em waves jee

8/18/2019 Em Waves Jee

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CHAPTER 32

Maxwell’s Equations andElectromagnetic Waves

1* ! A parallel-plate capacitor in air has circular plates of radius 2.3 cm separated by 1.1 mm. Charge is flowing

onto the upper plate and off the lower plate at a rate of 5 A. (a) Find the time rate of change of the electric field

between the plates. (b) Compute the displacement current between the plates and show that it equals 5 A.

(a) Use Equ. 23-25: E = Q/ε 0 A; dE /dt = (dQ/dt )/ε 0 A

(b) Use Equ. 32-3:φ

e = EA

dE /dt = I /ε 0 A = 3.40×1014 V/m.s I d =

ε 0 A(dE /dt ) = I = 5 A

2 ! In a region of space, the electric field varies according to t2000sin)C/ N(0.05= E 0, where t is in

seconds. Find the maximum displacement current through a 1-m2 area perpendicular to E.

Use Equs. 23-14 and 3-3 I d = (8.85×10 –12 ×0.05×2000) A = 8.85×10 –10 A

3 !! For Problem 1, show that at a distance r from the axis of the plates the magnetic field between the plates is

given by B = (1.89×10 –3 T/m)r if r is less than the radius of the plates.1. Use Equ. 32-4; I = I d ; apply cylindrical symmetry

2. Evaluate B(r )

2π rB = µ 0 I d (r 2/ R

2); B = µ 0 I d r /2π R

2

B(r ) = (1.89×10 –3 T/m)r

4 !! (a) Show that for a parallel-plate capacitor the displacement current is given by I d = C dV /dt , where C is the

capacitance and V the voltage across the capacitor. (b) A parallel plate capacitor C = 5 nF is connected to an

emf E = E 0 cos ω t , where E 0 = 3 V and ω = 500π . Find the displacement current between the plates as a

function of time. Neglect any resistance in the circuit.

(a) Use Equs. 25-10 and 32-3; E = V /d

(b) dV /dt = – E 0 sin ω t

I d = ε 0 A(dE /dt ) = (ε 0 A/d )(dV /dt ) = C dV /dt

I d = –(23.6 µ A) sin 500π t

5* !! Current of 10 A flows into a capacitor having plates with areas of 0.5 m2. (a) What is the displacement

current between the plates? (b) What is dE /dt between the plates for this current? (c) What is the line integral

of !d ⋅ B around a circle of radius 10 cm that lies within and parallel to the plates?(a) See Problem 1

(b) dE /dt = I d /ε 0 A (see Problem 1)

(c) Use Equ. 32-4; I d enclosed = I d (π r 2/ A)

I d = 10 A

dE /dt = 2.26×1012 V/m.s∫ B .d ! = µ 0 I d (π r

2/ A) = 7.90×10 –7 T.m

6 !! A parallel-plate capacitor with circular plates is given a charge Q0. Between the plates is a leaky dielectric

having a dielectric constant of κ and a resistivity ρ . (a) Find the conduction current between the plates as a

function of time. (b) Find the displacement current between the plates as a function of time. What is the total


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Chapter 32 Maxwell’s Equations and Electromagnetic Waves

(conduction plus displacement) current? (c) Find the magnetic field produced between the plates by the leakage

discharge current as a function of time. (d ) Find the magnetic field between the plates produced by the

displacement current as a function of time. (e) What is the total magnetic field between the plates during

discharge of the capacitor?

(a) If Q is the charge on the capacitor plates, then the discharge current I = – dQ/dt. Also, I = V / R = VA/d ρ ,

where we have used Equ. 26-8. From the definition of capacitance, I = AQ/Cd ρ . The differential equation dQ/dt + α Q = 0 has the solution Q = Q0 e

– α t . Here α = A/Cd ρ = 1/ε 0κρ . Thus, I = – dQ/dt = (Q0/ε 0κρ ) e / t 0 ρ κ ε − .

(b) From Problem 32-4, I d = C dV /dt = dQ/dt = – I . The total current is zero.

(c) B = µ 0 I r /2π R2 (see Problem 32-3); B = (µ 0Q0r /2π R

2ε 0κρ ) e

/ t 0 ρ κ ε − .

(d ) Since I d = – I , Bd = –(µ 0Q0r /2π R2ε 0κρ ) e

/ t 0 ρ κ ε − .

(e) B = 0.

7 !! The leaky capacitor of Problem 6 is charged such that the voltage across the capacitor is given by V (t ) =

(0.01 V/s)t . (a) Find the conduction current as a function of time. (b) Find the displacement current. (c) Find

the time for which the displacement current is equal to the conduction current.

(a) I = V / R; R = d ρ / A

(b) I d = C dV /dt

(c) Set I = I d

I = (0.01 A/d ρ )t A

I d = 0.01κε 0 A/d

t = ε 0κρ

8 !! The space between the plates of a capacitor is filled with a material of resistivity ρ = 104 Ω-m and

dielectric constant κ = 2.5. The parallel plates are circular with a radius of 20 cm and are separated by 1 mm.

