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Lecture 14 : Electromagnetic Waves β’ Wave equations β’ Properties of the waves β’ Energy of the waves Recap (1) β’ In a region of space containing no free charge (ππ = 0) or current (π½π = 0), Maxwellβs equations can be written as π΅. π¬ = π ππ© π΅×π¬=β ππ π΅. π© = π ππ¬ π΅ × π© = ππ πΊ π ππ Recap (2) β’ Work needs to be performed when creating an πΈ-field (e.g. by assembling charges), or a π΅-field (e.g. by establishing a current against an inductance) β’ This creates a potential energy stored in the electromagnetic field with energy densities 1 π0 πΈ 2 2 and π΅2 , 2π0 respectively Wave equations β’ We will now explore how Maxwellβs equations lead to wave motion in the form of light β an electromagnetic wave Wave equations β’ A wave is a propagating disturbance that carries energy from one point to another β’ A wave may be characterized by its amplitude, wavelength and speed and can be longitudinal or transverse speed Wave equations β’ To discover EM waves, consider evaluating π» × π» × πΈ β’ Substituting in Maxwellβs equations, this equals π» × β π ππ‘ π»×π΅ = π β ππ‘ ππΈ π0 π0 ππ‘ = ππ΅ β ππ‘ = π2 πΈ βπ0 π0 2 ππ‘ β’ Also by vector calculus, π» × π» × πΈ = π» π». πΈ β π» 2 πΈ β’ Using Maxwellβs equation π». πΈ = 0, this just equals βπ» 2 πΈ β’ Putting the two sides together, we find π΅π π¬ = ππ π¬ ππ πΊ π π ππ Wave equations β’ The equation represents a travelling wave β’ To see why, consider a 1D wave travelling along the π₯-axis with displacement π¦ π₯, π‘ = π΄ sin(ππ₯ β ππ‘) in terms of its amplitude π΄, wavelength π = 2π , π time period π = 2π , π velocity π£ = π π = π π Wave equations β’ The displacement π(π, π) of a wave travelling with speed π satisfies the wave equation ππ π πππ = π ππ π ππ πππ β’ To check this, substitute in π¦ π₯, π‘ = π΄ sin(ππ₯ β ππ‘) where π£ = β’ On the left-hand side: π2 π¦ ππ₯ 2 β’ On the right-hand side: β’ Using 1 π£2 = π2 , π2 = βπ 2 π΄ sin(ππ₯ β ππ‘) π2 π¦ ππ‘ 2 = βπ2 π΄ sin(ππ₯ β ππ‘) we recover the wave equation! π π Wave equations β’ The equation ππ π πππ = π ππ π ππ πππ describes general wave travel Wave equations β’ The relation π»2πΈ = π2 πΈ π0 π0 2 ππ‘ implies that each component of the electric field is a wave travelling at speed π = π ππ πΊπ =π β’ Evaluating, we obtain π = 3 × 108 m/s : the speed of light Properties of the waves β’ Letβs find out more about the properties of electromagnetic waves β’ To simplify the case, suppose that the electric field is along the π¦-axis, such that πΈπ¦ π₯, π‘ = π΄ sin(ππ₯ β ππ‘) and πΈπ₯ = πΈπ§ = 0. This is known as a linearly polarized wave. π¦ π₯ Properties of the waves β’ We can deduce the π΅-field using the relation π» × πΈ = β’ π₯-component: β ππ΅π₯ ππ‘ β’ π¦-component: β ππ΅π¦ β ππ‘) ππΈπ§ ππ¦ β = ππΈπ₯ ππ§ β ππΈπ¦ ππ§ = 0, hence π΅π₯ = 0 ππΈπ§ ππ₯ = 0, hence π΅π¦ = 0 ππΈπ¦ ππ΅π§ ππΈ = β π₯ ππ‘ ππ₯ ππ¦ 1 = πΈπ¦ (π₯, π‘) π£ β’ π§-component: β π΄ sin(ππ₯ π£ ππ‘ = ππ΅ β ππ‘ = ππ΄ cos(ππ₯ β ππ‘), hence π΅π§ π₯, π‘ = β’ The magnetic field is perfectly in phase with the electric field and oriented at 90°! Properties of the waves β’ The form of a linearly-polarized electromagnetic wave: β’ The changing πΈ-field causes a changing π΅-field, which in its turn causes a changing πΈ-field Properties of the waves β’ For an electromagnetic wave travelling in a medium with relative permittivity ππ and relative permeability ππ we can π use Maxwellβs Equations to show that speed π = πΊπ ππ β’ The quantity ππ ππ is known as the refractive index Why does this produce a spectrum of light? Energy of the waves β’ Waves transport energy from one place to another β’ Energy is stored in the electric field 1 with density π0 πΈ 2 and the 2 magnetic field with density 1 π΅2 2π0 β’ For a linearly polarized wave, π΅ = 1 πΈ/π£, where π£ = , hence 1 π0 πΈ 2 2 = π0 π0 1 1 π0 π£ 2 π΅2 = π΅2 2 2π0 β’ The energy contained in the electric and magnetic fields is equal! Energy of the waves β’ What is the rate of energy flow? β’ The energy flowing past area π΄ in time π‘ is contained in a volume π΄π£π‘ π£×π‘ Energy density π Propagation speed π£ π΄ β’ If the energy density is π, this energy is ππ΄π£π‘ hence the energy flow per unit area per unit time is equal to ππ£ β’ For linearly-polarized electromagnetic 1 1 waves, ππ£ = π΅π§ 2 π£ = πΈπ¦ π΅π§ π0 π0 β’ The energy flow can be represented π by the Poynting vector (π¬ × π©) ππ Energy of the waves β’ The Poynting vector, 1 π0 πΈ × π΅ = πΈ × π», is oriented in the direction of travel, perpendicular to the πΈ- and π΅-fields πΈ×π» Energy of the waves β’ As an example, consider current πΌ flowing along a cylindrical conductor driven by a potential difference π π0 πΌ π΅= 2ππ πΈ= π πΏ β’ The energy flux is equal to the 1 Poynting vector πΈ×π΅ π0 β’ This is non-zero at the surface of the conductor, flowing inward π β’ If the curved surface area is π΄, the power flowing in is π = 1 1 π π0 πΌ πΈπ΅π΄ = 2πππΏ = ππΌ π0 πΏ π0 πΏ 2ππ β’ We recover Joule heating! Summary β’ Maxwellβs equations predict that the πΈ- and π΅-fields satisfy linked 1 wave equations with speed π0 π0 β’ For a linearly-polarized wave, the πΈ- and π΅-fields are in phase and oriented at ππ° β’ The energy densities stored in the πΈ- and π΅-fields is equal, and the energy flow per unit area per unit time is equal to the Poynting vector πΈ × π»
