Solution MC Hwk (20) - MC HW 20 PDF

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Solution for Multiple-Choice Homework 20 EM Waves Solution to Multiple-Choice Homework Problem 20.1(Intensity of Sun at Earth) Problem: The total energy output of the sun is 3.8 × 1026 J per second. The radius of the earth’s orbit is 1.5 × 1011 m. What is the intensity of the sun at the radius of the earth? Select One of the Following: (a) 120, 000 mJ2 s (b) 12000 mJ2 s (c-Answer) 1300 mJ2 s (d) 5200 mJ2 s (e) Toby Solution The power P = 3.8 × 1026 J/s is given. Intensity is energy per second per area, I=

3.8 × 1026 J/s P J = = 1300 2 4πr2 4π (1.5 × 1011 m)2 m s

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 20.2(Intensity to Tear Hydrogen Apart) Problem: The average electric field of the proton at the electron in a hydrogen atom is 6 × 1011 N C . If the electric component of a light wave is equal to this field, the light wave will tear the atom apart. How intense must a light wave be to tear apart a hydrogen atom? Select One of the Following: (a) This is impossible since light and electric fields are different things. W (b) 5.4 × 105 m 2 W (c-Answer) 4.8 × 1020 m 2 W (d) 8 × 108 m 2

(e) Toby Solution The intensity of a light wave is related to the magnitude of the electric component by I=

N )2 (6 × 1011 C W E02 = 4.8 × 1020 2 = Tm m −7 8 2µ0 c m 2(4π × 10 )(3 × 10 s ) A

which is pretty intense, but still technically achievable. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 20.3(Intensity of Electric Field in Supernova Light)

Problem: A supernova releases about 1foe = 1 × 1044 J in one second. This is about equal to the total energy the sun will produce, ever. Suppose you are at a distance of the earth’s orbit about the sun, 1.5 × 1011 m, from the supernova and suppose all the energy goes into producing light, what is the magnitude of the electric field in the light wave from the supernova at a distance of the distance from the earth to the sun (d=1.5 × 1011 m)? 1

Select One of the Following: . (a) 5.7 × 104 N C N . (b-Answer) 5.2 × 1011 C

(c) 3.8 × 1015 NC . (d) 9.1 × 1020 N . C (e) Toby. Solution The intensity of an electromagnetic wave is

1 I = ε0 cE02 2 where E0 is the amplitude of the electric field Intensity is energy (U ) per unit area (A) per unit time, I=

W U 1 × 1044 J U = 3.537 × 1020 2 = = 4π (1.5 × 1011 m)2 1s At m 4πd 2 t

where the area is the area of a sphere A = 4πd 2 . Now solve the intensity for the electric field, s r 2(3.537 × 1020 mW2 ) 2I N = 5.2 × 1011 = E0 = C2 m) 8 −12 C ε0 c (8.85 × 10 2 )(3 × 10 s Nm

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 20.4(Displacement Current Tabletop Accelerator)

Problem: A table top particle accelerator creates an electric field of 1.0 × 1011 NC sin(ωt) by using a laser to produce a plasma. If the laser uses visible light, then the linear frequency is about 5 × 1014 Hz. Compute the magnitude of the displacement current through a circular loop with radius 1µm whose normal is parallel to the electric field at time t = 0. This radius is about equal to the wavelength. Select One of the Following: (a) 0 (b) 30A (c) 150A (d-Answer) 9000A (e) Toby Solution (a) Compute the Flux: The electric flux is φe = EA = E0 πR2 sin(2πf t). (b) Compute the Displacement Current: The displacement current is given by Id = ε0

