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Alternating Current Electromagnetic Waves

Alternating Current Electromagnetic Waves. Y(x). Y(x) = Y P sin (kx - f ). Y p. Y p = amplitude  = wavelength = phase k = wavenumber [k = 2  / ]. x. f/ k. -Y p. l. Sinusoidal Function of Distance.

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Alternating Current Electromagnetic Waves

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  1. Alternating Current Electromagnetic Waves

  2. Y(x) Y(x) = YP sin (kx - f) Yp • Yp = amplitude •  = wavelength • = phase k = wavenumber [k = 2/] x f/k -Yp l Sinusoidal Function of Distance A sinusoidal function of distance is characterized by its: amplitude, wavelength, and phase constant

  3. V(t) V(t) = VP sin (wt - f) Vp • Vp = amplitude • f = frequency [f = w/2] • = phase T = period [T = 1/f = 2/w] t f/ -Vp Sinusoidal Function of Time A sinusoidal function of time is characterized by its: amplitude, frequency, and phase constant wis theangular frequency(angular speed) [radians per second]. f is thefrequency[cycles per second, or Hertz (Hz)] w = 2p f

  4. Use of a rotating vector to generate a sinusoidal function of time. Plot the y component of the vector, as a function of time, as the vector rotates with constant angular speed. • V(t) Vp  t f/ -Vp V(t) = VP sin (wt - f)

  5. f = 60 sec-1 = 60 Hz  = 2f Alternating Voltages and Currents When we plug a lamp into a wall socket, the voltage and current supplied vary sinusoidally, with a frequency f of 60 cycles per second

  6. Root Mean Square (rms) Values The average value of I2 is ½ I2MAX The root mean square (rms) value of the current is defined as :

  7. P = I2R instantaneous power P = I2rmsR average power Alternating Voltages and Currents The power dissipated in the resistor is: The average power dissipated in the resistor is:

  8. The AC generator has a maximum voltage of VMAX = 24 V • and frequency f = 60 Hz. The resistor has R = 265 . • Find: • The rms voltage • The rms current • The average dissipated power • The maximum instantaneous value of the dissipated power

  9. An alternating current in the primary circuit creates an alternating magnetic flux, that is concentrated in the iron core. The alternating magnetic flux induces an alternating emf, and hence an alternating current, in the secondary circuit Transformers A transformer is used to change the voltage in an electric circuit

  10. But Ɛ  V, then: Since P = S  Transformers

  11. Transformers Since energy is conserved, the power in the primary circuit equals the power in the secondary circuit

  12. Bug Zapper If the zapper operates at 4000 V, and the primary coil plugs to a standard 120 V outlet, and has 27 turns: How many turns does the secondary coil have? What is the ratio of the current in the primary, to the current in the secondary circuit? (IP/IS)

  13. Plane Electromagnetic Waves Ey Bz c x

  14. Plane Electromagnetic Waves

  15. Plane Electromagnetic Waves E and B are perpendicular to each other, and to the direction of propagation of the wave. The direction of propagation is given by the right hand rule: Curl the fingers from E to B, then the thumb points in the direction of propagation. Electromagnetic waves propagate in vacuum with speed c, the speed of light.

  16. Plane Electromagnetic Waves Ey E(x, t) = EP sin (kx-t) B(x, t) = BP sin (kx-t) Bz ˆ j ˆ z c x Waves are in phase, but fields oriented at 900 Speed of wave is c At all times E = c B

  17. F(x) v x For electromagnetic waves the speed of propagation is: and c = f  • Moving wave • F(x, t) = FP sin (kx - t) •  = 2 f •  = angular frequency • f = frequency • k = 2   • k = wavenumber • = wavelength • v =  / k = f 

  18. The distance between Earth and the Sun is 1.50x1011 m. How long does it take for the light to cover this distance?

  19. Visible Region of the Electromagnetic Spectrum Find the frequency of red light with wavelength 700 nm. Find the wavelength of light with frequency 7.5x1014 Hz

  20. The figure represents green-blue light with  = 500 nm Ey Bz a P b c x Plane Electromagnetic Wave • What is the distance from a to b? • How long does it take for the wavefront to move from a to b? • What does an electric field sensor detect at point P • as a function of time?