The voltage across the plates is given by V 0 cos ω t , with V 0 = 40 V and ω = 120π rad/s. (a) What is the

displacement current density? (b) What is the conduction current between the plates? (c) At what angular

frequency is the total current 45° out of phase with the applied voltage?(a) I d = C dV /dt ; C = κε 0 A/d ; V = (40 V)cos 102π t

(b) I = AV /d ρ

(c) δ = 45° when ωκε 0 A/d = A/d ρ ; ω = 1/κε 0ρ

I d = –(41.9 µ A) sin 120π t ; J d = –(334 µ A/m2) sin 120π t

I = (0.503 A) cos 120π t

f = ω /2π = 719 kHz

9* !!! In this problem, you are to show that the generalized form of Ampère’s law (Equation 32-4) and the Biot–

Savart law give the same result in a situation in which they both can be used. Figure 32-12 shows two charges

+Q and – Q on the x axis at x = – a and x = +a, with a current I = – dQ/dt along the line between them. Point P is

on the y axis at y = R. (a) Use the Biot–Savart law to show that the magnitude of B at point P is

a+ R

1

R

Ia = B

22π

µ

2

0

(b) Consider a circular strip of radius r and width dr in the yz plane with its center at the origin. Show that the

flux of the electric field through this strip is

dr r ar aQ

=dA E x )+(2/3-22

0ε

(c) Use your result for part (b) to find the total flux φ e through a circular area of radius R. Show that

R + -1Q=

22e0

a

aφ ε

(d ) Find the displacement current I d , and show that


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R + I=

22d

a

a I + I

(e) Then show that Equation 32–4 gives the same result for B as that found in part (a).

(a) Use Equ. 29-11 to find B at point P. Note that sin θ 1 = sin θ 2 =a + R

R

22, so

a + R

1

R2

I = B

22

0

π

µ .

(b) E x = [2kQ/( R2 + a

2)] cos θ 1 =

)a + r (

aQk 23/222

. dA = 2π r dr , so E x dA = dr r )a + r (

aQ3/222

0ε .

(c) ε 0φ e =

−

−∫ a + R

a 1Q =

a

1 +

a + R

1 aQ = dr r E 2

2222 x

R

0

π .

(d ) I d = ε 0(d φ e/dt ). Only Q depends on t , and dQ/dt = – I . So I d =

−−

a + R

a 1 I

22 and I + I d =

a + R

a I

22.

(e) !d ⋅∫ B = 2π RB = µ 0( I + I d ); so a + R1

R2 I = B

22

0

π µ ; Q.E.D.

10 !! Theorists have speculated on the possible existence of magnetic monopoles, and there have been several, as

yet unsuccessful, experimental searches for such monopoles. Suppose magnetic monopoles were found and that

the magnetic field at a distance r from a monopole of strength qm is given by .r /4/2

m0 q)(= B π µ How would

Maxwell's equations have to be modified to be consistent with such a discovery?

Two changes would be required. Equ. 32-6b should read qdA B0

S mn = µ ∫ and Equ. 32-6c should read

∫ ⋅

C d ! E = – d /dt ∫ S Bn dA – I m /ε 0, where I m is the current associated with the motion of the magnetic poles.

11 !! Show that the normal component of the magnetic field B is continuous across a surface. Do this by

applying Gauss's law for B (∫ Bn dA = 0) to a pillbox Gaussian surface that has a face on each side of thesurface.

The figure shows the end view of a pillbox surrounding a small area dA

of the surface. The normal components of the magnetic field, Bn, are

shown with different magnitudes. When performing the surface integral

the normal to the surface is outward, as shown in the figure. It the

follows from Equ. 32-6b that Bn must be continuous across the surface.

12 ! Which waves have greater frequencies, light waves or X rays?

X rays.

13* ! Are the frequencies of ultraviolet radiation greater or less than those of infrared radiation?

f uv > f ir

14 ! What kind of waves have wavelengths of the order of a few meters?

FM radio waves. (See Problem 15.)


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15 ! Find the wavelength for (a) a typical AM radio wave with a frequency of 1000 kHz and (b) a typical FM

radio wave of 100 MHz.

(a), (b) λ = c/ f (a) λ = 300 m (b) 3 m

16 ! What is the frequency of a 3-cm microwave?

f = c/λ f = 10 GHz

17* ! What is the frequency of an X ray with a wavelength of 0.1 nm?

f = c/λ f = 3×108/10 –10 = 3×1018 Hz

18 ! The detection of radio waves can be accomplished with either a dipole antenna or a loop antenna. The

dipole antenna detects the (pick one) [electric] [magnetic] field of the wave, and the loop antenna detects the

[electric] [magnetic] field of the wave.

The dipole antenna detects the electric field, the loop antenna detects the magnetic field of the wave.

19 ! A transmitter uses a loop antenna with the loop in the horizontal plane. What should be the orientation of a

dipole antenna at the receiver for optimum signal reception?

The dipole antenna should be in the horizontal plane and normal to the line from the transmitter to the receiver.

20 !! The intensity of radiation from an electric dipole is proportional to (sin2 θ )/r

2, where θ is the angle between

the electric dipole moment and the position vector r. A radiating electric dipole lies along the z axis (its dipole

moment is in the z direction). Let I 1 be the intensity of the radiation at a distance r = 10 m and at angle θ = 90°.Find the intensity (in terms of I 1) at (a) r = 30 m, θ = 90°; (b) r = 10 m, θ = 45°; and (c) r = 20 m, θ = 30°.(a), (b), (c) I (θ ) = I 1 (100/r

2) sin

2θ (a) I = I 1/9 (b) I = I 1/2 (c) I = I 1/16

21* !! (a) For the situation described in Problem 20, at what angle is the intensity at r = 5 m equal to I 1? (b) At

what distance is the intensity equal to I 1 at θ = 45°?(a) 1/r 12 = (sin2 θ )/r 2

(b) 1/r 12 = (sin

2 45°)/r 2

sin2 θ = 1/4; θ = 30°

r 2 = 100/2 = 50 m

2; r = 7.07 m

22 !! The transmitting antenna of a station is a dipole located atop a mountain 2000 m above sea level. The

intensity of the signal on a nearby mountain 4 km distant and also 2000 m above sea level is 4×10 –12 W/m2.What is the intensity of the signal at sea level and 1.5 km from the transmitter? (See Problem 20.)

