ELECTROMAGNETIC WAVES

Transcription

1 LCTROMAGNTIC WAVS 3 3 IDNTIFY: Sine the speed is onstant, distane x = t ST U: The speed of light is = 3 yr = 356 s x 34 m XCUT: (a) t = = = s (b) x= t = (3 )(6 yr)(356 s/yr) = 5 m = 5 km VALUAT: The speed of light is very great The distane between stars is very large ompared to terrestrial distanes 3 IDNTIFY: Sine the speed is onstant the differene in distane is Δ t ST U: The speed of eletromagneti wes in air is = 3 XCUT: A total time differene of 6 μs orresponds to a differene in distane of Δ t = = 6 (3 )(6 s) m VALUAT: The time delay doesn t depend on the distane from the transmitter to the reeiver, it just depends on the differene in the length of the two paths 33 IDNTIFY: Apply = fλ ST U: = 3 3 m s 4 XCUT: (a) f = = = 6 Hz λ 5 m 3 m s (b) f = = = 6 Hz λ 5 m 3 m s 3 () f = = = 6 Hz 6 λ 5 m 3 m s 6 (d) f = = = 6 Hz 9 λ 5 m VALUAT: f inreases when λ dereases π 34 IDNTIFY: = fλ and k = λ ST U: = XCUT: (a) f = UVA: 5 Hz to 93 Hz UVB: 93 Hz to Hz λ π (b) k = UVA: 5 rad/m to 96 rad/m UVB: 96 rad/m to 4 rad/m λ VALUAT: Larger λ orresponds to smaller f and k 35 IDNTIFY: = fλ = B k = π / λ ω = π f ST U: Sine the we is treling in empty spae, its we speed is = XCUT: (a) f = = = 694 Hz 9 λ 43 m 6 (b) = B = (3 )(5 T) = 35 V/m 3-

2 3- Chapter 3 π π rad 4 5 () k = = = 45 rad/m ω = ( π rad)(694 Hz) = 436 rad/s 9 λ 43 m = 5 os( kx ωt ) = (35 V/m)os([45 rad/m] x [436 rad/s] t ) B B 6 5 os( kx ωt = ) = (5 T)os([45 rad/m] x [436 rad/s] t ) VALUAT: The os( kx ωt) fator is ommon to both the eletri and magneti field expressions, sine these two fields are in phase 36 IDNTIFY: = fλ = B Apply qs(3) and (39) ST U: The speed of the we is = XCUT: (a) f = = = 69 Hz 9 λ 435 m 3 V/m (b) 9 B = = = T 3 π 5 () k = = 44 rad/m ω = π f = 434 rad/s If (, z t ) = i ˆ os( kz +ω t ), then λ B (, z t ) = ˆ j B os( kz +ω t ), so that B will be in the kˆ diretion ˆ 3 5 ( z, t ) = i ( V/m)os([44 rad/s) z + [434 rad/s] t ) and B ˆ 5 ( z, t ) = j (9 T)os([44 rad/s) z + [434 rad/s] t ) VALUAT: The diretions of and B and of the propagation of the we are all mutually perpendiular The argument of the osine is kz + ωt sine the we is treling in the z-diretion Wes for visible light he very high frequenies 3 IDNTIFY and ST U: The equations are of the form of qs(3), with x replaed by z B is along the y-axis; dedue the diretion of 4 5 XCUT: ω = π f = π(6 Hz) = 33 rad/s 5 π π f ω 33 rad/s k = = = = = rad/m λ 3 4 B = 5 T 4 5 = B = (3 )(5 T) = 4 V/m B is along the y-axis B is in the diretion of propagation (the +z-diretion) From this we an dedue the diretion of, as shown in Figure 3 is along the x-axis Figure 3 = ˆos( i kz ωt) = 5 ˆ 5 (4 V/m) i os[( rad/m) z (33 rad/s) t] B = B ˆos( j kz ωt) = 4 5 ( 5 T) ˆj os[( rad/m) z (33 rad/s) t] VALUAT: and B are perpendiular and osillate in phase 3 IDNTIFY: For an eletromagneti we propagating in the negative x diretion, = os( kx+ ωt) ω = π f π and k = T = λ f = B ST U: The we speified in the problem has a different phase, so = sin( kx+ ωt) = 35 V/m, 5 k = 99 rad/m and ω = 59 rad/s XCUT: (a) B = = 5 μt

