Ch 30: Electromagnetic WavesWorksheetSee all chapters
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Concept #1: Energy Carried by Electromagnetic Waves

Transcript

Hey guys in this video we're going to talk about the energy carried by electromagnetic waves and a quantity related to the energy called the intensity let's get to it. Now there are two types of energy carried by electromagnetic waves as the name would apply there is electric energy and there is magnetic energy now the way that energy is typically discussed in the context of electromagnetic waves is always by discussing the energy density remember little U lower case U the energy density is the energy per unit volume the energy density carried by an electromagnetic wave is split up into those two categories you have electric energy which is one half epsilon not E squared whatever the electric field strength happens to be and the magnetic energy which is one half B squared over Mu not whatever the magnetic field strength happens to be right obviously more magnetic field strength more electric field strength more energy, so the total carry the total energy density carried by an electromagnetic wave is just the sum of those two parts of energy that it carries. One half epsilon not E squared plus one half B squared over Mu not now the thing to remember is that the electric field strength and the magnetic field strength are always related to one another in an electromagnetic wave we've seen instances before where we talked about individual charges producing an electric field or a moving charge producing a magnetic field or a moving charge in a magnetic field in an electric field etc. Those were disconnected electric and magnetic fields when you're talking specifically about an electromagnetic wave where you have paired electric and magnetic fields producing light then the electric field strength the magnetic field strength are always related to each other by the speed of light and if you use a substitution for what speed of light is the definition of the speed of light which we saw before which depends upon the permitivity and the permeability you can rewrite this summation as simply epsilon not E squared this is the simplest way to represented it just know what the electric field strength is of your electromagnetic wave square it multiply it by the permitivity of free space by epsilon not. Now it's useful to define a quantity that is known as the pointing flux which is the power carried per unit area by a wave by an electromagnetic wave now the thing is that because electromagnetic waves oscillate the pointing flux oscillates as well remember that in an electromagnetic wave let me minimise myself you're going to have an electric field that's oscillating up and down right, so here you have no electric field so you have no pointing flux here you have a maximum electric field so you have a maximum pointing flux here you have no electric field minimum pointing flux here you have maximum electric field again it's negative but the pointing flux depends upon the electric field squared so you have the maximum pointing flux again. Now the intensity of the wave which is the average of the pointing flux. That is a very very important thing to know, that tells you on average how much power is delivered or how much power is carried by a wave per unit area on average and with the sinusoidal waves the average of E squared this is something that we talked about or something that you saw during alternating currents that the average of E itself is going to be 0 but the average of E squared is going to be one half E squared so that's where you get this one half from so that intensity is just one half times epsilon not which is the permittivity of free space times C which is the speed of light times E squared max. Now I specifically put maximum indicating these positions right that is the amplitude of the oscillation.

Alright guys let's do a quick example and then be done with this a laser outputs light at an average of 0.5 nano watts if the light is emitted from the laser and its cylindrical beam of radius 0.5 millimetres what's the intensity of the light what is the maximum value of the electric field. The intensity is just going to be the average power per unit area in this case the beam is emitted cylindricaly so the area is just going to be that front circular cross-section of the cylinder so the area is just going to be that area of the cylinder and its radius is 0.5 millimetres. The intensity is going to be that average power divided by that area the average power we are told is 0.5 nanowatts nano is 10 to the -9 the area of a circle is just Pi R squared so Pi times 0.5 millimetres milli is 10 to the -3 don't forget squared and that whole thing equals 0.00064 watts per meter squared. Power is in units of watts and area is in units of meters squared now if we want to find the maximum electric field which is the next part of the question we simply need to use this equation. We can also say that the intensity is one half epsilon not C times E max squared I want to multiply the two up divide both the epsilon not and the C over and then I need to take the square root. So E max is the square root of two times I right I multiply the two up I divided epsilon not n C. This whole thing is 2 times 0.00064 divided by 8.85 times 10 to the -12, 3 times 10 to the 8. If you don't remember those numbers that's fine you can just look them up you are not expected to remember them. The speed of light you will be expected to remember, but not epsilon not right this whole thing is just 0.69 newtons per coulomb. Alright guys that wraps up our discussion on the energy carried by these electromagnetic waves. Thanks for watching.

Practice: A 100 W lightbulb actually emits around only 10 W of light. What is the intensity of the light 1 cm away if the light is emitted perfectly spherically? What is the magnitude of the electric field emitted by the lightbulb? What about the magnetic field?

