MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Spring 2014 Practice Problem Set 11 Solutions

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1 MASSACHUSES INSIUE OF ECHNOLOGY Department of Physics 8 Spring 4 Practice Problem Set Solutions Problem : Electromagnetic Waves and the Poynting Vector We have been studying one particular class of electric and magnetic fields solutions called plane sinusoidal traveling waves One special example is an electromagnetic wave traveling in the positive x -direction with the speed of light c is described by the functions E(x, y,z,t) = E y (x,t)ĵ = E sin π (x ct) y, ĵ B(x, y,z,t) = B z (x,t) ˆk = c E sin π (x ct) y, ˆk Suppose the plane wave front (infinite yz -plane) traveling at speed c in the positive x -direction passes through a rectangular volume of space that has area A perpendicular to the direction of propagation and length cδt, corresponding to the distance the electromagnetic wave travels in time Δt Inside this volume of space the electric and magnetic fields store energy and that energy changes in time as the wave passes through the volume he energy per time per area transported by this electromagnetic wave is called the Poynting vector and is defined as S = µ E B he power transmitted by the Poynting vector through a surface is the flux given by the expression P(t) = S(t) ˆn da surface he flux of the Poynting vector on a surface describes the energy that flows across the surface Question : What the Poytning vector associated with this wave? Answer:

2 S = E B = E µ µ y, sin π = µ c E sin π y, (x ct) (x ct) ĵ c E sin π y, î (x ct) ˆk So the energy per time per area is transported in the direction of propagation of the wave Question : What is the time-averaged Poynting vector field on the fixed plane x = over one period? he definition of a time average of a periodic function f (t) over one period is Recall that c = / µ ε f (t) = f (t)dt Answer: Set x = in your expression for the Poynting vector hen S(,t) = µ c E y, sin ( πct / )î Note that sin ( πct / ) = sin (πct / ) Use the trigonometric identity that sin (πct / ) = (/ )( cos(4πct / ) he time averaged Poynting vector on the plane x = is S(,t) = = E y, µ c dt he integration is straightforward, µ c E (/ )( cos(4πct / )dtî y, î E y, cos(4πct / )dt µ c î S(,t) = E y, µ c t î E y, µ c = E y, µ c î E y, µ c sin(4πc / ) î (4πc / ) sin(4πct / ) î (4πc / ) he second term vanishes because c = and sin(4π) = herefore

3 S(,t) = E y, î µ c Recall that c = / (µ ε ) so / (µ c) = c / (µ c ) = ε µ c / µ = c ε In conclusion the time averaged Poynting vector over one period on the plane x = is given by S(,t) = c ε E î y, Because the integral was only over the time, it doesn t matter what plane we choose to time average the Poynting vector, our result holds for all planes and so S(x,t) = c ε E î y, Question 3: What is the time-average (over one period) of the energy density stored in the electric and magnetic fields at the point (x, y,z,t) Recall that u elec = (/ )ε E and u mag = B / µ How is that related to the magnitude of the time-averaged Poynting vector? Answer: We begin by time averaging the electric field energy density u elec = u dt = elec ε E sin (π(x ct) / )dt y, Once again the integral sin (π(x ct) / )dt = o see this explicitly at arbitrary plane with fixed values of x, make the change of variables u = (π(x ct) / ) hen du = πcdt / dt = ( / πc)du he limits of the integral go from u = πx / to u = (π(x c ) / ) = (πx / π) because c = Our time-averaged integral is then Once again we have that u elec = ε E y, 4πc πx/ π sin (u)du = ε E y, 4π πx/ πx/ π sin (u)du πx/ 3