Use I = I 1 (r 12 /r

2) sin

2θ r

2 = 6.25 m

2; sin θ = 0.6; I = 3.69 pW/m

2

23 !!! A radio station that uses a vertical dipole antenna broadcasts at a frequency of 1.20 MHz with total power

output of 500 kW. The radiation pattern is as shown in Figure 32-8, i.e., the intensity of the signal varies assin

2 θ , where θ is the angle between the direction of propagation and the vertical, and is independent of

azimuthal angle. Calculate the intensity of the signal at a horizontal distance of 120 km from the station. What

is the intensity at that point as measured in photons per square centimeter per second?

1. From Problem 20, I (r,θ ) = I 0 sin2

θ /r 2; find I 0

2. Evaluate I (120 km, 90°)

∫ ∫ ===π π

π φ θ θ θ

0

2

0

002

tot kW7.59;3

8sin)I(r, I I d d r P

I = 4.14 µ W/m2


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3. E /photon = hf ; N /cm2.s = I /10

4hf N /cm

2.s = 5.21×1017

24 !!! At a distance of 30 km from a radio station broadcasting at a frequency of 0.8 MHz, the intensity of the

electromagnetic wave is 2×10 –13 W/m2. The transmitting antenna is a vertical dipole. What is the total power radiated by the station?

Use the result derived in Problem 32-23 Ptot = (8π /3)r 2

I (r ) = 1.51 mW

25* !!! A small private plane approaching an airport is flying at an altitude of 2500 m above ground. The airport's

flight control system transmits 100 W at 24 MHz, using a vertical dipole antenna. What is the intensity of the

signal at the plane's receiving antenna when the plane's position on a map is 4 km from the airport?

From Problem 23, I = (3P/8π )(sin2 θ )/r

2; θ = tan

–1(2.5/4.0); I = 0.151 µ W/m

2

26 ! An electromagnetic wave has an intensity of 100 W/m2. Find (a) the radiation pressure Pr , (b) E rms, and

(c) Brms.

(a) Use Equ. 32-14

(b) From Equs. 32-8 and 32-9 I = ε 0cE rms2

(c) Use Equ. 32-7

Pr = 0.333 µ Pa

E rms = 194 V/m

Brms = 0.647 µ T

27 ! The amplitude of an electromagnetic wave is E 0 = 400 V/m. Find (a) E rms, (b) Brms, (c) the intensity I , and

(d ) the radiation pressure Pr .

(a) E rms = E 0/21/2

(b) Brms = E rms/c

(c) Use Equ. 32-9

(d ) Use Equ. 32-14

E rms = 283 V/m

Brms = 0.943 µ T

I = 212 W/m2

Pr = 0.708 µ Pa

28 ! The rms value of the electric field in an electromagnetic wave is E rms = 400 V/m. (a) Find Brms, (b) the

average energy density, and (c) the intensity.

(a) Brms = E rms/c

(b) From Equ. 32-9 uav = E rms Brms/µ 0c

(c) Use Equ. 32-9

Brms = 1.33 µ T

uav = 1.41 µ J/m3

I = 424 W/m2

29* ! Show that the units of E = cB are consistent; that is, show that when B is in teslas and c is in meters per

second, the units of cB are volts per meter or newtons per coulomb.

[c][ B] = [m/s][N/A.m] = [N/C] = [V/m].

30 ! The root-mean-square value of the magnitude of the magnetic field in an electromagnetic wave is Brms =0.245 µ T. Find (a) E rms, (b) the average energy density, and (c) the intensity.

(a), (b), (c) Proceed as in Problem 32-28 (a) E rms = 73.5 V/m (b) uav = 47.8 nJ/m3 (c) I = 14.3 W/m

2

31 !! (a) An electromagnetic wave of intensity 200 W/m2 is incident normally on a rectangular black card with

sides of 20 cm and 30 cm that absorbs all the radiation. Find the force exerted on the card by the radiation. (b)

Find the force exerted by the same wave if the card reflects all the radiation incident on it.

(a) Use Equ. 32-14 and F = Pr A F = 40 nN


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(b) With reflection the force is doubled F = 80 nN

32 !! Find the force exerted by the electromagnetic wave on the reflecting card in part (b) of Problem 31 if the

radiation is incident at an angle of 30° to the normal. Note that only normal the component of the radiation pressure exerts a force on the card. Multiply the result of

Problem 32-31 by cos 30°. F = 69.3 nN.33* !! An AM radio station radiates an isotropic sinusoidal wave with an average power of 50 kW. What are the

amplitudes of E max and Bmax at a distance of (a) 500 m, (b) 5 km, and (c) 50 km?

(a) I = Pav/4π r 2 = cε 0 E rms

2 = cε 0 E max

2/2; Bmax = E max/c

(b) 5 km = 10×500 m; E max ∝ 1/r (c) 50 km = 100×500 m

E max = 3.46 V/m; Bmax = 11.5 nT

E max= 0.346 V/m; Bmax = 1.15 nT

E max = 0.0346 V/m; Bmax = 0.115 nT

34 !! The intensity of sunlight striking the earth's upper atmosphere (called the solar constant) is 1.35 kW/m2. (a)

Find E rms and Brms due to the sun at the upper atmosphere of the earth. (b) Find the average power output of the

sun. (c) Find the intensity and the radiation pressure at the surface of the sun.