3 letromagneti Wes 3-3 ω 4 π 5 (b) f = = 95 Hz λ = = 36 m = 36 nm T = = 5 s This welength is too short π k f to be visible 4 () = fλ = (95 Hz)(36 m) = 3 This is what the we speed should be for an eletromagneti we propagating in vauum ω π ω VALUAT: = fλ = = is an alternative expression for the we speed π k k 39 IDNTIFY and ST U: Compare the ( y, t ) given in the problem to the general form given by q(3) Use the diretion of propagation and of to find the diretion of B (a) XCUT: The equation for the eletri field ontains the fator sin( ky ωt) so the we is treling in the +y-diretion The equation for ( y, t ) is in terms of sin( ky ωt) rather than os( ky ωt); the we is shifted in phase by 9 relative to one with a os( ky ωt) fator 5 (b) ( yt, ) = (3 V/m) kˆ sin[ ky (65 rad/s) t] Comparing to q(3) gives ω = 65 rad/s π π π(99 ) 4 ω = π f = so λ = = = m λ ω (65 rad/s) () B must be in the +ydiretion (the diretion in whih the we is treling) When is in the z-diretion then B must be in the xdiretion, as shown in Figure 39 π ω 65 rad/s Figure 39 k = = = = λ 99 5 = 3 V/m 3 4 rad/m 5 3 V/m 3 3 = = = T Then B 99 Using q(3) and the fat that B is in the 3 ˆ 3 B = (3 T) isin[(4 rad/m) y (65 rad/s) t] ˆ i diretion when is in the ˆ k diretion, VALUAT: and B are perpendiular and osillate in phase 3 IDNTIFY: Apply qs(3) and (39) f = / λ and k = π / λ ST U: The we in this problem has a different phase, so B (, z t) = B sin( kx+ ωt) XCUT: (a) The phase of the we is given by kx + ωt, so the we is treling in the x diretion 4 π π f k (3 rad m)(3 m s) (b) k = = f = = = 659 Hz λ π π () Sine the magneti field is in the + y -diretion, and the we is propagating in the x -diretion, then the eletri field is in the + z -diretion so that B will be in the x -diretion =+ B kˆ = B ˆ sin( kx+ωt) k 9 4 = ( (35 T))sin ((3 rad/m) x+ (44 rad/s) t) kˆ 4 =+ (4 V m)sin (3 rad/m) x+ (44 rad/s) t kˆ ( ) VALUAT: and B he the same phase and are in perpendiular diretions 3 IDNTIFY and ST U: = fλ allows alulation of λ k = π / λ and ω = π f q(3) relates the eletri and magneti field amplitudes 99 XCUT: (a) = fλ so λ = = = 36 m 3 f 3 Hz y

4 3-4 Chapter 3 π π rad (b) k = = = 4 rad/m λ 36 m 3 6 () ω = π f = ( π)(3 Hz) = 5 rad/s (d) q(3): = B = (99 )(4 T) = 44 V/m VALUAT: This we has a very long welength; its frequeny is in the AM radio braodast band The eletri and magneti fields in the we are very weak 3 IDNTIFY: = B 4 ST U: The magneti field of the earth is about T XCUT: 3 35 V/m B = = = T 3 VALUAT: The field is muh smaller than the earth's field 33 IDNTIFY and ST U: v= fλ relates frequeny and welength to the speed of the we Use q(3) to alulate n and K v XCUT: (a) λ = = = 3 m 4 f 5 Hz 99 (b) λ = = = 56 m 4 f 5 Hz 99 () n = = = 3 v (d) n= KKm K so K = n = (3) = 9 VALUAT: In the material v< and f is the same, so λ is less in the material than in air v< always, so n is always greater than unity 34 IDNTIFY: Apply q(3) = B v= fλ Apply q(39) with μ = Kmμ in plae of μ ST U: K = 364 K m = 5 XCUT: (a) v = (3 m s) = = 69 m s KK (364)(5) m v 69 m s 6 (b) λ = = = 6 m f 65 Hz 3 V m () 4 B = = = T v 69 m s 3 B ( V m)(4 T) (d) I = = = 55 W m Kmμ (5) μ VALUAT: The we trels slower in this material than in air 35 IDNTIFY: I = / A I = = B ST U: The surfae area of a sphere of radius r is A= 4π r = 5 C /N m (5)(5 W) XCUT: (a) I = = = 33 W/m A 4 π (3 m) (b) I (33 W/m ) 6 = = = 5 V/m B = = T = μt (5 C /N m )(3 ) VALUAT: At the surfae of the bulb the power radiated by the filament is spread over the surfae of the bulb Our alulation approximates the filament as a point soure that radiates uniformly in all diretions 36 IDNTIFY and ST U: The diretion of propagation is given by B XCUT: (a) S ˆ = i ˆ ( ˆ j) =k ˆ (b) S ˆ = ˆ j i ˆ =k ˆ () S ˆ = ( k ˆ ) ( i ˆ ) = ˆ j (d) S ˆ = i ˆ ( k ˆ ) = ˆ j VALUAT: In eah ase the diretions of, B and the diretion of propagation are all mutually perpendiular