4 πx/ π sin (u)du = πx/ πx/ π πx/ ( cos(u)) du = sin((πx / π) ((πx / π) πx / ) = π sin((πx / π) sin((πx / ) = u=πx/ π u u=πx/ π sin(u) u=πx/ u=πx/ = π sin((πx / ) where we used the fact that sin((πx / π) = sin((πx / ) and therefore the second term in the above expression is zero So the time-averaged electric field energy density is u elec = ε E y, 4πc πx/ π sin (u)du = ε E y, 4 πx/ By a similar argument the time-averaged energy density stored in the electric and magnetic fields u mag = u dt = mag = 4µ c E = y, 4 ε E y, B dt = µ µ c E sin π y, (x ct) dt By a similar argument the time-averaged energy density stored in the electric and magnetic fields is u total = u elec + u mag = ε E y, Recall that the time-averaged magnitude of the Poynting vector is S(x,t) = c ε E î y, hus we see that time-averaged energy density is equal to the magnitude of the time-averaged Poynting vector divided by the speed of light u total = S(x,t) Question 4: What is the time-averaged energy stored in the electric and magnetic fields in a rectangular volume of cross-sectional area A and length cδt, where the length cδt is aligned along the x -axis and cδt << he last assumption allows us to assume that S(x,t) is nearly uniform across the box c 4

5 Answer: U total = u total ΔV vol = u total AcΔt = c ε E y AΔt Question 5: What is the time-averaged rate of change of the total energy stored in the electric and magnetic fields in the rectangular volume of cross-sectional area A and length cδt? Answer: d dt U total c = lim ε Ey AΔt = c Δt Δt ε Ey A Question 6: What is the time-averaged flow of power through the surface of area A located on the surface given by x = x p, perpendicular to the direction of propagation? How does the power that through the rectangular surface (this flux is negative since it flows into the volume) compare to the time derivative of the energy stored in the fields inside the volume? Answer: From the introductory remarks, the power through a surface (energy is flowing across the surface) is given by P(t) = S(x,t) ˆn da surface herefore the time-averaged power is on the surface of area A located on surface given by x = x p, P(t) = surface S(x p,t) ˆn da = S(x p,t) A = c ε E y For simplicity (to avoid complicated integrals), let s assume the wave at time t has just entered the box and the box is empty In time Δt later, the wave has not left the volume, so no power flows out hus we conclude that the power that flows into the rectangular volume (this flux is negative since it flows into the volume) equals the time derivative of the energy stored in the fields inside the volume S(x p,t) da = d dt U total closed surface A 5

6 Problem : Poynting Vector and Radiation Pressure: A plane electromagnetic wave transports energy in the direction of propagation of the wave he power per square area is given by the Poynting vector E B S = µ he power that flows into the rectangular volume cross-sectional area A and length cδt appears as rate of change of the energy stored in the fields inside the volume P power = S A = d dt U total he electromagnetic wave also transports momentum, and hence can exert a radiation pressure on a surface due to the absorption and reflection of the momentum he momentum carried by an electromagnetic wave is related to the energy of the wave according to U = c p If the plane electromagnetic wave is completely absorbed by a surface of cross-sectional area A then the momentum Δp delivered to the surface in a time Δ t is given by Δ p = ΔU c he force that the wave exerts on the surface is then the rate of change of the momentum in time F = lim Δt Δ p Δt = lim Δt c ΔU Δt = c du dt Since the rate of change of energy is related to the power flowing across the surface, the force is F = c du dt = c P power = c S A he radiation pressure P pressure is then defined to be the force per area that the wave exerts on the surface abs P pressure F A = c S, perfectly absorbing When the surface completely reflects the wave, then the change in momentum is twice the absorbing case since the wave completely reverses direction, 6