(a) From Equs. 32-8 and 32-9 I = ε 0cE rms2

(b) P = 4π R2 I ; R = 1.5×1011 m

(c) I = P/4π r 2; r = 6.96×108 m; Pr = I /c

E rms = 713 V/m; Brms = E rms /c = 2.38 µ T

P = 3.82×1026 W I = 6.28×107 W/m2; Pr = 0.209 Pa

35 !! A demonstration laser has an average output power of 0.9 mW and a beam diameter of 1.2 mm. What is the

force exerted by the laser beam on (a) a 100% absorbing black surface? (b) a 100% reflecting surface?

(a) F = Pr A = (Pav /cA) A = Pav /c

(b) With reflection, F is doubled

F = 3 pN

F = 6 pN

36 !! A laser beam has a diameter of 1.0 mm and average power of 1.5 mW. Find ( a) the intensity of the beam,

(b) E rms, (c) Brms, and (d ) the radiation pressure.

(a) I = P/ A; A = π d 2/4

(b) From Equs. 32-8 and 32-9 I = ε 0cE rms2

(c) Brms = E rms /c

(d ) Pr = I /c

I = 1.91 kW/m2

E rms = 849 V/m

Brms = 2.83 µ T

Pr = 6.37 µ Pa

37* !! Instead of sending power by a 750-kV, 1000-A transmission line, one desires to beam this energy via an

electromagnetic wave. The beam has a uniform intensity within a cross-sectional area of 50 m2. What are the

rms values of the electric and the magnetic fields?

1. Determine the intensity; I = P/ A2. I = E rms

2/cµ 0; solve for E rms; Brms = E rms/c

I = 7.5×108

/50 W/m2

= 1.5×107

W/m2

E rms = 75.2 kV/m; Brms = 0.251 mT

38 !! A laser pulse has an energy of 20 J and a beam radius of 2 mm. The pulse duration is 10 ns and the energy

density is constant within the pulse. (a) What is the spatial length of the pulse? (b) What is the energy density

within the pulse? (c) Find the electric and magnetic amplitudes of the laser pulse.

(a) L = c∆t (b) u = U /V ; V = π r

2 L

L = 3.0 m

u = 531 kJ/m3


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(c) u = ε 0 E rms2 = 1/2ε 0 E 0

2; E 0 = (2u/ε 0)

1/2; B0 = E 0/c E 0 = 346 MV/m; B0 = 1.15 T

39 !! The electric field of an electromagnetic wave oscillates in the y direction and the Poynting vector is given

by

i])10(3 -x[10cos )mW /(100 = ),S(922 t t x ×

where x is in meters and t is in seconds. (a) What is the direction of propagation of the wave? (b) Find the

wavelength and the frequency. (c) Find the electric and magnetic fields.

(a), (b) See Section 15-2

(c) From Equs. 32-8 and 32-9, S = ε 0 E 2/c

(a) The wave propagates in the positive x direction.

(b) λ = 2π /k = 0.628 m; f = ω /2π = 477 MHz

E = (194 V/m) cos[10 x – (3×109)t ] j; B = ( E /c) k = (0.647 µ T) cos[10 x – (3×109)t ] k

40 !! A pulsed laser fires a 1000-MW pulse of 200-ns duration at a small object of mass 10 mg suspended by a

fine fiber 4 cm long. If the radiation is completely absorbed without other effects, what is the maximum angle

of deflection of this pendulum?

1. Find pi, the initial momentum of the pendulum

2. Use energy conservation; pi2/2m = mgL(1 – cos θ )

pi = U /c = (200/3×108 ) kg.m/s = 6.67×10 –7 kg.m/scos θ = 1 – ( pi

2/2m

2gL) = 0.994; θ = 6.1°

41* !! A 10- by 15-cm card has a mass of 2 g and is perfectly reflecting. The card hangs in a vertical plane and is

free to rotate about a horizontal axis through the top edge. The card is illuminated uniformly by an intense light

that causes the card to make an angle of 1° with the vertical. Find the intensity of the light.The physical arrangement is shown in the figure. Note that the force exerted by the

radiation acts along the dashed line. Let the force acting on an area dA = w dx be dF L.

The torque on an area dA = w dx about the pivot is d τ R = dF L x. Next, note that

dF L = 2( I /c)(cos θ ) dA, the factor 2 arising from the mirror reflection. The net torquedue to the radiation about the pivot is obtained by integrating d τ over the length of the

card. Thus τ R = ( IA ! /c)cos θ . The restoring torque due to the gravitational force mg is

(mg ! /2)sin θ . Equating these torques gives I = (mgc/2 A)tan θ . Substituting

appropriate numerical values one finds I = 3.42 MW/m2

42 !! A valuable 0.08-kg gem and a 105-kg spaceperson are separated by 95 m. Both objects are initially at rest.

The spaceperson has a 1.5-kW laser that can be used as a photon rocket motor to propel the person toward the

diamond. How long would it take the spaceperson to move 95 m using the laser rocket propulsion?