5 letromagneti Wes IDNTIFY: = B B is in the diretion of propagation ST U: = 3 = 4 V/m XCUT: B = = 33 T For in the +x-diretion, B is in the +z-diretion when B is in the +y-diretion VALUAT:, B and the diretion of propagation are all mutually perpendiular 3 IDNTIFY: The intensity of the eletromagneti we is given by q(39): I = = rms The total energy passing through a window of area A during a time t is IAt ST U: = 5 F/m XCUT: nergy = rmsat = (5 F m)(3 m s)( V m) (5 m )(3 s) = 59 μj VALUAT: The intensity is proportional to the square of the eletri field amplitude 39 IDNTIFY and ST U: Use q(39) to alulate I, q(3) to alulate B, and use I = /4π r to alulate 5 (a) XCUT: I = ; = 9 V/m, so I = W/m (b) = B so B = / = 3 T 5 3 () = I(4 πr ) = (5 W/m )(4 π)(5 m) = 4 W (d) VALUAT: The alulation in part () assumes that the transmitter emits uniformly in all diretions 3 IDNTIFY and ST U: I = / Aand I = rms 4 4 XCUT: (a) The erage power from the beam is ( W m )(3 = IA = m ) = 4 W (b) I W m = = = 4 V m (5 F m)(3 m s) rms VALUAT: The laser emits radiation only in the diretion of the beam 3: IDNTIFY: I = / A ST U: At a distane r from the star, the radiation from the star is spread over a spherial surfae of area A= 4π r 3 5 XCUT: = I(4 πr ) = (5 W m )(4 π)( m) = 5 J VALUAT: The intensity dereases with distane from the star as /r 3 IDNTIFY and ST U: = fλ, = B and I = B /μ 3 m s XCUT: (a) f = = = 4 Hz λ 354 m 54 V m = = = T (b) B 3 m s B (54 V m)( T) 6 () I = S = = = 3 W m μ μ VALUAT: Alternatively, I = 33 IDNTIFY: = IAand I = ST U: The surfae area of a sphere is A= = SA= (4 π r ) μ V m = = = 4 T 3 m s XCUT: B 4 π r μ (6 W)(3 m s) μ = = = V m πr π(5m) VALUAT: and B are both inversely proportional to the distane from the soure 34 IDNTIFY: The oynting vetor is S = B ST U: The eletri field is in the +y-diretion, and the magneti field is in the +z-diretion os φ = ( + os φ) XCUT: (a) S ˆ = ˆ B ˆ = ( ˆ j) k ˆ =i ˆ The oynting vetor is in the x-diretion, whih is the diretion of propagation of the we

6 3-6 Chapter 3 (, ) (, ) (b) S = x t B x t = B os ( kx+ ωt) = B ( + os(( ωt+ kx)) ) But over one period, the μ μ μ B osine funtion erages to zero, so we he S = This is q(39) μ VALUAT: We an also show that these two results also apply to the we represented by q(3) 35 IDNTIFY: Use the radiation pressure to find the intensity, and then = I(4 π r ) I ST U: For a perfetly absorbing surfae, prad = 3 XCUT: prad = Iso I = p rad = W/m Then 3 5 = I(4 πr ) = ( W m )(4 π)(5 m) = 5 W VALUAT: ven though the soure is very intense the radiation pressure 5 m from the surfae is very small 36 IDNTIFY: The intensity and the energy density of an eletromagneti we depends on the amplitudes of the eletri and magneti fields ST U: Intensity is I = / A, and the erage power is = I/, where I = The energy density is u = 36, W XCUT: (a) I = /A = π (5 m) (b) I = gives = W/m = I/ = ( W/m ) 3 = 34 a I = = ( W/m ) (5 C /N m )(3 ) = 3 N/C () u = so u, ( ) B = / = (3 N/C)/(3 m/s) = 4 = and = u =, so ( ) 5 C /N m (3 N/C) = = T J/m 3 (d) As was shown in Setion 34, the energy density is the same for the eletri and magneti fields, so eah one has 5% of the energy density VALUAT: Compared to most laboratory fields, the eletri and magneti fields in ordinary radiowes are extremely weak and arry very little energy 3 IDNTIFY and ST U: Use qs(33) and (33) dp S I XCUT: (a) By q(33) the erage momentum density is = = dv 3 dp W/m 5 = = kg/m s dv (99 ) 3 S I W/m 6 (b) By q(33) the erage momentum flow rate per unit area is 6 = = = a 99 VALUAT: The radiation pressure that the sunlight would exert on an absorbing or refleting surfae is very small 3 IDNTIFY: Apply qs(33) and (333) The erage momentum density is given by q(33), with S replaed by S = I 5 ST U: atm = 3 a I 5 W m 6 XCUT: (a) Absorbed light: rad 33 p = = = a Then 3 m s 6 33 a prad = = 3 atm 5 3 a atm (b) Refleting light: p p 5 6 a rad 5 = = 3 a atm I (5W m ) rad = = = 3 m s 65 atm 5 6 a Then