7 Δ p = ΔU c herefore the radiation pressure of a wave on a perfectly reflecting surface is ref P pressure F A = c S, perfectly reflecting Suppose a space station is placed in orbit the same distance away from the sun as the earth but on the opposite side of the sun from the earth in a circumpolar orbit (the orbit plane is perpendicular to the plane of the earth s orbit around the sun) he average earth-sun distance is r e,s = 5 m An astronaut of mass m = kg on a space walk loses contact with the space station he spacesuit has a cross sectional area of approximately A = m he mass of the sun is m s = 99 3 kg he gravitational constant is G = 667 N m kg a) What is the radiation pressure of the sun on the astronaut spacesuit Assume it is perfectly reflecting? Answer: We need to know the magnitude of the Poynting vector We can use the fact that the time averaged power output of the sun is P sun = 4 6 W and hence herefore ref P pressure S = P sun 4πr = 39 6 W es (4π)(5 m) = 4 3 W m - = c S = c S = (4 3 W m ) = 9 6 Pa (3 8 m s ) b) What is the force on the astronaut? Answer: ref F = P pressure A = c S A = (4 3 W m )(m ) = 8 5 N (3 8 m s ) c) How does this force compare to the force of gravity? Answer: he gravitational force is F grav = G mm s = (667 N m kg )( kg)(99 3 kg) = 6 N r e,s (5 m) So the gravitational force is more than ten thousand times stronger 7

8 Problem 3: Intensity of the Sun At the upper surface of the earth s atmosphere, the timeaveraged magnitude of the Poynting vector, referred to as the solar constant, is given by S = 35 3 W m a) If you assume that the sun s electromagnetic radiation is a plane sinusoidal wave, what are the magnitudes of the electric and magnetic fields? Answer: he magnitude of the time-averaged Poynting vector is related to the amplitude of the electric field by S = c ε E y, So E y, = S cε = (35 3 W m ) (3 8 m s )(885 C N m ) = 3 V m he amplitude of the magnetic field is B = E c = 3 V m 3 8 m s - = 34 6 he associated magnetic field is less than one tenth the earth s magnetic field b) What is the time-averaged power radiated by the sun? he mean sun-earth distance is r es = 5 m Answer: he time-averaged power radiated by the sun at the distance r e, s that the earth lies from the sun is given by surface S(t) ˆn da P sun = = S 4πr r =r es es = (35 3 W m )(4π)(5 m) = 38 6 W 8

9 Problem 4: Intensity of Double Slit Interference When coherent monochromatic laser light falls on two slits separated by a distance d, the emerging light will produce an interference pattern on a viewing screen a distance D from the center of the slits he geometry of the double slit interference is shown in the figure below Suppose the slits are located at the plane x = D he light that emerges from slit and slit at time t are in phase Suppose a screen is placed at the plane x = Suppose the transverse component of the electric field of the wave from slit at the point P is given by E = E sin( ω t) Let s assume that the plane wave from slit has the same amplitude E as the wave from slit Since the plane wave from slit has to travel an extra distance to the point P equal to the path length, this wave will have a phase shift φ relative to the wave from slit, E = E sin( ω t + φ ) Question : Show that the phase shift φ, the wavelength, the distance between the slits, and the angle related θ by π φ = d sin θ As a hint determine a relationship between, the ratio of the phase shift φ to π and the ratio of the path length Δ r = d sin θ to wavelength ; ie how are φ / π and Δr / related? Answer: he ratio of the phase shift φ to π is the same as the ratio of the path length Δ r = d sin θ to wavelength, 9

10 φ Δr = π herefore the phase shift φ is given by π φ = d sin θ Question : he total electric field at the point P is the superposition of the these two fields ( ( ) ( )) Etotal = E + E = E sin ω t + sin ω t + φ Use the trigonometric identity A + B A B sin A + sin( B) = sin cos o show that the total transverse component of the electric field is φ φ Etotal = E + E = E sin ω t + cos Answer: We first use the superposition principle to add the component sof the two electric fields yielding E total = E ( sin( ωt) + sin( ωt + φ) ) = E sin ωt + ωt + φ ωt (ωt + φ) cos hus the total transverse component of the electric field is (noting that cos( φ / ) = cos(φ / ) φ φ Etotal = E + E = E sin ω t + cos Question 3: he intensity of the light is equal to the time-averaged Poynting vector I = S = E B µ Because the amplitude of the magnetic field is related to the amplitude of the electric field by B = E c he intensity of the light is proportional to the time-averaged square of the electric field, I E total,

where the time-averaged value of the square of the sine function is sin φ ω t + = Determine the time averaged total electric field, ( E + E ) Answer: he intensity of the electric field time-averaged