dφe d = ε0 E0 πR2 sin(2πf t) = 2ε0 E0 π 2 R2 f cos(2πf t) dt dt

At time t = 0, the displacement current is Id = 2ε0 E0 π 2 R2 f = 2(8.85 × 10−12

C2 N )(1.0 × 1011 )π 2 (1 × 10−6 m)2 (5 × 1014 s−1 ) = 9000A C Nm2

Total Points for Problem: 7 Points

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Solution to Multiple-Choice Homework Problem 20.5(Angle to Tip Your Head for Given Transmission) Problem: Sunlight reflecting from water at 57◦ is completely polarized. At what angle must the transmission axis of a polarizer (say your sunglasses) make with the polarization direction of this light to decrease the intensity of the light by a factor of 4? For your information, but not at all useful in the problem, light polarized by reflection has its electric field perpendicular to the plane of incidence, so the polarization direction in along the surface of the water. Select One of the Following: (a-Answer) 60◦ (b) 45◦ (c) 75◦ (d) 5◦ (e) Toby Solution Apply Malus’ Law, I0 /4 = I0 cos2 θ. Solve for the angle arccos( 12 ) = θ = 60◦ . Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 20.6(Intensity of Electric Field in Commercial Laser) Problem: A commerically available hand-held green laser has a power of 4.4W and produces a beam of radius 0.44mm. The laser will cut through plastic and set wood on fire, so it makes a great Christmas present. What is the amplitude of the electric field in the beam? Select One of the Following: . (a) 5.5 × 109 N C . (b-Answer) 7.4 × 104 N C (c) 2.9 × 103 NC . . (d) 1.5 × 103 N C (e) Toby. Solution The intensity I of an electromagnetic wave is given by I=

E 02 P = A 2µ0 c

where E0 is the magnitude of the electric field, c is the speed of light, P is the power and A is the area. Next, solve for E0 r P 2µ0 c E0 = A Substituting the information for the laser yields s (4.4W)(2)(4π × 10−7 Tm )(3 × 108 ms ) A E0 = 2 π(0.00044m) N E0 = 7.4 × 104 C 3

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 20.7(Magnitude of Magnetic Component Given Electric Component) N Problem: The electromagnetic waves produced by Motorola cell phones have a peak electric field | ~ E| = 125 C . What is the magnitude of the amplitude of the magnetic component of the waves?

Select One of the Following: (a-Answer) 4.2 × 10−7 T (b) 3.1 × 10−5 T (c) 1.25 × 102 T (d) There is no simple relation between the electric and magnetic component, so this cannot be calculated with the information given (e) Toby Solution The magnetic component of an EM wave is related to the electric component by |B| =

125CN |E | = 4.2 × 10−7 T = c 3.0 × 108 ms

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 20.8(Intensity of Radiation to Produce One G) Problem: The Planetary Society’s light sail, that we discussed in lecture, has a mass of 100kg and a sail area of 600m2 . We found a very boring acceleration for sunlight in near-earth orbit. To make the sail accelerate faster we shine lasers on it from earth. Assume the laser completely illuminate the sail area, all 600m2 , and that the sail is perfectly reflecting. What must the intensity of the light be at the sail to produce an acceleration of g = 9.81 m s2 ? Don’t forget the 2 for perfectly reflecting. Select One of the Following: W (a) 1.2 × 108 m 2 W (b) 2.0 × 107 m 2

(c) 4.9 × 108 W m2 W (d-Answer) 2.5 × 108 m 2

(e) Toby Solution The radiation pressure on a perfectly reflecting surface is Pr = 2I/c. The force on the sail is the pressure multiplied by the area.Newton’s Second Law relates mass to acceleration, F = mg = Pr A,where A is the area of the sail. Substituting the pressure gives 2IA mg = c Solving for the intensity yields I=

)(3 × 108 ms ) mgc (100kg)(9.81 m W s2 = 2.5 × 108 2 = m 2A 2(600m2 )

Total Points for Problem: 3 Points 4

Solution to Multiple-Choice Homework Problem 20.9(EM Wave - Electric Component Perpendicular to Propagation) Problem: An electromagnetic wave propagates in the +ˆ z direction. What can be concluded about the direction of the electric field forming the wave? Select One of the Following: (a) There is no general relation between the direction of the electric component and the direction of propagation. (b) The electric component must point in the ±ˆ z direction. (c-Answer) The electric component must oscillate in the x − y plane. (d) The electric component must oscillate in the y − z plane. (e) Toby Solution The electric component is perpendicular to the direction of propagation, so it oscillates in the x − y plane. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 20.10(Light, sun, and the former planet Pluto) Problem: Pluto is Mickey’s dog. Forget that. Try Again. The (now former) planet Pluto at its farthest distance from the Sun is a distance of 7.35 × 1012 m from the Sun. How far is that in dog years? The real question is how long does it take light to reach Pluto from the Sun at this distance. This is a question asked by a student in the lecture quiz on the first lecture day. Select One of the Following: (a) 0.150s (b-Answer) 24500s = 6.8hours (c) 125, 000s = 35hours (d) 0s. Light reaches Pluto instantaneously. (e) Toby. Solution The speed of light is c = 2.99 × 108 m/s. Time equals distance divided by speed, so we have t= =

d c

7.35 × 1012 m 2.99 × 108 m/s

= 24.5 × 103 s This is almost seven hours. (NOTE: In the year 2006, Pluto was stripped of its planet status.) Total Points for Problem: 3 Points

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