  21. The Electromagnetic Spectrum infra -red ultra -violet Radio waves g-rays m-wave x-rays

  22. The Electromagnetic Spectrum c = f  Ultraviolet: 7.5x1014 – 1017 Hz Damaging to skin/organisms. X-Rays: 1017 – 1020 Hz Damaging. Radiography. Gamma Rays: 1020 Hz Damaging. Used in therapy and sterilization. Radio Waves: 106 – 109 Hz Radio/TV, from antennas. Microwaves: 109 – 1012 Hz Radar, telephone, cooking. Infrared: 1012 – 1014 Hz Heat, remote controls. Visible: 4.3x1014 – 7.5x1014 Hz Detected by our eyes.

  23. Energy in Electromagnetic Waves • Electric and magnetic fields contain energy. • Potential energy stored in the field: uE and uB • uE: ½ 0 E2electric field energy density • uB: (1/0) B2 magnetic field energy density • The energy is put into the oscillating fields • by the sources that generate them. • This energy can then propagate to locations • far away, at the velocity of light.

  24. Energy in Electromagnetic Waves • The energy density stored in electric and magnetic fields • uE: ½ 0 E2electric field energy density • uB: (1/0) B2 magnetic field energy density • was calculated for static fields (Econstant, Bconstant) Electromagnetic fields are constantly changing and therefore it is more appropriate to treat the energy in the field using average values.

  25. Intensity of an Electromagnetic Wave The amount of energy that a wave delivers per unit area, per unit time, is referred to as the Intensity of the wave In the figure, a beam of light of cross-sectional area A, shines on a surface. All the light energy contained in the volume V = A (ct) strikes the surface in the time t. The energy in the volume V is: U = u V. Then, the intensity of the wave is:

  26. Intensity of an Electromagnetic Wave The average intensity of an electromagnetic wave is: Using E = c B we have: Using Erms = (1/2) EMAX and Brms = (1/2) BMAX

  27. At a given instant in time, the electric field in a beam of light has a magnitude of 510 N/C. What is the magnitude of the magnetic field at that same time?

  28. Since Intensity is energy per unit area per unit time Intensity is power per unit area • A garage is illuminated by a light bulb dangling from a wire. • The bulb radiates uniformly in all directions and consumes 50 W. • Calculate: • The average intensity of light 1 m from the bulb • The rms values of E and B, 1 m from the bulb • (assume 5% of the dissipated power is converted into light)

  29. A small laser emits a cylindrical beam of light 1 mm in diameter • with and average power of 5 mW. • Calculate the average intensity of the beam • Compare with the 50 W bulb of the previous problem • (Hint: use a 1 m separation to compare)

  30. Wave Momentum and Radiation Pressure An electromagnetic wave not only carries energy U but also carries momentum p. The momentum of a wave is related to its energy by: p = U / c For an electromagnetic wave absorbed by an area A in time t, the total energy received is: U = uAV A c t Then, the momentum received by the surface is: p = U/c = uAV A t = IAV A t / c

  31. Wave Momentum and Radiation Pressure The momentum received by the surface is: p = IAV A t / c Then, the average force exerted by the light on the surface is: FAV = p / t = IAV A / c And the average pressure on the surface is:

  32. One a sunny day the average intensity of sunlight on earth is • about 1.0x103 W/m2. • Find: • The average radiation pressure due to sunlight • The average force exerted on a beach towel that is • 1 m by 2.5 m in size (assume the towel absorbs the light)

  33. Electromagnetic waves are absorbed as they propagate through matter. The absorption of radiation is described by: I(t) = I0 exp [-t] Where: I (t)  Intensity at depth t I0  Initial intensity t  depth   absorption coefficient Propagation of Electromagnetic Waves in Matter

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