7 () The momentum density is dp S 5 W m = = = dv (3 m s) 4 kg m s letromagneti Wes 3- VALUAT: The fator of in prad for the refleting surfae arises beause the momentum vetor totally reverses diretion upon refletion Thus the hange in momentum is twie the original momentum 39 IDNTIFY: Apply q(34) and (39) ST U: q(36) is S = B XCUT: S = = = = B= B= = = μ μ μ μ μ μ μ μ 3 VALUAT: We an also write S = ( B) = B S an be written solely in terms of or solely in terms of B 33 IDNTIFY: The eletri field at the nodes is zero, so there is no fore on a point harge plaed at a node ST U: The loation of the nodes is given by q(336), where x is the distane from one of the planes λ = / f λ 3 m s XCUT: Δ xnodes = = = = m = m There must be nodes at the planes, whih f (5 Hz) are m apart, and there are two nodes between the planes, eah m from a plane It is at m, 4 m, and 6 m from one plane that a point harge will remain at rest, sine the eletri fields there are zero VALUAT: The magneti field amplitude at these points isn t zero, but the magneti field doesn t exert a fore on a stationary harge 33 IDNTIFY and ST U: Apply qs(336) and (33) XCUT: (a) By q(33) we see that the nodal planes of the B field are a distane λ / apart, so λ / = 355 mm and λ = mm (b) By q(336) we see that the nodal planes of the field are also a distane λ / = 355 mm apart 3 () v= fλ = ( Hz)( m) = 56 VALUAT: The spaing between the nodes of is the same as the spaing between the nodes of B Note that v<, as it must 33 IDNTIFY: The nodal planes of and B are loated by qs(36) and (3) 3 ST U: λ = = = 4 m 6 f 5 Hz λ XCUT: (a) Δ x = = m (b) The distane between the eletri and magneti nodal planes is one-quarter of a welength, so is λ Δx m = = = m 4 VALUAT: The nodal planes of B are separated by a distane λ /and are midway between the nodal planes of 333 (a) IDNTIFY and ST U: The distane between adjaent nodal planes of B is λ / There is an antinodal plane of B midway between any two adjaent nodal planes, so the distane between a nodal plane and an adjaent antinodal plane is λ / 4 Use v= fλ to alulate λ v XCUT: λ = = = 5 m f Hz λ 5 m 43 3 = = m = 43 mm 4 4 (b) IDNTIFY and ST U: The nodal planes of are at x =, λ /, λ, 3 λ /,, so the antinodal planes of are at x = λ / 4, 3 λ/4, 5 λ/4, The nodal planes of B are at x = λ / 4, 3 λ/ 4, 5 λ/4,, so the antinodal planes of B are at λ /, λ, 3 λ /, XCUT: The distane between adjaent antinodal planes of and antinodal planes of B is therefore λ / 4 = 43 mm () From qs(336) and (33) the distane between adjaent nodal planes of and B is λ / 4 = 43 mm VALUAT: The nodes of oinide with the antinodes of B, and onversely The nodes of B and the nodes of are equally spaed

8 3- Chapter IDNTIFY: valuate the derivatives of the expressions for y and Bz that are given in qs(334) and (335) ST U: sin kx = k oskx, sinωt = ωosωt oskx =ksinkx, osωt =ωsinωt t t y XCUT: (a) = ( sin kxsin ωt) = ( k oskxsin ωt) and y ω y = k sin kxsinωt= sin kxsin ωt= μ t Bz Similarly: = ( B oskxos ωt) = ( + kb sinkxos ωt) and Bz Bz k B oskxos t ω = ω = B oskxos ωt = μ t y (b) = ( sinkxsin ωt) =k oskxsinωt y ω = oskxsinωt = ω oskxsinωt =ωb oskxsinωt y Bz =+ (B oskxos ωt) = t t Bz Similarly: = ( + B oskxos ωt) =kb sinkxosωt Bz ω ω = B sinkxosωt = B sinkxosωt B (, ) z y x t = μω sinkxos ωt = μ ( sinkxsin ωt) = μ t t VALUAT: The standing wes are linear superpositions of two treling wes of the same k and ω 335 IDNTIFY: The nodal and antinodal planes are eah spaed one-half welength apart λ = L, and the frequeny of these wes obeys the equation fλ = λ = we he L = (5/)( m) = 35 m (b) Solving for the frequeny gives f = /λ = (3 m/s)/( m) = 46 9 Hz () L = 355 m in this ase ( ) λ = L, so λ = L/5 = (355 m)/5 = 4 m f = /λ = (3 m/s)/(4 m) = 9 Hz VALUAT: Sine mirowes he a reasonably large welength, mirowe ovens an he a onvenient size for household kithens Ovens using radiowes would need to be far too large, while ovens using visible light would he to be mirosopi 336 IDNTIFY: valuate the partial derivatives of the expressions for y ( x, t ) and Bz ( x, t ) ST U: sin( kx ωt) = k os( kx ωt), sin( kx ωt) =ωos( kx ωt) os( kx ωt) =k sin( kx ωt), t os( kx ωt) = ωsin( kx ωt) t XCUT: Assume ˆsin( = j kx ωt) and B = B ˆsin( k kx ωt+ φ), with π < φ < π q (3) is ST U: welengths fit in the oven, so ( ) XCUT: (a) Sine ( ) L, y Bz = t This gives k os( kx ωt) =+ ωb os( kx ωt + φ), so φ =, and k = ωb, so ω π f B = B = B = fωb = B Similarly for q(34), z = μ y gives k π / ω t kb os( kx ωt + φ) = μω os( kx ωt), so φ = and kb = μω, so μω πf fω B = = = = k π / ω VALUAT: The and B fields must osillate in phase

