind the voltage by 90◦ or quarter of a cycle. ii. ZL = √ R2 + ω2L2 = √ (44)2 + (2π × 150 × 0.06)2 = 71.63  ZC = √ R2 + 1 ω2C2 = √ 102 + 1 (2π × 150 × 10−4)2 = 14.58  14.24 (a) Electric current is the time rate of flow of charge; in symbols I = dQ/dt . (b) The charge Q is an integral multiple of the unit of electron’s charge e, that is, Q = ne, where n is a number. The charge Q is said to be quantized (c) (i) Current density j = i A = 2.4 × 10−4 (5.6 × 10−3)(50 × 10−6) = 857 A/m2 (ii) Drift speed Vd = j ne = 857 (8.5 × 1028)(1.6 × 10−19) = 6.3 × 10−8 m/s (iii) Collisions with atoms and ions of the conductor makes possible large currents to pass. 14.25 By problem ω = 1√ L1C1 = 1√ L2C2 (1) When the combinations L1C1 and L2C2 are connected in series, the combi- nation will have inductance L and capacitance C given by L = L1 + L2 (2) C = C1C2 C1 + C2 (3) Now LC = (L1 + L2) C1C2 C1 + C2 (4) 662 14 Electromagnetism II From (1) we have L2 = L1C1 C2 (5) Substituting (5) in (4) we get on simplification LC = L1C1 = 1 ω2 It follows that ω = 1√ LC 14.26 Zc = 1 ωc = 1 2π fc ∴ C = 1 2π fzc = 1 2π × 1000 × 500 = 3.18 × 10−7 F = 0.318 μF ZL = ωL = 2π fL ∴ L = ZL 2π f = 100 2π × 5000 = 3.18 × 10−3 H = 3.18 mH 14.27 The maximum stored energy in the capacitor must equal the maximum stored energy in the inductor, from the principle of energy conservation. ∴ 1 2 q2 m C = 1 2 Li2 m (1) where im is the maximum current and qm is the maximum charge. Substitut- ing CV0 for qm and solving for im in (1) im = V0 √ C L = 100 √ 0.01 × 10−6 10 × 10−3 = 0.1 A 14.28 For resonance ω = 1√ LC ∴ C = 1 4π2 f 2L = 1 4π2(5 × 105)2 × 10−3 = 1.013 × 10−9 F Quality factor Q = ωL R ∴ R = 2π fL Q = 2π(500 × 103)(10−3) 150 = 20.944  ∴ Resistance to be included in series is 20.944 − 5.0 = 15.944 . 14.3 Solutions 665 14.3.2 Maxwell’s Equations and Electromagnetic Waves, Poynting Vector 14.33 Ez = 100 cos(6 × 108t + 4x) (by problem) (1) Ez = A cos(ωt + kx) (standard equation) (2) Comparison of (1) and (2) shows that ω = 6 × 108 and k = 4 v = ω k = 6 × 108 4 = 1.5 × 108 m/s Dielectric constant, K = c v = 3 × 108 1.5 × 108 = 2.0 14.34 ∮ E · ds = q/ε0 (Gauss’ law) E(2π rl) = q/ε0 ∴ E = λêr 2πr ε0 (1) ∮ B · ds = μ0 I (Ampere’s law) 2πrB = μ0 I ∴ B = μ0 I 2πr êφ (2) E ′ r = γ (Er − vBφ) (Lorentz transformation) (3) = γ ( λ 2πr ε0 − v μ0 I 2πr ) = γ 2π r ( λ ε0 − vμ0 I ) Thus E ′ r = 0 if vμ0 I = λ ε0 or v = λ Iμ0 ε0 = λc2 I 14.35 Maxwell’s equations in vacuum are ∇ · E = 0 (1) ∇ · B = 0 (2) ∇ × E = −∂B ∂t (3) ∇ × B = μ0ε0 ∂E ∂t (4) 666 14 Electromagnetism II Use the vector identify ∇ × (∇ × B) = ∇(∇ · B) − ∇2B ∴ ∇ × ( μ0ε0 ∂E ∂t ) = −∇2 B (∵ ∇ · B = 0 by(2) ∴ μ0ε0 ∂ ∂t (∇ × E) = −μ0ε0 ∂ ∂t ∂B ∂t = −∇2B ∴ ∇2B = μ0ε0 ∂2B ∂t2 14.36 ∇ × B = μ0 j (Ampere’s law) (1) Use the vector identity A · (A × B) = 0. Put A = ∇. ∴ ∇ · (∇ × B) = 0 ∴ ∇ · j = 0 (2) More generally, ∇ · j + ∂ρ ∂t = 0 (continuity equation) (3) and ∇ · E = ε0ρ (Gauss’ law) (4) Combining (3) and (4) ∇ · ( j + 1 ε0 ∂E ∂t ) = 0 14.37 ∇2B = μ0ε0 ∂2B ∂t2 (free-space wave equation) Compare with the standard three-dimensional wave equation ∇2 = 1 v2 ∂2 ∂t2 (5) ∴ v = 1√ μ0ε0 = 1√ (4π × 10−7)(8.854 × 10−12) = 2.998 × 108 m/s = c 14.38 ω = 2πν = 