Electromagnetic Waves - Electricity and Magnetism - Notes, Study notes of Electromagnetism and Electromagnetic Fields Theory

Download Electromagnetic Waves - Electricity and Magnetism - Notes and more Study notes Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! Physics 4B Lecture Notes 34-1 Chapter 34 - Electromagnetic Waves Problem Set #13 - due: Ch 34 - 2, 6, 8, 12, 16, 17, 20, 25, 28, 35, 45, 47 Since Maxwell's Equations summarize everything we know about electricity and magnetism, they should lead us to an understanding of the properties of electromagnetic waves. Lecture Outline 1. Producing and Detecting Electromagnetic Waves 2. Properties of Electromagnetic Waves 3. Maxwell's Equations and Waves in Free Space 4. The Electromagnetic Spectrum 5. The Poynting Vector 6. Radiation Pressure and Momentum Transfer 1 . Producing and Detecting Electromagnetic Waves laser and microwave generator & detector Electromagnetic waves can be produced by an oscillating circuit connected to an antenna as shown at the left. The oscillating charges in the antenna set up electric and magnetic fields. EM waves can be detected with an antenna as shown at the right. The charges in the antenna are forces to oscillate by the EM waves and the resulting voltages can be detected with a voltmeter. The waves are initially produced by the charges on the antenna. The charge itself produces an E-field and the motion of the charges (current) produces a B-field as shown at the left. Later when the current switches directions, the fields also switch directions near the antenna. However, the changing fields away from the antenna induce more fields according to Maxwell’s Laws of Electricity and Magnetism as shown at the right. The amazing thing is that these field are self sustaining. The changing magnetic field producing an electric field and the changing electric field produces a magnetic field. This is the nature of the EM waves shown at the left. L R C antenna EM Waves transformer antenna EM Waves V I I I I I Physics 4B Lecture Notes 34-2 2 . Properties of Electromagnetic Waves • They continue to travel after the source is turned off. • They travel through empty space. • They always travel at the same constant speed. • The electric field is always perpendicular to the magnetic field. • The velocity is perpendicular to both the electric field and the magnetic field. • The ratio of the peak electric field to the peak magnetic field equals the speed of the waves. There are a couple of ways to represent these waves that illustrate these properties. The electric and magnetic fields oscillate in amplitude in space and they travel to the right as time goes on. This shows the varying strength of the fields, but it doesn't illustrate the fact that these waves are for all practical purposes infinite in the x and y direction. This illustration indicates the infinite extent in the x and y directions but it is hard to visualize that things are waving. Regardless of how you visualize the waves they must be explained by Maxwell's Equations. 3 . Maxwell's Equations and Waves in Free Space Maxwell's Equations in empty space (free of charges and currents) are: Gauss's Law for Electricity r E • d r A ∫ = 0 Gauss's Law for Magnetism r B • d r A ∫ = 0 Faraday's Law of Induction r E • d r s ∫ = − dΦB dt Ampere's Law r B • d r s ∫ = µoεo dΦe dt According to Faraday's Law changing magnetic fields make electric fields and according to Ampere's Law changing electric fields make magnetic fields. This is the essence of the propagation of electromagnetic waves. The fields produce each other as they change in space and time. All the properties of EM waves listed above can be explained by applying Maxwell's Equations. x y z B E υ υ Physics 4B Lecture Notes 34-5 Example 2: Find the conditions under which E = Em sin(kz −ω t) is a solution to the wave equation. The wave equation is ∂2E ∂z2 = µoεo ∂2E ∂t2 . Taking the space derivatives, ∂E ∂z = kEm cos(kz − ωt) and ∂2E ∂z2 = −k2E m sin(kz − ωt) = −k 2E and the time derivatives, ∂E ∂t = −ω Em cos(kz −ωt) and ∂2E ∂z2 = −ω 2Em sin(kz −ω t) = −ω 2E . Substituting into the wave equation, −k 2E = −µoεoω 2E ⇒ ω k = 1 µ oεo . From example 1 we know this ratio must be the speed of the waves, v = 1 µ oεo = 3.00x108 m/s ≡ c , the speed of light. Light is an EM wave! In order for EM waves to exist, they must travel at this speed regardless of the frequency or wavelength. c = 1 µoε o The Speed of EM Waves Example 3: Show that any function of kz±ωt is a solution to the wave equation. The general form of the wave equation is, ∂2 f ∂z2 = 1 υ2 ∂2 f ∂t2 . Define f (u) ≡ f (kz ±ω t) . We need the derivatives with respect to z, ∂ f ∂z = ∂ f ∂u ⋅ ∂u ∂z = ∂ f ∂u ⋅ k and ∂2 f ∂z2 = ∂ ∂z ∂ f ∂u ⋅ k = k ∂u ∂z ⋅ ∂2 f ∂u2 = k2 ∂2 f ∂u2 and with respect to t, ∂ f ∂t = ∂ f ∂u ⋅ ∂u ∂t = ∂ f ∂u ⋅ ±ω( ) and ∂2 f ∂t2 = ∂ ∂t ∂ f ∂u ⋅ ±ω( ) = ±ω ∂u ∂t ⋅ ∂2 f ∂u2 = ω2 ∂2 f ∂u2 . Plugging into the wave equation, ∂2 f ∂z2 = 1 υ2 ∂2 f ∂t2 ⇒ k2 ∂2 f ∂u2 = 1 c2 ω2 ∂2 f ∂u2 ⇒ ω = ck . It works! If any function of kz±ωt is a solution to the wave equation why do we usually discuss solutions of the form E = Em sin(kz −ω t) and B = Bm sin(kz −ω t) ? Fourier's Theorem states f(kz −ω t) = ai sin(k iz −ω i t) i ∑ . Any function can be written as a linear combination of sine and cosine waves. Physics 4B Lecture Notes 34-6 Example 4: Find the ratio of the peak electric field to the peak magnetic field. Recall that earlier we started with Faraday's Law and considered a path in the x-z plane and got the relationship, ∂E ∂z = − ∂B ∂t . Using the waves E = Em sin(kz −ω t) and B = Bm sin(kz −ω t) we get, kEm cos(kz −ω t) = +ωBm cos(kz −ω t ) ⇒ Em Bm = ω k = c . Em = cBm The Ratio of the Peak Fields All the properties of electromagnetic waves that were stated earlier are consistent with Maxwell's Equations. 4 . The Electromagnetic Spectrum We might suspect that all waves that travel at the speed of light are in fact electromagnetic in nature. For the most part this turns out to be true. The only essential difference between different types of EM waves is the wavelength (or frequency). 300k300 300M 300G 3x10 3x10 3x10 3x10 14 17 20 23 1Mm 1km 1m 1mm 1µm 1nm 1pm 1fm frequency in Hz wavelength in meters vi si bl e am ra di o fm ra di o - T V infrared ultraviolet x-rays radio waves -raysγ water absorbs air absorbs Earth you atom nucleus Discuss: The origin of human vision and the safety of exposure to EM waves (move on to energy). 5 . The Poynting Vector Notice that r E × r B point in the direction of the velocity vector. The magnitude of r E × r B = EB = 1c E 2 = cB2 . Recall that the energy density in electric and magnetic fields is given by, u = 12 εoE 2 + 12µo B 2 . Therefor r E × r B must be related to the energy in the waves. The exact relationship can be found by finding the total energy deposited by an EM wave landing on surface in a time dt. This energy must equal the energy density in the waves times the volume they occupy, U = uvol ⇒ dU = udV = uAdz = ucAdt . Substituting for u, dU = 12 εoE 2 + 12µo B 2( )cAdt . The power per unit area is, dU Adt = 12 εocE 2 + c2µo B 2( ) . Using the result from above r E × r B = 1c E 2 = cB2 the power per unit area can be written as dU Adt = 12 εoc 2 r E × r B + 12µo r E × r B ( ) = 12 εoc2 + 12µo( ) r E × r B . The two terms in parentheses are equal because c = 1 µoε o so finally, dU Adt = 1µo r E × r B . The Poynting Vector is defined accordingly, A dz=cdt Physics 4B Lecture Notes 34-7 r S ≡ 1µo r E × r B The Definition of the Poynting Vector The magnitude of the Poynting vector is power per unit area and it points in the same direction as the velocity. The instantaneous power per unit area is, S = 1µo EB = 1 µo EmBm sin 2 (kz −ω t) . Often the average power per unit area is called the “intensity.” Since the average value of the square of the sine is one half, I = Em Bm 2µo = Em 2 2µoc = cBm 2 2µo Intensity of EM Waves Example 5: Sunlight strikes earth with an average intensity of 1400W/m2. Find the peak electric and magnetic fields. The intensity of EM waves is, I = Em 2 2µoc ⇒ Em = 2µocI = 1060 V m . The ratio of the peak fields gives, Em = cBm ⇒ Bm = Em c = 3.53µT . Discuss: hood of a car in the sun and issues related to solar energy. Example 6: Find the total power radiated by the sun. There are 1400W/m2 landing on the earth which is 1.50x1011m away. Since the sun radiates EM waves uniformly in all directions this intensity will be the same over the surface area of a sphere that is 1.50x1011m away. Using the definition of power and the definition of intensity, I ≡ dU Adt = P A = P 4πr2 . Solving for the power, P = I4πr2 = (1400)(4π)(1.50x1011)2 = 3.96x1026 W . 6 . Radiation Pressure and Momentum Transfer When EM waves are absorbed by a surface, not only do they deposit energy, but they transfer momentum as well. The illustration at the right shows particles with kinetic energy U = 12 mv 2 = 12 pv absorbed by a surface, the momentum transferred can be written as, p = 2U v . The relationship between the transferred momentum and deposited energy for EM wave differs from this result by a factor of two. This is because the definition of kinetic energy U = 12 mv 2 is invalid for objects moving near the speed of light. For EM waves the answer is, p = U c The Momentum Transfer for Complete Absorption A A U = 12 mv 2 p = 2U v p = U c U m wave