1. Find the acceleration a; ma = dp/dt = (dU /dt )/c2. x = 1/2at

2a = (1.5×10

3

/105×3×108

) m/s2

= 4.76×10 –8

m/s2

t = 6.32×104 s = 17.5 h

43 !! It has been suggested that spacecraft could be propelled by the radiation pressure from the sun. What must be

the surface mass density (kg/m2) of a perfectly reflecting sheet so that at a distance of one astronomical unit the

force due to radiation pressure is twice that due to the gravitational attraction between the reflecting sheet and the

sun? ( Note: One astronomical unit is the average radius of the earth's orbit.) How will the ratio of radiation force

to gravitational force change as the reflecting sheet accelerates away from the sun?


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Because the sheet is perfectly reflecting, the force on it due to the radiation is 2Pr A = 2 IA/c. Set the area of the

surface times twice the radiation pressure equal to twice the force of gravity. Note that at 1 AU, the intensity of the

sun’s radiation is 1.35 kW/m2 (see Problem 32-34). The force of gravity exerted by the sun is F g = GM Sm/ R

2,

where m = σ A and R = 1 AU. Here σ is the surface mass density. Solve for σ = m/ A . One obtains

σ = 1.35×103×(1.5×1011)2/(3×108 × 1.99×1030 ×6.67×10 –11) kg/m2 = 7.63×10 –4 kg/m2. Since both I and F g are

inversely proportional to r 2

, the ratio of the two forces is independent of the distance from the sun.44 !! Suppose a mass of 50 kg is attached to a perfectly reflecting sheet whose surface mass density is that

obtained in Problem 43. What must be the surface area of the sheet so that at a distance of one astronomical

unit the acceleration of the system away from the sun is 0.4 mm/s2? How does the acceleration vary with

distance from the sun?

1. Write the condition in terms of the known

parameters and the area A. R = 1.5×1011 m.2. Solve for A; I = 1.35 kW/m

2, σ = 7.63×10 –4 kg/m2

a+ A R

+ AGM

c

A I F )kg50(=

)kg50(2 =

2

Snet σ

σ −

4.5×10 –6 A – 0.295 = 3.052×10 –7 A + 0.02; A = 7.5×104 m2

3. Since F net is proportional to 1/r 2, a is also proportional to 1/r 2.

A blackbody is an object that is a perfect absorber; that is, it absorbs all radiation incident on it. It is also a perfect

radiator. The power radiated by a blackbody of area A at temperature T is given by the Stefan-Boltzmann law

(Equation 21-17 with e = 1),

Pr = σ AT 4

where σ = 5.6703 × 10 –8

W/m2 . K

4.

45* !! A very long wire of radius 4 mm is heated to 1000 K. The surface of the wire is an ideal blackbody

radiator.

(a) What is the total power radiated per unit length? Find (b) the magnitude of the Poynting vector S , (c) E rms,

and (d ) Brms at a distance of 25 cm from the wire.

(a) Use Equ. 21-20; assume T 0 = 293 K

(b) Use Equ. 32-9; I = Pnet /2π rL = S

(c) From Equs. 32-9 and 32-7, S = E rms2/µ 0c

(d ) Use Equ. 32-7

Pnet / L = 1415 W/m

S = 901 W/m2

E rms = 583 V/m

Brms = 1.94 µ T

46 !! A blackbody sphere of radius R is a distance 2×1011 m from the sun. The effective area of the body for absorption of energy from the sun is π R

2, but the area for radiation by the object is 4π R

2. The power output of

the sun is 3.83×1026 W. What is the temperature of the sphere?1. Find the power absorbed from the sun

2. Write the power radiated by the object3. Set Pabs = Prad and solve for T

Pabs = π R2×3.83×1026 /4π ×4×1022 W

Prad = 4π R

2

σ T

4

T = [3.83×1026/(16πσ ×4×1022)]1/4 K = 241 K

47 !! (a) If the earth were an ideal blackbody with infinite thermal conductivity and no atmosphere, what would

be the temperature of the earth? (b) If 40% of the incident sun's energy were reflected, what then would be the

temperature of the earth? (See Problem 46.)

(a) Proceed as in Problem 32-46; D = 1.5×1011 m(b) Repeat with Pabs reduced by 60%

T = 278 K

T = (278×0.61/4 ) K = 245 K


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48 ! Show by direct substitution that Equation 32-17a is satisfied by the wave function

ct)k(x E = t)(kx E = E y −− sinsin 00 ω

where c = ω /k .

Write the second partial derivatives of E y = E 0 sin (kx – ω t ) with respect to x and t . Note that c = ω /k .

t) (kx E =t

E t); (kx E k =

x

E 22

y2

2

2

y2

ω ω ω −−∂

∂−−∂

∂sinsin 00 . So

t

E

c

1 =

t

E

k =

x

E 2

y2

22

y2

2

2

2

y2

∂∂

∂∂

∂∂

ω .

49* ! Use the known values of µ 0 and ε 0 in SI units to compute µ ε 00 / 1=c 0 and show that it is approximately

3×108 m/s.Evaluate (ε 0 µ 0)

– 1/2 (8.85×10 –12×4π ×10 –7 ) – 1/2 = 3.00×108 m/s

50 !!! (a) Using arguments similar to those given in the text, show that for a plane wave, in which E and B are

independent of y and z,

t

B =

x

E y z

∂∂

∂∂

andt

E z00 ∂∂

∂∂

ε µ = x

B y

(b) Show that E z and B y also satisfy the wave equation.