2πc λ = 2π × 3 × 108 530 × 10−9 = 3.55 × 1015 σ = 1 ρ = 1 26.5 × 10−9 = 3.77 × 107 δ = √ 2 μ0σω = √ 2 4π × 10−7 × 3.77 × 107 × 3.55 × 1015 = 3.45 × 10−9 m = 3.45 nm 14.3 Solutions 667 14.39 E = E0 cos(kx − ωt) (1) ∂ Ey ∂x = −k E0 sin(kx − ωt) (2) B = B0 cos(kx − ωt) (3) ∂ BZ ∂t = ωB0 sin(kx − ωt) (4) But ∂ Ey ∂x = −∂ Ez ∂t (5) Combining (2), (4) and (5), we get E0 = ω k B0 = cB0 14.40 Maxwell’s equations for a non-ferromagnetic homogeneous isotropic medium can be written as ∇ · E = ρ ε (1) ∇ · B = 0 (2) ∇ × E = −∂B ∂t (3) ∇ × B = μσE + με ∂E ∂t (4) Taking the curl of (4) ∇ × (∇ × B) = μσ(∇ × E) + με ∂ ∂t (∇ × E) (5) where the time and space derivatives are interchanged as E is assumed to be a well-behaved function. Expression (3) can be substituted in (5) to obtain ∇ × (∇ × B) = −μσ ∂B ∂t − με ∂2B ∂t2 (6) Using the vector identity ∇ × (∇ × B) = ∇ (∇ · B) − ∇2B (7) By virtue of (2), ∇ · B = 0 and (6) becomes ∇2B = με ∂2B ∂t2 + μσ ∂B ∂t (8) 670 14 Electromagnetism II Substituting (2) in (1) u B = 1 2 μ H · H = 1 2 μ H2 = B2 2μ (3) In free space μ = μ0. Therefore in vacuum uB = B2 2μ0 14.44 uE = 1 2 ε0 E2 (energy density in E-field) (1) uB = 1 2 B2 μ0 (energy density in B-field) (2) The fields for the plane wave are E = Em sin(kx − ωt) (3) B = Bm sin(kx − ωt) (4) Substituting (3) in (1) and (4) in (2) uE = 1 2 ε0 E2 m sin2(kx − ωt) (5) uB = 1 2 B2 m μ0 sin2(kx − ωt) (6) Dividing (5) by (6) uE uB = ε0μ0 E2 m B2 m (7) But ε0μ0 = 1 c2 and Em = cBm ∴ uE uB = 1 or uE = uB 14.45 By prob. (14.41) the magnetic energy stored in a coaxial cable UB = μ0i2l 4π ln(b/a) (1) where i is the current and l is the length of the cable. Further, its capacitance is given by C = 2πε0l ln(b/a) (2) 14.3 Solutions 671 The electric energy stored in the cable UE = 1 2 ξ2C = 1 2 (i2 R2) ( 2πε0l ln(b/a) ) (3) By problem UE = UM . ∴ 1 2 (i2 R2) ( 2πε0l ln(b/a) ) = μ0i2l 4π ln(b/a) (4) ∴ R = 1 2π √ μ0 ε0 ln(b/a) = 1 2π √ 4π × 10−7 8.85 × 10−12 ln(b/a) = 376.7 2π ln(b/a). 14.46 As the kinetic energy of proton (20 MeV) is much smaller than its rest mass energy (938 MeV), non-relativistic calculations will be valid. In the classical picture a charged particle undergoing acceleration ‘a’ emits electromagnetic radiation. The electromagnetic energy radiated per second is given by P = q2a2 6πε0c3 (1) Now a = v2 R = 1 2 mv2 2 mR = 2K mR (2) Substituting (2) in (1) and putting q = e for the charge of proton P = 2e2 K 2 3πε0c3m2 R2 (3) The energy radiated per orbit is given by multiplying P by this time period 2π R/v = 2π R √ m/2K ∴ K = 4e2 K 3/2 3 √ 2ε0c3m3/2 R = 4(1.6 × 10−19)2(20 × 1.6 × 10−13)3/2 3 √ 2(8.85 × 10−12)(3 × 108)3(1.67 × 10−27)3/2(0.5) = 1.89 × 10−31 J = 1.18 × 10−12 eV which is quite negligible 14.47 Consider a sphere of radius r with its centre at the point source of power P . Then the intensity of radiation at distance r will be I = P 4πr2 . 