11 where the time-averaged value of the square of the sine function is sin φ ω t + = Determine the time averaged total electric field, ( E + E ) Answer: he intensity of the electric field time-averaged square of the electric field is ( ) = 4E cos φ E + E sin ωt + φ φ = E cos Let I max be the amplitude of the intensity hen the intensity of the light at the point P is φ = I I cos max Question 4: Show that the intensity is maximal when d sin θ = m, m =, ±, ±, ± 3, Answer: he intensity has a maximum when the argument of the cosine is an integer number of π multiples of π, φ = ± mπ Since the phase shift is given by φ = d sin θ, we have that π d sin θ = ± mπ hus we have the condition for constructive interference, d sin θ = ± m, m =, ±, ±, ± 3, Question 5: Graph the intensity pattern on the screen as a function of distance y from the point O for the case that D >> d and d >>

12 Question 6: Determine the average over all phases of the time averaged total electric field, ( E + E ) = phase mπ average ( ) dφ E + E Answer: he intensity of the electric field time-averaged square of the electric field is If we now average this over all phases, E + E mπ ( ) = E cos φ φ E E E cos d ( cos ) d (m sin ) E mπ mπ mπ mπ mπ π φ = + φ φ = π + φ = Question 7: Do intensities add, ie is it true that ( E + E ) = E + E? Or do intensities add when we average over all phases, ( E + E ) = E phase average + E? Answer: he time-average of the individual electric fields are given by ( ) E = E sin ω t = E, ( ) E = E sin ω t + φ = E If you only time average the total electric fields then ( E + E ) E + E But if you also average over all possible phases then ( E + E ) = E phase average + E

13 Problem 5: Superposition of wo ravelling Waves Suppose the electric field of an electromagnetic wave is given by the superposition of two waves E = E cos(kz ωt)î + E cos(kz + ωt)î You may find the following identities useful cos(kz ± ωt) = cos(kz)cos(ωt) sin(kz)sin(ωt) a) What is the associated magnetic field B(x, y,z,t)? Answer: We treat each contribution to the electric field separately hen we have that B = E c cos(kz ωt) ĵ E cos(kz + ωt) ĵ, c where we used that fact that that the amplitude of the magnetic field is related to the amplitude of the electric field by B = E / c, and the direction of the field satisfy dir( E B) = dir(propagation) he contribution E = E cos(kz ωt)î is propagating in the positive ˆk -direction, and the contribution E = E cos(kz + ωt)î is propagating in the negative ˆk -direction b) Use the identity cos(a)sin(a) = (/ )sin(a) to show that the energy per unit area per unit time (the Poynting vector) transported by this wave is S = (E / cµ )sin(kz)sin(ωt) ˆk Answer: We first use the identities cos(kz ± ωt) = cos(kz)cos(ωt) sin(kz)sin(ωt), to rewrite the electric field as E = E cos(kz ωt)î + E cos(kz + ωt)î = E cos(kz)cos(ωt)î he corresponding expression for the magnetic field is he Poynting vector is then B = E c cos(kz ωt) ĵ E c cos(kz + ωt) ĵ = E c sin(kz)sin(ωt) ĵ S= E B = E µ o µ cos(kz)cos(ωt)î E o c sin(kz)sin(ωt) ĵ S= 4E cos(kz)cos(ωt)sin(kz)sin(ωt) ˆk cµ 3