You are trying to estimate the efficiency of a lightbulb by measuring the power of the light emitted vs the power consumed by the bulb. You have a circular detector with a radius of 12 cm and you measure an intensity of 1.7 kW/m2 from a distance of 1.5 m. If the lightbulb pulls 150 W of power from an outlet, what is the efficiency of the lightbulb?
Two sources of sinusoidal electromagnetic waves have average powers of P1 = 100.0 W and P2 = 320 W and emit uniformly in all directions. At the same distance from the source what is the ratio of maximum electric fields E2 to that of E1 i.e... E2/E1
Two sources of sinusoidal electromagnetic waves have average powers of P1 = 100.0 W and P2 = 320 W and emit uniformly in all directions. For the electromagnetic wave with P 1 = 100.0 W find the ratio of intensity S1 10 m away from the source to the intensity S 1 100m away from the source, i.e. S1 (10m)/S1(100m) (Hint: Area of sphere 4πR2)
An 800-kHz signal is detected at a point 7.7 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 550 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to: A) 2.0 x 10-4 W/m2 B) 5.7 x 10-4 W/m2 C) 8.0 x 10-4 W/m2 D) 2.8 x 10-4 W/m2 E) 4.0 x 10-4 W/m2
An isotropic source of electromagnetic radiation emits light that, when measured at 10 cm from the source, carries an electric field amplitude of 1.0x103 N/C and a magnetic field amplitude of 3.33x10-6 T.  (a) What is the average Poynting flux, Sav, through a sphere of radius 10 cm? (b) How much light-energy is emitted by the source per second? (c) If the source is 80% efficient at converting electrical power into light-energy, how much power must be supplied to the source?
An open MRI scanner can produce a magnetic field strength of up to 3 T. If the space between the poles of the magnet (where the patient lies) is cylindrical, with a diameter of 2 m and a height of 70 cm, answer the following questions: (a) What is the magnetic energy density within this MRI? (b) What is the total magnetic energy contained within the MRI?
Calculate the total average power of the station's radio transmitter. a. <P> = 330 MW b. <P> = 485 MW c. <P> = 667 MW d. <P> = 802 MW e. <P> = 950 MW
A large electromagnet has a 22 T magnetic field between its poles. What is the magnetic energy density between the poles? A) 240 J/cm3 B) 30,000 J/cm3 C) 88 J/cm3 D) 190 J/cm3
The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 695 mW/m2. The wave is incident upon a rectangular area, 1.5 m by 2.0 m, at right angles. The total energy that traverses the area in a time interval of one minute is closest to: A) 130 J B) 250 J C) 160 J D) 220 J E) 190 J
A point source emits electromagnetic radiation uniformly in all directions. If the power output of the source is 960 W, what are the amplitudes of the electric and magnetic fields in the wave at a distance of 15.0 m from the source? (The surface area of a sphere that has radius R is 4πR2, ε0 = 8.854 x 1012 C2/(N•m2). μ0 = 4π x 10-7 T•m/A.) electric field amplitude = _______________ magnetic field amplitude = _______________
A laser beam has a wavelength of 633 nm and a power of 0.500 mW spread uniformly over a circle 1.20 mm in diameter. This beam falls perpendicularly on a perfectly reflecting piece of paper having twice the diameter of the laser beam and a mass of 1.50 mg. What are the amplitudes of the electric and magnetic fields in this laser beam?
A large electromagnet has a 22T magnetic field between its poles. What is the magnetic energy density between the poles?A) 88 J/cm3B) 190 J/cm3C) 30,000 J/cm3D) 240 J/cm3 
An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 200 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to:A) 3.8 x 10-5 W/m2B) 5.3 x 10-5 W/m2C) 7.5 x 10-5 W/m2D) 1.1 x 10-4 W/m2E) 2.8 x 10-5 W/m2
A point source of electromagnetic waves emits waves uniformly in all directions. At a distance of 10.0 m from the source the magnetic field amplitude for the waves is 4.0 x 10-8 T. (Note: c = 3.00 x 108 m/s, ε0 = 8.854 x 10-12 C2 / (N•m2), μ0 = 4π x 10-7 T•m/A)What is the electric field amplitude at this point, 10.0 m from the source?