672 14 Electromagnetism II I = P 4πr2 = c ε0 E2 0 2 ∴ E0 = [ P 2π r2cε0 ]1/2 = [ 40 2π(1.0)2(3 × 108 × 8.85 × 10−12) ]1/2 = 49 V/m 14.48 (a) I0 = P A = 1.2 × 10−3 4 × 10−6 = 300 W/m2 (b) I0 = cε0 E2 0 2 → E0 = [ 2I0 cε0 ]1/2 = [ 2 × 300 3 × 108 × 8.85 × 10−12 ]1/2 = 475 V/m (c) B0 = E0 c = 475 3 × 108 = 1.58 × 10−6 T = 1.58 μT 14.49 I = P A = P πr2 = 314 × 10−3 3.14 × (0.5 × 10−3)2 = 4 × 105 W/m2 14.50 The average value of the Poynting vector < s > over a period of oscillation of the electromagnetic wave is known as the radiant flux density, and if the energy is incident on a surface it is called irradiance. E = E0 cos(kx − ωt), B = B0 cos(kx − ωt) ∴ S = c2ε0E × B = c2ε0E0 × B0 cos2(kx − ωt) ∴ 〈S〉 = c2ε0 |E0 × B0| 〈 cos2(kx − ωt) 〉 But 〈 cos2(kx − ωt) 〉 = 1 T T∫ 0 cos2(kx − ωt)dt = 1 2 Further E⊥B and E0 = cB0. ∴ < s >= I = 1 2 cε0 E2 0 = 1 2 E2 0 μ0c 14.51 Ez = 50 sin [ 4π × 1014 ( t − x 3 × 108 )] I = cε0 E2 0 2 = 1 2 (3 × 108)(8.85 × 10−12)(502) = 0.066 W/m2 14.52 |E × H| = |E × B| μ0 = E2 μ0c (∵ B = E/c) E × H will be in the direction of S = (E × B)/μ0, which is the direction of propagation. 14.3 Solutions 675 14.58 Magnetic energy density uB = 1 2 B2 μ0 = 1 2 42 4π × 10−7 = 2 π × 107 J/m3 Energy stored in the magnetic field UB = u B × volume = uBπr2l = 2 × 107 × 32 × 12.5 = 2.25 × 109 J 14.59 If P0 is the power radiated by sun of radius r , then using the results of prob. (14.51) P0 = 〈S〉 4πr2 = 1 2μ0c E2 m4πr2 ∴ Em = 1 r √ P0μ0c 2π ∴ Em = 1 7 × 108 √ 4 × 1026 × 4π × 10−7 × 3 × 108 2π = 2.21 × 105 V/m Bm = Em c = 2.21 × 105 3 × 108 = 7.37 × 10−4 T = 7.37 G 14.60 By prob. (14.51) 〈S〉 = E2 m 2μ0c ∴ Em = √ 2μ0c 〈S〉 = √ 2 × 4π × 10−7 × 3 × 108 × 1300 = 990 V/m 14.61 Bm = Em c = 300 3 × 108 = 1 × 10−6 T 〈S〉 = E2 m 2μ0c = (300)2 2 × 4π × 10−7 × 3 × 108 = 119.4 W/m2 14.62 |E | |H | = E μ0 B = (Bc)μ0 B = μ0√ μ0ε0 = √ μ0 ε0 where we have used the equations B = μ0 H and c = 1√ μ0ε0 . √ μ0 ε0 = √ 4π × 10−7 8.85 × 10−12 = 377  Units of |E | |H | are Volt/metre Ampere/metre = Volt Ampere = Resistance 676 14 Electromagnetism II 14.63 Skin depth δ = √ 2 2π f σμ0μ = √ 2 2π × 20 × 103 × 6 × 107 × 4π × 10−7 × 1 = 4.6 × 10−4m = 0.46 mm 14.64 δ = √ 2 2π f σμ0μ = √ 1 π × 3 × 108 × 5.6 × 107 × 4π × 10−7 × 1 = 1.23 × 10−4 m = 0.123 mm = 3.88 μm 14.65 Let us begin with the ‘curl H ’ Maxwell’s equation ∇ × H = J + ∂D ∂t (1) Assume that all the fields and currents in this equation are sinusoidal at a single frequency ω. In that case (1) can be replaced by ∇ × H = J + jωD = J + jωεE (2) where J is the current density, D = εE is the displacement vector and j is imaginary (√−1 ) . The ‘curl E’ Maxwell’s equation is obtained in a similar way as (1). ∇ × E = −jωB (3) Applying the curl operation to each side of (3) ∇ × (∇ × E) = −jω∇ × B (4) Now ∇ × B can be replaced using (2) and the relation H = B/μ: ∇ × (∇ × E) = −jωμ(J + jωεE) (5) We now use the vector identity ∇ × ∇ × E = ∇(∇ · E) = −∇2E (6) Conductive materials do not contain any real charge density because any real charge that many exist will repel itself and move outwards until it resides on 14.3 Solutions 677 the material’s outer surface. Therefore ∇ ·D = 0 and so also ∇ ·E = 0. Thus the first term on the right by (6) vanishes. Therefore (5) becomes ∇2E = jωμ(J + jωεE) (7) Assume that the material under consideration obeys Ohm’s law, J = σE E, where σE is the conductivity. Then (7) becomes ∇2E = jωμ(σE + jωε)E (8) For simplicity assume that the given material is an excellent conductor, so that σE >> |ωε|. In that case, the displacement