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Lecture 14 : Electromagnetic Waves
Wave equations Properties of the waves Energy of the waves
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Recap (1) π΅ Γ π¬ =β π π© ππ π΅ . π¬ =π π΅ Γ π© = π π πΊ π π π¬ ππ π΅ . π© =π
In a region of space containing no free charge ( π π =0) or current ( π½ π = 0 ), Maxwellβs equations can be written as π΅ Γ π¬ =β π π© ππ π΅ . π¬ =π π΅ Γ π© = π π πΊ π π π¬ ππ π΅ . π© =π
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Recap (2) 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) This creates a potential energy stored in the electromagnetic field with energy densities π 0 πΈ 2 and π΅ 2 2 π 0 , respectively
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Wave equations We will now explore how Maxwellβs equations lead to wave motion in the form of light β an electromagnetic wave
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Wave equations A wave is a propagating disturbance that carries energy from one point to another A wave may be characterized by its amplitude, wavelength and speed and can be longitudinal or transverse speed
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Wave equations To discover EM waves, consider evaluating π» Γ π» Γ πΈ
Substituting in Maxwellβs equations, this equals π» Γ β π π΅ ππ‘ =β π ππ‘ π» Γ π΅ =β π ππ‘ π 0 π 0 π πΈ ππ‘ =β π 0 π 0 π 2 πΈ π π‘ 2 Also by vector calculus, π» Γ π» Γ πΈ = π» π» . πΈ β π» 2 πΈ Using Maxwellβs equation π» . πΈ =0, this just equals β π» 2 πΈ Putting the two sides together, we find π΅ π π¬ = π π πΊ π π π π¬ π π π
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Wave equations The equation represents a travelling wave
To see why, consider a 1D wave travelling along the π₯-axis with displacement π¦ π₯,π‘ =π΄ sin (ππ₯βππ‘) in terms of its amplitude π΄, wavelength π= 2π π , time period π= 2π π , velocity π£= π π = π π
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Wave equations The displacement π(π,π) of a wave travelling with speed π satisfies the wave equation π π π π π π = π π π π π π π π π To check this, substitute in π¦ π₯,π‘ =π΄ sin (ππ₯βππ‘) where π£= π π On the left-hand side: π 2 π¦ π π₯ 2 =β π 2 π΄ sin (ππ₯βππ‘) On the right-hand side: π 2 π¦ π π‘ 2 =β π 2 π΄ sin (ππ₯βππ‘) Using 1 π£ 2 = π 2 π 2 , we recover the wave equation!
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Wave equations The equation π π π π π π = π π π π π π π π π describes general wave travel
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Wave equations The relation π» 2 πΈ = π 0 π 0 π 2 πΈ π π‘ 2 implies that each component of the electric field is a wave travelling at speed π= π π π πΊ π =π Evaluating, we obtain π=3Γ m/s : the speed of light
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Properties of the waves
Letβs find out more about the properties of electromagnetic waves To simplify the case, suppose that the electric field is along the π¦-axis, such that πΈ π¦ π₯,π‘ =π΄ sin (ππ₯βππ‘) and πΈ π₯ = πΈ π§ =0. This is known as a linearly polarized wave. π¦ π₯
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Properties of the waves
We can deduce the π΅ -field using the relation π» Γ πΈ =β π π΅ ππ‘ π₯-component: β π π΅ π₯ ππ‘ = π πΈ π§ ππ¦ β π πΈ π¦ ππ§ =0, hence π΅ π₯ =0 π¦-component: β π π΅ π¦ ππ‘ = π πΈ π₯ ππ§ β π πΈ π§ ππ₯ =0, hence π΅ π¦ =0 π§-component: β π π΅ π§ ππ‘ = π πΈ π¦ ππ₯ β π πΈ π₯ ππ¦ =ππ΄ cos (ππ₯βππ‘) , hence π΅ π§ π₯,π‘ = π΄ π£ sin (ππ₯βππ‘) = 1 π£ πΈ π¦ (π₯,π‘) The magnetic field is perfectly in phase with the electric field and oriented at 90Β°!
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Properties of the waves
The form of a linearly-polarized electromagnetic wave: The changing πΈ -field causes a changing π΅ -field, which in its turn causes a changing πΈ -field
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Properties of the waves
For an electromagnetic wave travelling in a medium with relative permittivity π π and relative permeability π π we can use Maxwellβs Equations to show that speed π= π πΊ π π π The quantity π π π π is known as the refractive index Why does this produce a spectrum of light?
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Energy of the waves Waves transport energy from one place to another
Energy is stored in the electric field with density π 0 πΈ 2 and the magnetic field with density 1 2 π 0 π΅ 2 For a linearly polarized wave, π΅=πΈ/π£, where π£= 1 π 0 π 0 , hence π 0 πΈ 2 = 1 2 π 0 π£ 2 π΅ 2 = 1 2 π 0 π΅ 2 The energy contained in the electric and magnetic fields is equal!
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Energy of the waves What is the rate of energy flow?
The energy flowing past area π΄ in time π‘ is contained in a volume π΄π£π‘ 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£ For linearly-polarized electromagnetic waves, ππ£= 1 π 0 π΅ π§ 2 π£= 1 π 0 πΈ π¦ π΅ π§ The energy flow can be represented by the Poynting vector π π π ( π¬ Γ π© ) π£Γπ‘ Energy density π π΄ Propagation speed π£
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Energy of the waves The Poynting vector, 1 π 0 πΈ Γ π΅ = πΈ Γ π» , is oriented in the direction of travel, perpendicular to the πΈ - and π΅ -fields πΈ Γ π»
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Energy of the waves As an example, consider current πΌ flowing along a cylindrical conductor driven by a potential difference π The energy flux is equal to the Poynting vector 1 π 0 πΈ Γ π΅ This is non-zero at the surface of the conductor, flowing inward If the curved surface area is π΄, the power flowing in is π= 1 π 0 πΈπ΅π΄= 1 π 0 π πΏ π 0 πΌ 2ππ 2πππΏ=ππΌ We recover Joule heating! π΅= π 0 πΌ 2ππ πΈ= π πΏ π πΏ
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Summary Maxwellβs equations predict that the πΈ - and π΅ -fields satisfy linked wave equations with speed π 0 π 0 For a linearly-polarized wave, the πΈ - and π΅ -fields are in phase and oriented at ππΒ° 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 Γ π»
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