9 letromagneti Wes IDNTIFY and ST U: Take partial derivatives of qs(3) and (34), as speified in the problem y Bz XCUT: q(3): = t Taking t of both sides of this equation gives y Bz B = q(34) says z μ y = Taking t t t of B y z y B z y y both sides of this equation gives = μ, so = But = (The order in t t μ t t x Bz Bz Bz Bz whih the partial derivatives are taken doesn't hange the result) So = and = μ t μ t, as was to be shown VALUAT: Both fields, eletri and magneti, satisfy the we equation, q(3) We he also shown that both fields propagate with the same speed v = / μ 33 IDNTIFY: The erage energy density in the eletri field is the magneti field is ub, = ( B ) μ kx ωt = ST U: ( os ( )) u = ( ) and the erage energy density in, XCUT: (, ) os( ) y x t = kx ωt u = y = os ( kxωt) and u, = 4 Bz = B os( kx ωt), so u = B Bz B os ( kx ωt) μ = μ and ub, = B = B, so 4μ u, = 4 B =, so u, = B, whih equals u B, μ μ VALUAT: Our result allows us to write u = u, = and u = ub, = B μ 339 IDNTIFY: The intensity of an eletromagneti we depends on the amplitude of the eletri and magneti fields Suh a we exerts a fore beause it arries energy ST U: The intensity of the we is I = / A=, and the fore is F = A where = I/ XCUT: (a) I = /A = (5, W)/[4π(55 m) ] = 6 W/m I (6 W/m ) (b) I =, so = = = 3 N/C (5 C /N m )(3 ) B = / = (3 N/C)/(3 9 m/s) = T () F = A = (I/)A = (6 W/m )(5 m)(4 m)/(3 m/s) = N VALUAT: The fields are very weak ompared to ordinary laboratory fields, and the fore is hardly worth worrying about I 34 IDNTIFY: = fλ = B I = For a totally absorbing surfae the radiation pressure is ST U: The we speed in air is = XCUT: (a) f = = = Hz λ 34 m 35 V/m 9 (b) B = = = 45 T 3 3 () I = = (54 C /N m )(3 )(35 V/m) = 4 W/m 3 IA (4 W/m )(4 m ) (d) F = (pressure) A= = = 94 N 3 VALUAT: The intensity depends only on the amplitudes of the eletri and magneti fields and is independent of the welength of the light

10 3- Chapter 3 34 (a) IDNTIFY and ST U: Calulate I and then use q(39) to alulate and q(3) to alulate B 3 3 W XCUT: The intensity is power per unit area: I = = = 3 65 W/m A π (5 m) I =, so = μi μ = (4π T m/a)(99 )(65 W/m ) = V/m V/m 6 B = = = 34 T 99 VALUAT: The magneti field amplitude is quite small (b) IDNTIFY and ST U: qs(4) and (3) give the energy density in terms of the eletri and magneti field values at any time For sinusoidal fields erage over and B to get the erage energy densities XCUT: The energy density in the eletri field is u = = os( kx ωt) and the erage value of os ( kx ωt) is The erage energy density in the eletri field then is u is 6 3, B B (34 T) 6 3 ub = The erage value is ub, = = = 9 J/m μ 4μ 4(4π T m/a) = = (54 C / N m )( V/m) = 9 J/m The energy density in the magneti field VALUAT: Our result agrees with the statement in Setion 34 that the erage energy density for the eletri field is the same as the erage energy density for the magneti field () IDNTIFY and ST U: The total energy in this length of beam is the total energy density 6 3 u = u, + ub, = J/m times the volume of this part of the beam XCUT: U = u LA= ( J/m )( m) π (5 m) = J VALUAT: This quantity an also be alulated as the power output times the time it takes the light to trel L = L 3 m m: U = = (3 W) = J, whih heks IDNTIFY: Use the gaussian surfae speified in the hint ST U: The we is in free spae, so in Gauss s law for the eletri field, Q enl = and da A = Gauss s law for the magneti field says B da A = XCUT: Use a gaussian surfae suh that the front surfae is ahead of the we front (no eletri or magneti Qenl fields) and the bak fae is behind the we front, as shown in Figure 34 da A = x A= =, so x = ε B da A = Bx A= and B x = VALUAT: The we must be transverse, sine there are no omponents of the eletri or magneti field in the diretion of propagation Figure IDNTIFY: I = / A For an absorbing surfae, the radiation pressure is I prad = ST U: Assume the eletromagneti wes are formed at the enter of the sun, so at a distane r from the enter of the sun I = /(4 π r )