current term, masked by the conduction current, can be neglected, yielding ∇2E = jωμσEE (9) Since E = J/σE we can write ∇2J = jωμσEJ (10) Suppose the current flows through this material in the z-direction. The cur- rent density is independent of x and y. In that case (10) simplifies to ∂2 Jz ∂z2 = jωμσE Jz (11) which has the solution J x = Ae−(1+j)z/δ + Be(1+j)z/δ (12) where A and B are constants, and δ given by δ = √ 2 ωμσE (13) is known as the skin depth. Thus the magnitude of current density decreases with depth. This effect is of practical importance as it affects resistive losses accompanying a high-frequency current flow in an electronic circuit. 14.66 The average value of the Poynting vector is 〈S〉 = 1 2μ0 Em Bm = 1 2μ0 Em ( Em c ) = E2 m 2μ0c = ( 10−3 )2 2 × 4π × 10−7 × 3 × 108 = 1.327 × 10−9 W/m2 = 1.327 × 10−13. W/cm2. 680 14 Electromagnetism II Thus ∇ × (∇ × F) = −6x2zî + 6xz2k̂ (1) − ∇2 F + ∇(∇ · F) = −(6x2z + 2z3)î + 2z3 î + 6xz2k̂ = −6x2zî + 6xz2k̂ (4) Comparing (1) and (4), the identity ∇ × (∇ × E) = −∇2 E + ∇(∇ · E) is verified. 14.70 The electric field of the cable is radial and is given by E = Er = V r ln(b/a) (1) where a and b are the radii of the inner and outer cable. The corresponding magnetic intensity is tangential and is given by H = Hφ = I 2πr (2) As the angle between E and H is 90◦, the Poynting vector |S| = |E × H| = EH (3) So that S = Sz = VI 2πr2 ln(b/a) (4) and the direction of S is that of the current in the positive conductor. The power flow is confined to the space between the conductors and for any plane perpendicular to the axis of the conductor P = b∫ a Sz2πrdr = b∫ a VI ln(b/a) dr r (5) where we have substituted Sz from (4). But b∫ a dr r = ln(b/a) (6) ∴ P = VI (7) This is the entire power transmitted by the cable. It follows that the Poynting theorem indicates that the entire flow of energy resides in the space between the conductors. 14.3 Solutions 681 If the resistance of the cable cannot be neglected then V is no longer constant. An axial component of E is necessary to maintain the flow of current to compensate for the Ohmic energy loss. 14.71 The current density in the wire is (I/πa2)êz . Therefore the electric field in the wire, including on the surface of the wire, will be (I/πa2σE)êz . The magnetic field intensity by Ampere’s theorem is (I/2πa2)êφ . The Poynting vector at the surface is given by S = E × H = ( I πa2σE êz ) × ( I 2πa êφ ) = − I 2 2π2a3σE êr (1) Now Poynting’s theorem is − ∫ s (E × H) · ds = ∂ ∂t ∫ V ε |E|2 2 dV + ∂ ∂t ∫ V μ |H|2 2 dV + ∫ V E · J dV (2) Since the fields are constant in time, the first two terms on the right of (2) which contain time derivative ∂/∂t vanish. The power dissipated in the wire is then P = ∫ E · JdV = − ∫ (E × H) · ds = − I 2 2π2σEa3 (2πaL) = I 2L πσEa2 = I 2 R 14.72 Given B = − me ne2 ∇ × J (1) Use the vector identity ∇ × (∇ × B) = −∇2B + ∇(∇ · B) (2) Use Maxwell’s equations ∇ · B = 0 (3) ∇ × B μ0 = J + ε0 ∂E ∂t (4) Here ∂D ∂t = ε0 ∂E ∂t = 0 (5) so that (4) becomes ∇ × B = μ0J (6) Using (3) and (6) in (2) 682 14 Electromagnetism II μ0∇ × J = −∇2B (7) Using (1) in (7) ∇2B = μ0ne2B me 14.73 High-frequency resistance Rs = √ ωμ 2σ (1) Direct current