14 Now use the identity cos(a)sin(a) = sin(a) to rewrite the Poynting vector as S = E sin(kz)sin(ωt) ˆk cµ c) What is the time-average of the Poynting vector? Briefly explain your answer Answer: Note the time average of the Poynting vector is given by herefore he integral is then S = = E µ c E S Sdt µ c sin(kz)sin(ωt)dt ˆk sin(kz) sin(ωt)dt ˆk sin(ωt)dt = ω cos(ωt) =, where we used the fact that ω = 4π and cos(ω ) = cos(4π ) = hus S = his result may seem surprising at first If we rewrite the z -component of the Poynting vector as where S z (t) = A(t)sin(kz), A(t) = E µ c sin(ωt) is the time dependent amplitude hen the spatial dependence sin(kz) remains fixed with respect to time ie there is no propagation Only the amplitude A(t) varies sinusoidally with time In the graph below, we plot S z (t) = E sin(ωt)sin(kz) vs z for various values of µ c ωt =, ± π /, ± π / 6, ± π / 4, ± π / 3, ± π / 4

15 We see that the nodal structure does not change If we time average S z (t) we see that the sinusoidal amplitude averages to zero over one cycle 5

16 Problem 6: Diffraction A monochromatic light with a wavelength of = 6 nm passes through a single slit which has a width of 8 mm Question: What is the distance between the slit and the screen if the first minimum in the diffraction pattern is at a distance mm from the center of the screen? Answer: he general condition for destructive interference is sin θ = m m = ±, ±, ± 3, a For small θ, we employ the approximation sinθ tan θ = y / L, which yields y m L a he first minimum corresponds to m = If y = mm, then L = ay m = (8 4 m)( 3 m) = 33 m ()(6 9 m) Question : Calculate the width of the central maximum Answer: he width of the central maximum is (see figure below) w = y = ( 3 m) = mm 6

17 Problem 7 Second-Order Bright Fringe A monochromatic light is incident on a single slit of width 8 mm, and a diffraction pattern is formed at a screen that is 8 m away from the slit he second-order bright fringe is at a distance 6 mm from the center of the central maximum What is the wavelength of the incident light? Solution: he general condition for destructive interference is sinθ = m a y L where small-angle approximation has been made hus, the position of the m-th order dark fringe measured from the central axis is y m = m L a Let the second bright fringe be located halfway between the second and the third dark fringes hat is, y b = ( y + y ) = L a + 3L a = 5 L a he approximate wavelength of the incident light is then = y a b 5L = ()(8 3 m)(6 3 m) = 64 7 m (5)(8 m) 7

18 Problem 8: Phase Difference In the double-slit interference experiment shown in the figure, suppose d = mm and L = m, and the incident light is monochromatic with a wavelength =6 nm (a) What is the phase difference between the two waves arriving at a point P on the screen when θ = 8? Answer: φ = π δ d sinθ = π = (34) ( 4 m)(sin8 ) = 46rad 6 7 m (b) What is the phase difference between the two waves arriving at a point P on the screen when y = 4mm? Answer: We use the approximation that sinθ y / L hen φ = πd sinθ d y L = (34) ( 4 m)(4 3 m) = 349rad (6 7 m)(m) (c) If the phase difference between the two waves arriving at point P is φ=/3 rad, what is the value of θ? Answer: φ = 3 rad = π d sinθ φ θ = sin π d = 38 4 rad = 8 (d) If the path difference is δ = / 4, what is the value of θ? Answer: δ δ = d sinθ θ = sin d = sin 4d = 5 3 rad = 86 (e) In the double-slit interference experiment, suppose the slits are separated by d = cm and the viewing screen is located at a distance L = m from the slits Let the incident 8

19 Answer: light be monochromatic with a wavelength = 5nm Calculate the spacing between the adjacent bright fringes on the viewing screen Because y b = m L / d, the spacing between adjacent bright fringes is Δy b = y b (m +) y b (m) = (m +) L d m L d = L d = (5 7 m)(m) = 6 5 m = 6 µm ( m) (f) What is the distance between the third-order fringe and the center-line on the viewing screen? Answer: Δy b = y b (3) y b () = (3) L d = 3 L d = 3 (5 7 m)(m) = 8 4 m = 8 µm ( m) 9