11 6 39 W XCUT: (a) At the sun s surfae: I = = = 64 W m and 4πR 4 π(696 m) I 64 W m rad = = = a letromagneti Wes 3- p 3 m s (b) Halfway out from the sun s enter, the intensity is 4 times more intense, and so is the radiation pressure: I = 6 W/m and p rad = 5 a At the top of the earth s atmosphere, the measured sunlight intensity is 6 4 W m 5 a =, whih is about, times less than the values above VALUAT: (b) The gas pressure at the sun s surfae is 5, times greater than the radiation pressure, and 3 halfway out of the sun the gas pressure is believed to be about 6 times greater than the radiation pressure Therefore it is reasonable to ignore radiation pressure when modeling the sun s interior struture I = I = A ST U: 3 3 W XCUT: I = = = W/m A 36 m I ( W/m ) = = = 4 N/C (54 C /N m )(3 ) 344 IDNTIFY: VALUAT: This value of is similar to the eletri field amplitude in ordinary light soures 345 IDNTIFY: The same intensity light falls on both refletors, but the fore on the refleting surfae will be twie as great as the fore on the absorbing surfae Therefore there will be a net torque about the rotation axis ST U: For a totally absorbing surfae, F = A= (/) I A, while for a totally refleting surfae the fore will be twie as great The intensity of the we is I = One we he the torque, we an use the rotational form of Newton s seond law, τ net = Iα, to find the angular aeleration A XCUT: The fore on the absorbing refletor is FAbs = p A= (/) I A= = A For a totally refleting surfae, the fore will be twie as great, whih is The net torque is therefore τ net = FRefl ( L/) FAbs( L/) = AL/4 Newton s nd law for rotation gives τ net = Iα A L/4 = m( L/) α ( 5 C /N m )(5 m) (5 N/C) 3 Solving for α gives α = A/( ml) = = 39 rad/s ()(4 kg)( m) VALUAT: This is an extremely small angular aeleration To ahieve a larger value, we would he to greatly inrease the intensity of the light we or derease the mass of the refletors 346 IDNTIFY: For light of intensity Iabs inident on a totally absorbing surfae, the radiation pressure is Iabs Irefl prad,abs = For light of intensity I refl inident on a totally refleting surfae, prad,refl = ST U: The total radiation pressure is prad = prad,abs + prad,refl Iabs = wi and Irefl = ( w) I Iabs Irefl wi ( wi ) ( wi ) XCUT: (a) prad = prad,abs + prad,refl = + = + = I I (b) (i) For totally absorbing w= so prad = (ii) For totally refleting w= so prad = These are just equations 33 and ( 9)(4 W m ) 6 () For w= 9 and I = 4 W/m, prad = = 53 a For w = and 3 3 ( )(4 W m ) 6 I = 4 W m, prad = = a 3 VALUAT: The radiation pressure is greater when a larger fration is refleted 34 IDNTIFY and ST U: In the wire the eletri field is related to the urrent density by q(5) Use Ampere s law to alulate B The oynting vetor is given by q(3) and the equation that follows it relates the energy flow through a surfae to S

12 3- Chapter 3 XCUT: (a) The diretion of is parallel to the axis of the ylinder, in the diretion of the urrent From q(5), = ρj = ρi/ πa ( is uniform aross the ross setion of the ondutor) (b) A ross-setional view of the ondutor is given in Figure 34a; take the urrent to be oming out of the page Apply Ampere s law to a irle of radius a A B dl = B( π a) Ienl = I Figure 34a μi A B dl = μienlgives B( π a) = μi and B = π a The diretion of B is ounterlokwise around the irle () The diretions of and B are shown in Figure 34b Figure 34b The diretion of S = B μ is radially inward ρi μi S = B= μ μ πa πa ρi S = 3 π a (d) VALUAT: Sine S is onstant over the surfae of the ondutor, the rate of energy flow is given by S I li times the surfae of a length l of the ondutor: = SA = ρ ρ S( πal) = ( πal) 3 π a = πa But ρl R = π a, so the result from the oynting vetor is RI = This agrees with = I R, the rate at whih eletrial energy is being dissipated by the resistane of the wire Sine S is radially inward at the surfae of the wire and has magnitude equal to the rate at whih eletrial energy is being dissipated in the wire, this energy an be thought of as entering through the ylindrial sides of the ondutor 34 IDNTIFY: The intensity of the we, not the eletri field strength, obeys an inverse-square distane law ST U: The intensity is inversely proportional to the distane from the soure, and it depends on the amplitude of the eletri field by I = S = XCUT: Sine I =, I A point at m ( m) from the soure is 5 times loser to the soure than a point that is m from it Sine I /r and ( m)/( m) = /5, we he I = 5 I Sine I, we he = 5 = (5)(5 N/C) = 5 N/C VALUAT: While the intensity inreases by a fator of 5 = 5, the amplitude of the we only inreases by a fator of 5 Reall that the intensity of any we is proportional to the square of its amplitude dφb 349 IDNTIFY and ST U: The magnitude of the indued emf is given by Faraday s law: = To alulate dt dφ B / dt we need db / dt at the antenna Use the total power output to alulate I and then ombine q(39) and (3) to alulate B The time dependene of B is given by q(3) XCUT: Φ B = Bπ R, where R = 9 m is the radius of the loop (This assumes that the magneti field is db uniform aross the loop, an exellent approximation) = π R dt B = B os( kx ωt) so db = Bωsin( kx ωt) dt The imum value of db dt is Bω, so = πr B ω R= = = = 6 9 m, ω π f π(95 Hz) 59 rad/s R