resistance per metre R = 1 σ A = 1 σπr2 (2) Further the skin thickness δ = √ 2 μσω (3) For metals assume μ = μ0. Combining (1), (2) and (3) Rs R = πr2 δ √ μ0 2 = π(10−3)2 6.6 × 10−5 √ 4π × 10−7 2 = 3.77 × 10−5 14.74 When a charge q moves in a magnetic field, it experiences a magnetic force Fm = qv × B (1) When an electric conductor is physically moved across a magnetic field, the free electrons in the conductor will experience a force on them in the direc- tion of the force. The flow of electrons implies the existence of a potential difference between the ends of the conductor. The situation is the same as if an electric field had been set up in the conductor which is expressed by the relation Em = Fm/q = v × B V/m (2) Equation (2) implies that every moving magnetic field is accompanied by an electric field. From (2) and the definition of the emf ξ of a source, the instant emf of the source is 14.3 Solutions 685 and ∇ × H = J + ∂D ∂t (2) Define a new electric potential function φ(r, t) such that E = −∇φ − ∂A ∂t (3) The reason for redefining the scalar potential in this fashion is that (3) is consistent with Faraday’s law, ∇ × E = − (∂B/∂t), as can be verified by substitution. On the other hand the electrostatic definition, E = −∇V is inconsistent with Faraday’s law and therefore cannot be used in electrody- namics. However, the relation B = ∇ × A, continues to be correct. Substi- tuting (3) in (1), we obtain ∇2φ + ∂ ∂t (∇ · A) = −ρ ε (4) Substituting B = ∇ × A in (2), we get ∇2A − ∇(∇ · A) = −μJ − με ∂E ∂t (5) It is convenient to choose a Lorentz gauge given by ∇ · A = − 1 c2 ∂ϕ ∂t (Lorentz condition) (6) With the use of (6), (4) and (5) are simplified to ∇2ϕ − 1 c2 ∂2ϕ ∂t2 = −ρ ε (7) ∇2A − 1 c2 ∂2A ∂t2 = −μJ (8) 14.77 ∇ × E = −∂B ∂t (Faraday’s law) (1) Writing the curl in rectangular form gives ∂ Ey ∂z = −∂ Bx ∂t Integrating in time and choosing the constant of integration as zero, we obtain Bx = 1 v f (z − vt) (2) 686 14 Electromagnetism II Notice that the variation of B is exactly the same as the variation of E , except that Ey and Bx are at right angles to each other and perpendicular to the direction of propagation. From (2) and the relation Hx = Bx/μ0 we find Hx = 1 μ0v f (x − vt) (3) so that Ey = μ0vHx (4) Using the relation v = √ 1/μ0ε0, we can write (4) in the form Ey = Z0 Hx with Z0 = √ μ0 ε0 = 376.6 . 14.78 (a) E(z, t) = E0[x̂ sin(kz − ωt) + ŷ cos(kz − ωt)]. This is a plane polar- ized wave polarized in the xy-plane and propagating in the positive z-direction. (b) The magnetic lines will suffer refraction in passing from one magnetic medium to another. (i) The continuity of B lines is first specified as a necessary condi- tion. Figure 14.8 shows a bundle of B lines in passing through the interface between two magnetic media characterized by μ1 and μ2. (ii) Since div B = 0, it is required that the magnetic flux associated with the flux lines be constant in passing through the interface. φ = B1 ds1 = B2 ds2 Fig. 14.8 The refraction of magnetic lines where ds1 and ds2 are the cross-section of the flux lines in medium 1 and 2, respectively. Dividing by ds, the corresponding area on the interface, we get B1 ds1 ds = B2 ds2 ds 14.3 Solutions 