A point source of electromagnetic waves emits waves uniformly in all directions. At a distance of 10.0 m from the source the magnetic field amplitude for the waves is 4.0 x 10-8 T. (Note: c = 3.00 x 108 m/s, ε0 = 8.854 x 10-12 C2 / (N•m2), μ0 = 4π x 10-7 T•m/A)What is the intensity of the wave at this point, 10.0 m from the source?
Sinusoidal electromagnetic waves from a radio station pass perpendicularly through an open window that has area 0.200 m2. At the window, the magnetic field of the wave has amplitude Bmax = 7.00 x 10-10 T. How much energy does the wave carry through the window in 2.00 minutes? (Note: ε0 = 8.854 x 10-12 C2 / N•m2 and μ0 = 4π x 10-7 T•m/A)
The energy density of an electromagnetic wave isA) equally divided between the magnetic and the electric fields.B) entirely in the magnetic field.C) 1/4 in the electric field and 3/4 in the magnetic field.D) entirely in the electric field.E) 1/4 in the magnetic field and 3/4 in the electric field.
If the magnetic field in a traveling electromagnetic wave has a maximum value of 16.5 nT. What is the maximum value of the electric field associated with this wave? (c = 3.00 x 108 m/s)A) 5.5 x 10-17 V/mB) 4.95 V/mC) 0.495 V/mD) 55.0 x 10-16 VmE) 55.0 x 10-15 Vm
Sinusoidal electromagnetic waves from a radio station process pass perpendicularly through an open window that has area 0.400 m2. The intensity of the wave is constant over the area of the window. The wave carries 0.800 J of energy through the window in 5.00 s. (Note: ε0 = 8.854 x 10-12 C2/N•m2 and μ0 = 4π x 10-7 T•m/A) What is the intensity of the wave at the window?
Sinusoidal electromagnetic waves from a radio station process pass perpendicularly through an open window that has area 0.400 m2. The intensity of the wave is constant over the area of the window. The wave carries 0.800 J of energy through the window in 5.00 s. (Note: ε0 = 8.854 x 10-12 C2/N•m2 and μ0 = 4π x 10-7 T•m/A) At the window, what is the amplitude of the magnetic field of the wave?
Sinusoidal electromagnetic waves from a radio station process pass perpendicularly through an open window that has area 0.400 m2. The intensity of the wave is constant over the area of the window. The wave carries 0.800 J of energy through the window in 5.00 s. (Note: ε0 = 8.854 x 10-12 C2/N•m2 and μ0 = 4π x 10-7 T•m/A) The window is replaced b y a mirror that has area 0.400 m2. What is the force that the wave exerts on the mirror, if the wave is totally reflected?
A cylindrical laser beam has diameter 8.00 mm. The average energy density in the beam is 8.00 x 10-3 J/m3. What is the amplitude of the magnetic field in the beam?
A sinusoidal electromagnetic wave has intensity I = 100 W/m 2 and an electric field amplitude E. What is the electric field amplitude of a 50 W/m 2 electromagnetic wave with the same wavelength?(a) 4E(b) 2E(c) 2√2E(d) √2E(e) E / (2√2) (f) E / √2(g) E / 4(h) E / 2
A cylindrical laser beam has diameter 8.00 mm. The average energy density in the beam is 8.00 x 10-3 J/m3. What is the power output of the laser?
The power radiated by a star is 8.0 x 1030 W. What is the intensity of the radiation from this star at a distance of 1.0 x 1012 m from the star?(a) 8.0 x 1018 W/m2(b) 8.0 x 106 W/m2(c) 2.5 x 106 W/m2(d) 6.4 x 105 W/m2(e) 3.2 x 105 W/m2(f) none of the above answers
The graph shows the intensity of transmitted light of different colors (wavelengths) as a function of depth x in meters for absorption in water. Approximately what is the absorption coefficient α for UV (ultraviolet) LIGHT, given the exponential transmission I(x) = I0e−αx ? A. 2 m-1 B. 10-6 m-1 C. 0.003 m-1 D. 50 m-1 E. 0.025 m-1
A laser beam has a wavelength of 633 nm and a power of 0.500 mW spread uniformly over a circle 1.20 mm in diameter. This beam falls perpendicularly on a perfectly reflecting piece of paper having twice the diameter of the laser beam and a mass of 1.50 mg. What acceleration does the laser beam give to the paper?