(a) In Figure 32-11 replace B z by E z. For ∆ x small E z( x2) = E z( x1) + (∂ E z/∂ x)∆ x. Now take the line integralaround the rectangular area ∆ x∆ z. This gives –(∂ E x/∂ x)∆ x∆ z. The magnetic flux through this same area is

B y∆ x∆ z, so from Faraday’s law we have ∂ E z/∂ x = ∂ B y/∂t. Now, in Figure 32-10, replace E y by B y and proceed asin the text (see p. 1013). One obtains the equation ∂ B y/∂ x = µ 0ε 0(∂ E z/∂t ).(b) Use the result obtained in part (a) to write the second partial derivatives of E x with respect to x and t .

t

E

c

1 =

t

E

t =

t x

B =

t

B

x =

x

E =

x

E

x 2 z

2

2

z

0

y2

y

2

z2

z

∂∂

∂∂

∂∂

∂∂∂

∂

∂∂∂

∂∂

∂∂

∂∂

ε µ 0 . Taking the second partial derivatives of

B y with respect to x and t , it likewise follows that B y satisfies the wave equation.

51 ! True or false:

(a) Maxwell's equations apply only to fields that are constant over time.

(b) The wave equation can be derived from Maxwell's equations.

(c) Electromagnetic waves are transverse waves.

(d ) In an electromagnetic wave in free space, the electric and magnetic fields are in phase.

(e) In an electromagnetic wave in free space, the electric and magnetic field vectors E and B are equal in

magnitude.

( f ) In an electromagnetic wave in free space, the electric and magnetic energy densities are equal.

(a) False (b) True (c) True (d ) True (e) False ( f ) True

52 ! (a) Show that if E is in volts per meter and B is in teslas, the units of the Poynting vector S = E× B/µ 0 are


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watts per square meter. (b) Show that if the intensity I is in watts per square meter, the units of radiation

pressure Pr = I /c are newtons per square meter.

(a) Note that B has the dimensions of µ 0[I]/[L], where [I] represents current (see, for example, Equ. 29-12).

The dimension of the electric field is [V]/[L], where [V] denotes potential difference, i.e., volts. Thus

[E][B]/[µ 0] = [V][I]/[L]2 = [P]/[L]

2, where P denotes power. Hence, the SI units of S are watts per square

meter.(b) Pr = I /c. The dimensions of I are [F][L]/[T][L]

2 = [F]/[T][L]. Then I /c has the dimensions

([F]/[T][L])/([L]/[T])= [F]/[L]2. If I is in watts per square meter and c in meters per second, the Pr is in newtons

per square meter.

53* !! A loop antenna that may be rotated about a vertical axis is used to locate an unlicensed amateur radio

transmitter. If the output of the receiver is proportional to the intensity of the received signal, how does the

output of the receiver vary with the orientation of the loop antenna?

The current induced in a loop antenna is proportional to the time-varying magnetic field. For maximum signal,

the antenna’s plane should make an angle θ = 0° with the line from the antenna to the transmitter. For any other angle, the induced current is proportional to cos θ . The intensity of the signal is therefore proportional to cos θ .

54 !! An electromagnetic wave has a frequency of 100 MHz and is traveling in a vacuum. The magnetic field is

given by B( z, t ) = (10 –8

T) cos (kz – ω t )i. (a) Find the wavelength, and the direction of propagation of this

wave. (b) Find the electric vector E( z, t ). (c) Give Poynting's vector, and find the intensity of this wave.

(a) λ = f /c; k = 2π /λ ; ω = 2π f

(b) Use Equs. 32-10 and 32-7

(c) Use Equs. 32-10 and 32-9

λ = 3 m; the wave propagates in the z direction

E( z, t ) = [(3 V/m) cos (2.09 z – 6.28×108 t )] – jS = (23.9 mW/m

2) cos

2(2.09 z – 6.28×108 t ) k;

I = 11.9 mW/m2

55 !! A circular loop of wire can be used to detect electromagnetic waves. Suppose a 100-MHz FM station

radiates 50 kW uniformly in all directions. What is the maximum rms voltage induced in a loop of radius 30 cm

at a distance of 105 m from the station?

1. I = P/4π R2 = B0

2c/2µ 0; evaluate B0

2. Maximum induced E when the plane of the loop is

perpendicular to B. E max = Aω B0; E rms = 0.707E max

B0 = 5.774×10 –11 TE rms = 0.707(π ×0.09)(6.28×108)(5.774×10 –11) V= 7.25 mV

56 !! The electric field from a radio station some distance from the transmitter is given by E = (10 –4

N/C) cos

106t, where t is in seconds. (a) What voltage is picked up on a 50-cm wire oriented along the electric field

direction? (b) What voltage can be induced in a loop of radius 20 cm?

(a) V = EL since E does not depend on x

(b) B0 = E 0/c; E

= ω B0 A = ω E 0π R

2

/c

V = (50 µ V) cos 106t

E

= 41.9 nV

57* !! A circular capacitor of radius a has a thin wire of resistance R connecting the centers of the two plates. A

voltage V 0 sin ω t is applied between the plates. (a) What is the current drawn by this capacitor? (b) What is the

magnetic field as a function of radial distance r from the centerline within the plates of this capacitor? (c) What

is the phase angle between current and applied voltage?

(a) I = I c + I d ; I c = V / R; For I d , use Equ. 32-4;

I d = ε 0 A(dE /dt ) = (ε 0 A/d )(dV /dt ); I c = V / R

I c = (V 0/ R)sin ω t ; I c = (V 0ε 0πω a2/d )cos ω t

I = V 0[(1/ R)sin ω t + (ε 0πω a2/d )cos ω t ]


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(b) Use Equ. 32-4; here I d ′ = I d (r 2/a2)(c) δ = tan

–1( I d / I c)

B(r ) = (µ 0V 0/2π r )[(1/ R)sin ω t + (ε 0π r 2ω /d )cos ω t ]

δ = tan –1

(π a2ε 0ω R/d )

58 !! A 20-kW beam of radiation is incident normally on a surface that reflects half of the radiation. What is the

force on this surface?