687 which from Fig. 14.8 may be written as B1 cos θ1 = B2 cos θ2 (1) which may be written as B1 · n̂ = B2 · n̂ which shows that the normal component of the B vector is the same on both sides of the boundary. (iii) Next we apply Ampere’s circuital law to the path across the inter- face, Fig. 14.8. Assuming that no current exists in the interface, for the path considered ∮ H · dl = 0 Breaking the integral into individual parts of the path ∮ H · dl = b∫ a H · dl + c∫ b H · dl + d∫ c H · dl + a∫ d H · dl = 0 In the limit the path shrinks approaching the interface c∫ b H · dl = a∫ d H · dl = 0 ∴ b∫ a H · dl + d∫ c H · dl = 0 Thus Ht1 = Ht2 i.e. H1 × n̂ = H2 × n̂ (2) This implies that the tangential component of the H vector is the same on both sides of the boundary. (c) Dividing (2) by (1) H1 sin θ1 B1 cos θ2 = H2 sin θ2 B2 cos θ2 ∴ 1 μ1 tan θ1 = 1 μ2 tan θ2 ∴ tan θ1 tan θ2 = μ1 μ2 (law of refraction) 690 14 Electromagnetism II For the transmitted wave the relations are lT = x sin φ + z cos φ (5) ⎧ ⎨ ⎩ ET = ETe−jkTlT(cos φ ex − sin φez) HT = ET η2 e−jkTlT ey (6) The boundary conditions at z = 0 require that the tangential E that is Ex and tangential Hy be continuous. Setting Ex (z = 0) in region 1 equal to Ex (z = 0) in region 2, E0 cos θ e−jkIx sin θ + ER cos θR e−jkIx sin θR = ET cos φe−jkTx sin φ (7) If this equation is to hold for all values of x then kI sin θ = kI sin θR = kT sin φ (8) It follows that θR = θ , that is, the angle of incidence is equal to angle of reflection as in a plane mirror. Further sin φ sin θ = kI kT = √ ε1√ ε2 = n1 n2 (9) where n is the index of refraction of the material. Equation (9) then gives Snell’s law (n1 sin θ = n2 sin φ) which holds irrespective of the nature of polarization. Using (9) in (7) and cancelling the exponential terms we have E0 cos θ + ER cos θ = ET cos φ (10) Using (8), the boundary conditions on Hy yield E0 η1 − ER η1 = ET η2 (11) Solving (10) and (11) we get ER E0 = η2 cos φ − η1 cos θ η2 cos φ + η1 cos θ (12) ET E0 = 2η2 cos θ η2 cos φ + η1 cos θ . (13) Substituting η1 = √ μ1/ε1, η2 = √ μ2/ε2, μ1 = μ2 = μ0 for non- magnetic dielectric and using (9) in (12), we obtain 14.3 Solutions 691 R = ( E1 E0 )2 = ( n2 cos θ − n1 cos φ n2 cos θ + n1 cos φ )2 14.81 (a) Reflectance R = 0 if n2 cos θ − n1 cos φ = 0 (1) Given tan θ = n2 n1 (Brewster’s law of polarization) (2) sin θ sin φ = n2 n1 (Snell’s law of refraction) (3) Combining (2) and (3), we have (Fig. 14.11) sin φ = cos θ (4) or φ = 90◦ − θ (5) Fig. 14.11 R and T against n1/n2 Eliminating n2 between (1) and (2) n2 cos θ − n1 cos φ = n1(sin θ − cos φ) = n1(sin θ − sin θ) = 0 where we have used (5). (b) tan θ = n2 n1 = 1.5 1 = 1.5 ∴ θ = 56.31◦ 14.82 (a) R = (n1 − n2) 2 (n1 + n2) 2 = ( n1 n2 − 1 )2 ( n1 n2 + 1 )2 = (x − 1)2 (x + 1)2 (1) where x = n1 n2 T = 1 − R = 4x (x + 1)2 (2) (b) Setting R = T yields the quadratic equation x2 − 6x + 1 = 0, whose solution is x = 3 + 2 √ 2 or 5.828. Thus for n1/n2 = 5.828, we get R = T = 0.5. 