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Alternating Current Electromagnetic Waves. Sinusoidal Function of Distance A sinusoidal function of distance is characterized by its: amplitude, wavelength,

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Presentation on theme: "Alternating Current Electromagnetic Waves. Sinusoidal Function of Distance A sinusoidal function of distance is characterized by its: amplitude, wavelength,"— Presentation transcript:

1 Alternating Current Electromagnetic Waves

2 Sinusoidal Function of Distance A sinusoidal function of distance is characterized by its: amplitude, wavelength, and phase constant Y(x) = Y P sin (kx -  ) Y p = amplitude = wavelength  = phase k = wavenumber [k =  / ]  k Y(x) x YpYp -Y p

3 Sinusoidal Function of Time  is the angular frequency (angular speed) [radians per second]. f is the frequency [cycles per second, or Hertz (Hz)]  f A sinusoidal function of time is characterized by its: amplitude, frequency, and phase constant V(t) = V P sin (  t -  ) V p = amplitude f = frequency [f =   = phase T = period [T = 1/f =  /  // V(t) t VpVp -V p

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. 

5 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 f = 60 sec -1 = 60 Hz  = 2  f

6 Root Mean Square (rms) Values The average value of I 2 is ½ I 2 MAX The root mean square (rms) value of the current is defined as :

7 Alternating Voltages and Currents The power dissipated in the resistor is: The average power dissipated in the resistor is: P = I 2 R instantaneous power P = I 2 rms R average power

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

9 Transformers A transformer is used to change the voltage in an electric circuit 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

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

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? (I P /I S )

13 Plane Electromagnetic Waves x EyEy BzBz c

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15 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 Waves are in phase, but fields oriented at 90 0 Speed of wave is c At all times E = c B E(x, t) = E P sin (kx-  t) B(x, t) = B P sin (kx-  t) ˆ z ˆ j x EyEy BzBz c

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

18 The distance between Earth and the Sun is 1.50x10 11 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 7.5 x Hz

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

21 The Electromagnetic Spectrum Radio waves  -wave infra -red  -rays x-rays ultra -violet

22 The Electromagnetic Spectrum Radio Waves: 10 6 – 10 9 Hz Radio/TV, from antennas. Microwaves: 10 9 – Hz Radar, telephone, cooking. Infrared: – Hz Heat, remote controls. Visible: 4.3 x – 7.5 x Hz Detected by our eyes. Ultraviolet: 7.5 x – Hz Damaging to skin/organisms. X-Rays: – Hz Damaging. Radiography. Gamma Rays:  Hz Damaging. Used in therapy and sterilization. c = f

23 Energy in Electromagnetic Waves Electric and magnetic fields contain energy. Potential energy stored in the field: u E and u B u E : ½  0 E 2 electric field energy density u B : (1/  0 ) B 2 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 u E : ½  0 E 2 electric field energy density u B : (1/  0 ) B 2 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 E rms = (1/  2) E MAX and B rms = (1/  2) B MAX

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: a)The average intensity of light 1 m from the bulb b)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. a)Calculate the average intensity of the beam b)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 = u AV A c  t Then, the momentum received by the surface is:  p =  U/c = u AV A  t = I AV A  t / c

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

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33 One a sunny day the average intensity of sunlight on earth is about 1.0x10 3 W/m 2. Find: a)The average radiation pressure due to sunlight b)The average force exerted on a beach towel that is 1 m by 2.5 m in size (assume the towel absorbs the light)

34 Propagation of Electromagnetic Waves in Matter Electromagnetic waves are absorbed as they propagate through matter. The absorption of radiation is described by: I(t) = I 0 exp [-  t] Where: I (t)  Intensity at depth t I 0  Initial intensity t  depth   absorption coefficient


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