13 letromagneti Wes 3-3 Calulate the intensity I at this distane from the soure, and from that the magneti field amplitude B : 3 55 W 4 ( B ) I = = = W/m I 3 4πr 4 π(5 m) = B μ = μ = μ Thus B 4 μi (4π T m/a)( W/m ) = = = T 9 = πrb ω = π(9 m) (4 T)(59 rad/s) = 36 V VALUAT: An indued emf of this magnitude is easily deteted 35 IDNTIFY: The nodal planes are one-half welength apart ST U: The nodal planes of B are at x = λ/4, 3λ/4, 5λ/4,, whih are λ/ apart XCUT: (a) The welength is λ = /f = (3 m/s)/( 6 Hz) = m So the nodal planes are at ( m)/ = 364 m apart (b) For the nodal planes of, we he λ n = L/n, so L = nλ/ = ()( m)/ = 9 m VALUAT: Beause radiowes he long welengths, the distanes involved are easily measurable using ordinary meterstiks 35 IDNTIFY and ST U: Find the fore on you due to the momentum arried off by the light xpress this fore in terms of the radiated power of the flashlight Use this fore to alulate your aeleration and use a onstant aeleration equation to find the time (a) XCUT: prad = I/ and F = prad A gives F = IA/ = / 9 a = F/ m= /( m) = ( W)/[(5 kg)(3 )] = 444 x 9 4 Then x x = vxt+ axt gives t = ( x x)/ a x = (6 m)/(444 ) = 49 s = 36 h VALUAT: The radiation fore is very small In the alulation we he ignored any other fores on you (b) You ould throw the flashlight in the diretion away from the ship By onservation of linear momentum you would move toward the ship with the same magnitude of momentum as you ge the flashlight = IAand I = = B ST U: The power arried by the urrent i is = Vi 5 Vi (5 V)( A) 4 XCUT: I = = and = = = = 64 V m A A A ( m ) (3 m s) 35 IDNTIFY: B 4 64 V m 4 5 = = = T 3 m s 5 (5 V)( A) 6 VALUAT: I = Vi/ A= = 5 W/m This is a very intense beam spread over a m large area 353 IDNTIFY: The orbiting satellite obeys Newton s seond law of motion The intensity of the eletromagneti wes it transmits obeys the inverse-square distane law, and the intensity of the wes depends on the amplitude of the eletri and magneti fields ST U: Newton s seond law applied to the satellite gives mv /R = GmM/r, where M is the mass of the arth and m is the mass of the satellite The intensity I of the we is I = S =, and by definition, I = /A XCUT: (a) The period of the orbit is hr Applying Newton s nd law to the satellite gives mv /R = GmM/r, m( π r/ T) GmM whih gives = Solving for r, we get r r ( 66 N m /kg )( 59 kg)( 36 s) /3 4 GMT r = 66 m = = 4π 4π The height above the surfae is h = 66 m 63 6 m = m The satellite only radiates its energy to the lower hemisphere, so the area is / that of a sphere Thus, from the definition of intensity, the intensity at the ground is (b) I = S =, so I = /A = /(πh ) = (5 W)/[π( m) ] = 95 5 I (95 W/m ) /3 Then = = = (5 C /N m )(3 ) B = = = 5 W/m 6 5 / ( N/C)/(3 ) 93 T t = d = = / ( m)/(3 ) 63 s 6 N/C