692 14 Electromagnetism II 14.83 Using Maxwell’s equations ∇ · E = 0 (1) ∇ · B = 0 (2) ∇ × E = −∂ B ∂t (3) (a) E = E0ei(ωt−k.r+φ) (4) B = B0ei(ωt−k.r+φ) (5) Let the wave be propagated in the z-direction. Then ∇ · E = −i k · E = 0 and ∇ · B = −i k · B = 0 This shows that both E and B are perpendicular to k, which is the direc- tion of propagation. Therefore both E and B are transverse oscillations. (b) B and E are in phase (c) Using (3) ∇ × E = −i k × E = −iωB ∴ B = k × E ω = 1 c k × E k ∴ B = 1 c ŝ × E (6) where ŝ = k/k is a unit vector in the direction of propagation. The three vectors E, B and k form a right-handed rectangular coordinate system. From (6) we obtain B = E/c. 14.84 E = σ 2εrε0 ∴ σ = 2ε0εr E = 2 × 8.85 × 10−12 × 6 × 2 × 103 = 2.124 × 10−7C/m2 14.3.3 Phase Velocity and Group Velocity 14.85 Using the result of prob. 14.93 1 Vg − 1 Vp = Vp − Vg VgVp = Vp − Vg c2 − = ω c dn dω 14.3 Solutions 695 Now v = c/n (2) ∴ dv dω = dv dn dn dω = − c n2 dn dω (3) Substituting (2) and (3) in (1) and using ω = kv and rearranging we get vg = c n + ω(dn/dω) 14.92 Vg = distance time = 50 1 × 10−6 = 5 × 107 m/s Vg = c √ 1 − (λ/2a)2 ∴ 5 × 107 = 3 × 108 √ 1 − (λ/5)2 Solving for λ, we find the free-space wavelength λ = 4.93 cm. Vp = c2 Vg = (3 × 108)2 5 × 107 = 1.8 × 109 m/s 14.93 Vg = dω/dk (1) Rewriting (1), 1/Vg = dk/dω ∴ 1 Vg = d dω ( ω Vp ) = 1 Vp − ω V 2 p dVp dω (2) Substituting Vp = c/n in (2) 1 Vg = 1 Vp − ωn2c c2 ( − 1 n2 dn dω ) = 1 Vp + ω c dn dω (3) 14.94 Vg = ∂ω ∂k ∴ 1 Vg = ∂k ∂ω = ∂ (2π/λ) ∂(2πν) = ∂ (1/λ) ∂ν But n = c vp = c νλ → 1 λ = nν c ∴ Vg = ∂ν ∂ ( 1 λ ) = ∂ν ∂ (nν/c) = c∂ν ∂(n ν) 696 14 Electromagnetism II 14.95 vg = dω dk = d(ωh̄) d(kh̄) = dE dp = d ( p2/2m ) dp = 1 2m dp2 dp = 2p 2m = mv m = v 14.96 By prob. (14.85) Vp − Vg Vp = −nλ dn dλ (1) where Vp = c n (2) Re-arranging (1) with the aid of (2) and writing μ for n we find Vg = c [ 1 μ + λ d μ d λ ] (3) μ = 1.420 + 3.60 × 10−14 λ2 (by problem) (4) Substituting λ = 500 nm = 5 × 10−7 m in (4) μ = 1.564 (5) Differentiating μ with respect to λ in (4) dμ dλ = −7.2 × 10−14 λ3 or λ dμ dλ = −7.2 × 10−14 λ2 = −0.288 (6) Substituting (5) and (6) in (3), we find Vg = 0.35 c. 14.3.4 Waveguides 14.97 (a) Assuming the dominant mode, for a = 2.5 cm, and b = 2.5 cm, for the rectangular waveguide, and c = 3×108 m/s, the velocity of electromag- netic waves in free space, the phase velocity is given by Vp = c√ 1 − (λ/2a)2 = 3 × 108 √ 1 − (4/5)2 = 5 × 108 m/s (b) Vg = c √ 1 − (λ/2a)2 = 3 × 108 √ 1 − (4/5)2 = 1.8 × 108 m/s (c) λg = λ√ 1 − (λ/2a)2 = 4√ 1 − (4/5)2 = 6.67 cm 14.3 Solutions 697 14.98 λg = λ√ 1 − (λ/2a)2 (1) λg = 3λ (by problem) (2) Combining (1) and (2) and solving for λ with a = 3 cm we find the free- space wavelength λ = 4 √ 2 cm. 14.99 (a) λg = λ√ 1 − (λ/2a)2 = 8√ 1 − (8/10)2 = 13.33 cm (b) The cut-off wavelength λc = 2a = 2 × 5 = 10 cm. 14.100 N = 8πν2dνV c3 = 8πdλ · V λ4 Mean λ = 5500Å = 5.5 × 10−7 cm dλ = (6000 − 5000)Å = 10−7 cm V = (0.5)3 cm3 N = 8π × 10−7 × (0.5)3 (5.5 × 10−7)4 = 3.43 × 1018 14.101 The cut-off frequency of the TMmn or TEmn mode is ωmn = c [(πm a )2 + (πn b )2 ]1/2 The lowest value we can have for νmn is for the choice m = 1, a = 15 and n = 0, that is, for TM10 wave. ∴ ν10 = c 2 × 1 15 = 3 × 108 30 = 107 Hz = 104 kHz This is much above the range of AM waves (530–1600 kHz). Hence AM waves cannot propagate in the tunnel. 