Note that for the 10 kW that is reflected, the force on the surface is (2×10 kW)/c; for the 10 kW that isabsorbed, the resulting force on the surface is (10 kW)/c. Thus the total force is F = (30×103/3×108 ) N = 0.1mN.

59 !! Show that the relation between the momentum carried by an electromagnetic wave and the energy,

Equation 32-13, can also be derived using the Einstein–Planck relation, E = hf ; the de Broglie equation, p =

h/ λ ; and c = f λ .

h = E / f ; p = h/λ = E / f λ = E /c

60 !! The electric fields of two harmonic waves of angular frequency ω 1 and ω 2 are given by E1 = E 1,0 cos (k 1 x –

ω 1t ) j and E2 = E 2,0 cos (k 2 x – ω 2t + δ ) j. Find (a) the instantaneous Poynting vector for the resultant wave motion

and (b) the time-average Poynting vector. If E2 = E 2,0 cos (k 2 x + ω 2t + δ ) j, find (c) the instantaneous Poynting

vector for the resultant wave motion and (d ) the time-average Poynting vector.

(a) Since E1 and E2 propagate in the x direction, E× ×× × B = S i. Therefore B = B k and

B = (1/c)[ E 1,0 cos (k 1 x – ω 1t ) + E 2,0 cos (k 2 x – ω 2t + δ )] k. The Poynting vector is

S = (1/µ 0c)[ E 1,02cos

2(k 1 – ω 1t ) + 2 E 1,0 E 2,0 cos (k 1 x – ω 1t ) cos (k 2 x – ω 2t + δ ) + E 2,0

2cos

2(k 2 x – ω 2t + δ )] i.

(b) The time average of the cross product term is zero for ω 1 ≠ ω 2, and the time average of cos2 (ω t ) = 1/2.So Sav = (1/2µ 0c)( E 1,0

2 + E 2,0

2) i.

(c) In this case B2 = – B k since the wave with k = k 2 propagates in the – i direction. The magnetic field is then

B = (1/c)[ E 1,0 cos (k 1 x – ω 1t ) – E 2,0 cos (k 2 x + ω 2t + δ )] k. The Poynting vector is now

S = (1/µ 0c)[ E 1,02cos

2(k 1 – ω 1t ) – E 2,0

2cos

2(k 2 x – ω 2t + δ )] i.

(d ) Sav = (1/2µ 0c)( E 1,02 – E 2,0

2) i.

61* !! At the surface of the earth, there is an approximate average solar flux of 0.75 kW/m2. A family wishes to

construct a solar energy conversion system to power their home. If the conversion system is 30% efficient and

the family needs a maximum of 25 kW, what effective surface area is needed for perfectly absorbing

collectors?

Write the expression for P; P = IAε A = P/ I ε = 111 m2

62 !! Suppose one has an excellent radio capable of detecting a signal as weak as 10 –14

W/m2. This radio has a

2000-turn coil antenna having a radius of 1 cm wound on an iron core that increases the magnetic field by a

factor of 200. The radio frequency is 140 KHz. (a) What is the amplitude of the magnetic field in this wave?

(b) What is the emf induced in the antenna? (c) What would be the emf induced in a 2-m wire oriented in thedirection of the electric field?

(a) Use Equ. 32-9; S av = I = (c/2µ 0) B02; Find B0

(b) Use Faraday’s law; E = K m NAω B0 cos (ω t )

(c) E = EL = cB0 L sin (ω t )

B0 = 9.153×10 –15 TE = (1.01 µ V) cos (8.80×105 t )E = (5.49 µ V) sin (8.80×105 t )

63 !! A 654-nm laser whose beam diameter is 0.4 mm points upward. A small, perfectly reflecting spherical

particle having a diameter of 10 µ m and a density of 0.2 g/cm3 is supported against gravity by the radiation


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pressure from the laser beam. Determine the power output of this laser.

1. Find upward force due to the laser beam

2. Set F up = mg = (4π /3)gR3ρ and solve for I

3. Plas = IA beam

F up = 2 IA/c = (2π ×25×10 –12/3×108 ) I = 5.236×10 –19 I I = 1.962 MW/m

2

Plas = 0.247 W

64 !!! A long, cylindrical conductor of length L, radius a, and resitivity ρ carries a steady current I that isuniformly distributed over its cross-sectional area. (a) Use Ohm's law to relate the electric field E in the

conductor to I , ρ , and a. (b) Find the magnetic field B just outside the conductor. (c) Use the results for parts

(a) and (b) to compute the Poynting vector S = E×××× B/µ 0 at r = a (the edge of the conductor). In what direction isS? (d ) Find the flux ∫ nS dA through the surface of the conductor into the conductor, and show that the rate of energy flow into the conductor equals I

2 R, where R is the resistance. (Here S n is the inward component of S

perpendicular to the surface of the conductor.)

(a) V = IR = I ρ L/ A = I ρ L/π a2 = EL. So E = I ρ /π a

2.

(b) B = µ 0 I /2π a (see Equ. 29-12).

(c) S

= E

× ×× × B

/µ 0 = –( I

2

ρ /2π

2a

3

) r

; using the right-hand rule to determine the direction of B

one finds that thedirection of S is radially inward.