14 3-4 Chapter 3 () = I/ = (95 5 W/m )/(3 m/s) = 35 3 a (d) λ = /f = (3 m/s)/(554 6 Hz) = 9 m VALUAT: The fields and pressures due to these wes are very small ompared to typial laboratory quantities 354 IDNTIFY: For a totally refletive surfae the radiation pressure is I Find the fore due to this pressure and mm sun express the fore in terms of the power output of the sun The gritational fore of the sun is Fg = G r 3 ST U: The mass of the sum is M sun = 99 kg G = 66 N m /kg XCUT: (a) The sail should be refletive, to produe the imum radiation pressure I (b) Frad = A, where A is the area of the sail I =, where r is the distane of the sail from the sun 4π r A A A mm sun Frad = = F rad = Fg so = G 4πr πr π r r 3 πgmm sun π(3 )(66 N m /kg )(, kg)(99 kg) A = = 6 39 W 6 A = 64 m = 64 km () Both the gritational fore and the radiation pressure are inversely proportional to the square of the distane from the sun, so this distane divides out when we set F = F rad g VALUAT: A very large sail is needed, just to overome the gritational pull of the sun 355 IDNTIFY and ST U: The gritational fore is given by q() xpress the mass of the partile in terms of its density and volume The radiation pressure is given by q(33); relate the power output L of the sun to the intensity at a distane r The radiation fore is the pressure times the ross setional area of the partile mm 4 3 XCUT: (a) The gritational fore is Fg = G The mass of the dust partile is m= ρv = ρ r 3 πr Thus 3 4ρGπMR Fg = 3r I (b) For a totally absorbing surfae prad = If L is the power output of the sun, the intensity of the solar radiation L a distane r from the sun is L I = Thus p rad = 4π r 4π r The fore F rad that orresponds to p rad is in the diretion of propagation of the radiation, so Frad = prad A, where A = π R is the omponent of area of the partile L LR perpendiular to the radiation diretion Thus Frad = ( R ) π = 4π r 4r () Fg = Frad 3 4ρGπMR LR = 3r 4r 4ρ G π M L 3 R= and R= L 3 4 6ρGπM 6 3(39 W) R = 3 3 6(99 )(3 kg/m )(663 N m / kg ) π (99 kg) R = = 9 m 9 μm VALUAT: The gritational fore and the radiation fore both he a dependene on the distane from the sun, so this distane divides out in the alulation of R F rad LR 3r 3L (d) = F 3 = rad is proportional to R and F g is proportional to R 3, so this Fg 4r 4ρGπmR 6ρGπMR ratio is proportional to /R If R < μm then Frad > Fg and the radiation fore will drive the partiles out of the solar system v 356 IDNTIFY: The eletron has aeleration a = R 9 9 ST U: ev = 6 C An eletron has q = e= 6 C r

15 letromagneti Wes 3-5 XCUT: For the eletron in the lassial hydrogen atom, its aeleration is 9 v mv (36 ev)(6 J ev) a = = = = 93 Then using the formula for the rate of energy 3 R mr (9 kg)(59 m) emission given in roblem 35: 9 d q a (6 C) (93 m s ) = = = 464 J s = 9 ev s This large value of d 3 3 dt 6π 6 π(3 m s) dt would mean that the eletron would almost immediately lose all its energy VALUAT: The lassial physis result in roblem 35 must not apply to eletrons in atoms v 35 IDNTIFY: The orbiting partile has aeleration a = R 3 ST U: K = mv An eletron has mass m e = 9 kg and a proton has mass m p = 6 kg qa C(ms) N m J d XCUT: (a) W 3 = = = = = 3 6 π (C N m )(m s) s s dt (b) For a proton moving in a irle, the aeleration is 6 9 v mv (6 ev) (6 J ev) 5 a = = = = 53 m s The rate at whih it emits energy beause of R mr (6 kg) (5 m) 9 5 d q a (6 C) (53 m s ) 3 5 its aeleration is = = = 33 J s = 3 ev s 3 3 dt 6π 6 π(3 m s) 5 ( d dt)(s) 3 ev Therefore, the fration of its energy that it radiates every seond is 6 39 = = 6 ev () Carry out the same alulations as in part (b), but now for an eletron at the same speed and radius That means the eletron s aeleration is the same as the proton, and thus so is the rate at whih it emits energy, sine they also he the same harge However, the eletron s initial energy differs from the proton s by the ratio of their masses: 3 me 6 (9 kg) e = p = (6 ev) = 33 ev Therefore, the fration of its energy that it radiates every m p (6 kg) 5 ( d dt)(s) 3 ev seond is 54 = = 33 ev 6 9 (6 ev)(6 J/eV) VALUAT: The proton has speed v = = = 339 The eletron mp 6 kg has the same speed and kineti energy 3 kev The partiles in the aelerator radiate at a muh smaller rate than the eletron in roblem 356 does, beause in the aelerator the orbit radius is very muh larger than in the atom, so the aeleration is muh less 35 IDNTIFY and ST U: Follow the steps speified in the problem kcx ( xt, ) = e sin( kx ωt) XCUT: (a) y y C C ( C)e k x = k sin( kcx ωt) + ( + kc)e k x os( kcxωt) y C C ( C)e k x = + k sin( kcx ωt) + ( kc)e k x os( kcxωt) kcx kcx + ( k )e os( k x ωt) + ( k )e sin( k x ωt) C C C C y C Ce k x C = k os( kcxωt) y = e k x ω os( kx C ωt) x t y μy C C = gives k x k x C C C Setting kc ρt μ ωμ =, or kc = ω ρ ρ C k e os( k x ωt) = e ωos( k x ωt) This will only be true if (b) The energy in the we is dissipated by the ir heating of the ondutor y ρ ( Ω m) 5 () y = kcx=, x= = = = 66 m 6 e kc ωμ π ( Hz) μ VALUAT: The lower the frequeny of the wes, the greater is the distane they an penetrate into a ondutor A dieletri (insulator) has a muh larger resistivity and these wes an penetrate a greater distane in these materials

16