14.102 The cut-off frequency will be least for TE10 waves. Of course TM10 waves do not exist. ν10 = c 2π π a = c 2a = 3 × 108 2 × 0.05 = 3 × 109 Hz = 3 GHz Note that we take the higher dimension (5 cm) for the lower value of cut-off frequency. 700 14 Electromagnetism II where m and n are positive integers. Finally, we get the result Ez = K sin (m π x a ) sin (n π y b ) e−jkz (5) where K = (B)(D) is another constant. Note that Ez vanishes if either m = 0 or n = 0. In that case TMm0 or TM0n wave does not exist. The other field components are given from equations which are obtained by manipulating Maxwell’s equations. Ex = − j ω2με − k2 ( k ∂ Ez ∂x + ωμ ∂ Hz ∂y ) (6) Ey = − j ω2με − k2 ( −k ∂ Ez ∂y + ωμ ∂ Hz ∂x ) (7) Using (5) in (6) and (7) and putting Hz = 0 Ex = − j K kmπ (ω2με − k2)a cos mπx a sin nπx b e−jkz (8) Ey = − j K knπ (ω2με − k2)b sin mπx a cos nπy b e−jkz (9) The solutions (5), (8) and (9) represent an infinitely large family of waves, characterized by different values of the integers m and n. They differ from one another by the values of the integers m and n. They also differ in their velocity as well as field configuration. (b) Combining (3) and (4) we get k2 = ω2 c2 − (mπ a )2 − (nπ b )2 (10) where we have substituted με = 1/c2. The cut-off frequency is obtained by setting k = 0. ωmn = πc [(m a )2 + (n b )2 ]1/2 (11) 14.105 (a) For TE waves there is no Ez. Here, we must find boundary conditions on Hz that cause the tangential component of Ez to vanish. The given equation is ( ∂2 ∂x2 + ∂2 ∂y2 ) Hz = (k2 − ω2με) Hz (1) 14.3 Solutions 701 The general solution is Hz = (A cos kx x + B sin kx x)(C cos ky y + D sin ky y)e−jkz (2) ∴ ∂ Hz ∂x = (−k2 A sin kx x + Bk2 cos kx x) (C cos ky y + D sin ky y)e−jkz For ∂ Hz ∂x = 0 at x = 0, it is necessary that B = 0. Similarly ∂ Hz ∂y = (A cos kx x + B sin kx x)(−Cky sin ky y + D cos ky y)e−jkz For ∂ Hz ∂y = 0, D = 0. Therefore (2) becomes Hz = K cos(kx x) cos(ky y) e−jkz (3) where K is the product of A and B is another constant. Imposing boundary conditions at x = a and y = b, we have ∂ Hz ∂x = −K kx sin kx x cos ky y = 0 yielding kx a = mπ (4) and ∂ H ∂y = −K ky cos kx x sin ky y = 0 yielding kyb = nπ (5) ∴ Hz = K cos (mπx a ) cos (nπy b ) (6) Substituting (2) in (1) we obtain k2 x + k2 y = ω2με − k2 (7) which is identical with (3) of prob. (14.104). (b) Substituting the values of kx and ky from (4) and (5) in (7) and setting k = 0 gives the cut-off frequency. 702 14 Electromagnetism II ωmn = πc [(m a )2 + (n b )2 ]1/2 (8) which is identical with (11) of prob. (14.104). (c) Thus the features which are identical for the TM and TE modes are (i) ω − k plot (ii) Phase (iii) Group velocity (iv) Cut-off frequency However the important difference is that when either m = 0 or n = 0, the TM mode fails to exist. On the other hand Hz does not vanish for m = 0 or n = 0 (see (6)). Because of this fact the TE10 is the mode which has the lowest cut-off frequency. Here we have assumed that a > b. The cut-off frequency for TE10 mode is given by ω10 = π c a (9) the free space wavelength being 2a. The small dimension (b) has no bearing on the cut-off frequency for this mode. The other advantage is that single mode operation is feasible over a wide range of frequencies.