(d ) ∫ nS dA = S ×2π aL = I 2ρ L/π a2 = VI = I 2 R. Here we have taken the positive direction for S n to be radiallyinward.

65* !!! A long solenoid of n turns per unit length has a current that slowly increases with time. The solenoid has

radius R, and the current in the windings has the form I (t ) = at . (a) Find the induced electric field at a distance r


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given by

cmr

P =a

2π 4

AS

where PS is the power output of the sun and is equal to 3.8×1026

W, A is the surface area of the sail, m is thetotal mass of the spacecraft, r is the distance from the sun, and c is the speed of light. (b) Show that the velocity

of the spacecraft at a distance r from the sun is found from

−

r

1

r

1

mc

P +v=v

0

20

2

π 2

AS

where v0 is the initial velocity at r 0. (c) Compare the relative accelerations due to the radiation pressure and the

gravitational force. Use reasonable values for A and m. Will such a system work?

(a) This is similar to Problem 32-44. In this case the mass of the sail, which we take to be perfectly reflecting,

is included in the mass of the object. We therefore neglect the product σ A in the solution of that problem and

replace the mass of 50 kg by m. The intensity of the sun’s radiation is I = PS /4π r 2 and the force due to radiation

pressure is 2 I /c. Thus the net force is F net = ( APS /2π c – GM Sm)/r 2, and the acceleration is then a = ( AP/2π c –

GM Sm)/mr 2. (Note: The answer given in the Problem assumes that the sails reflect no radiation and neglects the

gravitational force of the sun on the mass m.)

(b) Since a is a function of r , the velocity must be found by integration. Note that a = dv/dt = (dv/dr )(dr /dt ) =

v(dv/dr ). Thus v dv = a dr.

( )∫ ∫ ∫

−===−=

v

v

r

r

r

r r r

K r

dr K adr vvvdv

0 0 0

112/1

02

20

2 , where

K = ( AP/2π c – GM Sm)/m. Then v2

= v02

+[( AP/π c – 2GM Sm)/m](1/r 0 – 1/r )(c) Judging by the result obtained in Problem 32-44, this scheme is not likely to work effectively. For any

reasonable mass, the surface mass density of the sail would have to be extremely small and the sail would have

to be huge. Moreover, unless struts are built into the sail, it will collapse in the attempt to accelerate the mass.

68 !!! Novelty stores sell a device called a radiometer (Figure 32-13), in which a balanced vane spins rapidly. A

card is mounted on each arm of the vane. One side of each card is white and the other is black. Assume that the

mass of each card is 2 g, that the light-collecting area for each card is 1 cm2, and that each arm of the vane has a

length of 2 cm. (a) If a 100-W light bulb produces 50 W of electromagnetic energy and the bulb is 50 cm from

the radiometer, find the maximum angular acceleration of the vane. (Estimate the moment of inertia of the vane

by assuming all the mass of each card is at the end of the arms.) (b) How long will it take for the vane to

accelerate to 10 rev/min if it starts from rest and is subject to the maximum angular acceleration at all times?

(c) Can the radiation pressure account for the rapid motion of the radiometer?

(a) 1. Find Pr = P/4π r 2c

2. Find F net, τ , and I (assume 4 vanes)

3. Find α = τ / I

(b) ω = α t ; ω = (20π /60) rad/s ≈1 rad/s(c) No

Pr = 53.1 nN/m2

F net ≈ 5.31 pN, τ ≈1×10 –13 N.m, I ≈ 3×10 –6 kg.m2

α ≈ 3×10 –8 rad/s2

t ≈ 3×107 s ≈ 1 year


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69* !!! When an electromagnetic wave is reflected at normal incidence on a perfectly conducting surface, the

electric field vector of the reflected wave at the reflecting surface is the negative of that of the incident wave.

(a) Explain why this should be. (b) Show that the superposition of incident and reflected waves results in a

standing wave. (c) What is the relationship between the magnetic field vector of the incident and reflected

waves at the reflecting surface?

(a) At a perfectly conducting surface, E = 0. Therefore, the sum of the electric field of the incident andreflected wave must add to zero, and so Ei = – Er .

(b) Let E i = E 0 y cos(ω t – kx). Then E r = – E 0 y cos(ω t + kx). Using cos(α + β ) = cos α cos β – sin α sin β ,

E i + E r = 2 E 0 y sin(ω t ) sin(kx), which is the expression for a standing wave.

(c) Using E× ×× × B = µ 0S and S the direction of propagation of the wave, we see that for the incident wave

Bi = B z cos(ω t – kx). Since both S and E y are reversed for the reflected wave, Br = B z cos(ω t + kx). So the

magnetic field vectors are in the same direction at the reflecting surface and add at that surface; i.e., B = 2 Bi.

70 !!! An intense point source of light radiates 1 MW isotropically. The source is located 1.0 m above an infinite

perfectly reflecting plane. Determine the force that acts on the plane.

Let the point source be a distance a above the plane. Now consider a ring of radius r and thickness dr in the

plane and centered at the point directly below the light source. The intensity anywhere along this infinitesimal

ring is P/4π (r 2 + a

2), and the element of force dF on this ring of area 2π r dr is then given by

a + r

a

)a + r c(

dr r P =dF

2222, where we have taken into account that only the normal component of the incident

radiation contributes to the force on the plane, and that the plane is a perfectly reflecting one. Integrating dF

from

r = 0 to r = ∞ one obtains F = P/c = 3.33 mN.

em waves jee

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