20 Problem 9: Generating Infinite Plane Waves Open the following applet that we worked with during Week 4D in class ry pulling the sheet and observe the fields that are generated In the problem below we will calculate the electric and magnetic fields associated with an oscillating infinite charged sheet Consider an infinite charged sheet that has a uniform charge density σ and is being pulled in the positive y -direction at a velocity v (see two views of the moving sheet below) a) What is a vector description of the surface current density? b) Use Ampere s Law to find the direction and magnitude of the magnetic field that is generated by the current sheet c) If the sheet position oscillates as y(t) = y sin(ωt), the magnetic field generated by the surface current will propagate along the x-axis What is the direction and magnitude of the magnetic field as a function of (x,t), for both regions x > and x <? d) What is the direction and magnitude of the electric field as a function of (x,t), for both regions x > and x <? e) What is the Poynting vector associated with this infinite plane wave in the regions regions x > and x <? f) What is total time-average powerd per unit area radiated away? g) What is the source of energy associated with the radiation? h) Calculate the time-averaged power per unit area generated by the electric force on the moving charges to determine the efficiency of the power generation in the electromagnetic radiation

21 Solution: a) What is a vector description of the surface current density? he surface current density is given by K = σ v = σv y ĵ b) Use Ampere s Law to find the direction and magnitude of the magnetic field that is generated by the current sheet We draw an Amperian loop as shown in the rotated view of the infinite sheet with the surface current density pointing out of the page So we need to do Ampere s law around the loop he current enclosed is the product of the surface current density K with the length of the Amperian loop l herefore Ampere s Law becomes he magnetic field is then B d s = µ I enc Bl = µ Kl = µ σv y l B = µ σv y / B = µ σv o y ˆk; x > µ o σv y ˆk; x < c) If the sheet position oscillates as y(t) = y sin(ωt), what are is the direction and magnitude of the magnetic field as a function of (x,t)? Suppose the sheet is now oscillating according to y(t) = y sin(ωt) hen the y-component of the velocity of the sheet is given by the expression v y (t) = ω y cos(ωt) hen the magnetic field propagates according to

22 B(x,t) = µ σω y cos(ωt kx) ˆk; x > µ σω y cos(ωt + kx) ˆk; x < d) What is the direction and magnitude of the electric field as a function of (x,t), for both regions x > and x <? he associated electric field is then E(x,t) = cµ σω y cos(ωt kx) ĵ; x > cµ σω y cos(ωt + kx) ĵ; x < e) What is the Poynting vector associated with this infinite plane wave in the regions regions x > and x <? E(x,t) cµ σ ω y cos (ωt kx) B(x,t) î; x > S(x,t) = = 4 µ cµ σ ω y cos (ωt + kx) î; x < f) What is total time-averaged power per unit area radiated away? Because the power is radiated in both positive a negative x-directions, the total time averaged power per unit area radiated away is

23 P total Area = S = cµ σ ω y 4 two sides = cµ σ ω y 4 g) Where is the source of energy associated with the radiation? It is coming from the energy required to move the charged sheet h) Calculate the time-averaged power per unit area generated by the electric force on the moving charges to determine the efficiency of the power generation in the electromagnetic radiation he electric field exerts a force on the charges, and they are moving, so P A = Fv A = q Ev A = σ Ev = σ cµ σω y cos (ωt kx) = cµ σ ω y 4 his is the same as the time-averaged power radiated, so this is % efficient process! 3

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics: Final Exam Review Session Problems Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics: Final Exam Review Session Problems Solutions Department of Physics: 8 Problem 1: Spherical Capacitor 8 Final Exam Review Session Problems Solutions A capacitor consists of two concentric spherical shells The outer radius of the inner shell is a =