HW14 Solutions (due Thurs, May 7) 1. T&M 30.P.003

Which of the following statements are true? (Select all that

apply.)

a) the electric and magnetic fields of an electromagnetic wave in

free space are in phase.

b) Electromagnetic waves are transverse waves.

c) None of these statements are true.

d) The electromagnetic wave equation can be derived from

Maxwell's equations.

e) Maxwell's equations apply only to electric and magnetic fields

that are constant over time.

f) In an electromagnetic wave in free space, the electric and

magnetic energy densities are equal. Solution:

a) True, b) True, c) False, d) True, e) False, f) True. 2. T&M 30.P.031

The amplitude of an electromagnetic wave is E0 = 400 V/m. Find

(a) Erms, (b) Brms, (c) the intensity I (d) the radiation pressure Pr Solution: Picture the Problem The rms values of the electric and magnetic fields are found from their amplitudes by dividing by the square root of two. The rms values of the electric and magnetic field strengths are related according to Brms = Erms/c. We can find the intensity of the radiation using I = ErmsBrms/µ0 and the radiation pressure using Pr = I/c. (a) Relate Erms to E0:

V/m283

V/m 8.2822

V/m400

2

0

rms

=

===E

E

(b) Find Brms from Erms:

nT 943T9434.0

m/s10998.2

V/m8.282

8

rms

rms

==

!==

µ

c

EB

(c) The intensity of an electromagnetic wave is given by:

0

rmsrms

µ

BEI =

Substitute numerical values and evaluate I:

( )( )

22

27

W/m212W/m3.212

N/A104

T9434.0V/m8.282

==

!=

"#

µI

(d) Express the radiation pressure in terms of the intensity of the wave:

c

IP =r

Substitute numerical values and evaluate Pr:

nPa 708m/s10998.2

W/m3.212

8

2

r=

!=P

3. T&M 30.P.033

(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. Solution: Picture the Problem We can find the force exerted on the card using the definition of pressure and the relationship between radiation pressure and the intensity of the electromagnetic wave. Note that, when the card reflects all the radiation incident on it, conservation of momentum requires that the force is doubled. (a) Using the definition of pressure, express the force exerted on the card

APFrr

=

by the radiation: Relate the radiation pressure to the intensity of the wave:

c

IP =r

Substitute for Pr to obtain: c

IAF =r

Substitute numerical values and evaluate Fr:

( )( )( )

nN40

m/s10998.2

m30.0m20.0W/m200

8

2

r

=

!=F

(b) If the card reflects all of the radiation incident on it, the force exerted on the card is doubled:

nN80r=F

4. T&M 30.P.037

An electromagnetic plane wave has an electric field that is

parallel to the y axis, and has a Poynting vector given by S(x,t)

= (100 W/m2) cos2(kx - ωt) (in the z direction). x is in meters, k

= 10.0 rad/m, ω = 3.00 x 109 rad/s, and t is in seconds.

(a) What is the direction of propagation of the wave?

(b) Find the wavelength and the frequency of the wave.

(c) Find the electric and magnetic fields of the wave as

functions of x and t. (Use x and t as necessary.)

Solution: Picture the Problem We can determine the direction of propagation of the wave, its wavelength, and its frequency by examining the argument of the cosine function. We can find E from cE

0

2 µ=S!

and B from B = E/c. Finally, we can

use the definition of the Poynting vector and the given expression for S!

to find E

!andB!

. (a) Because the argument of the cosine function is of the form tkx !" , the wave propagates in the +x direction.

(b) Examining the argument of the cosine function, we note that the wave number k of the wave is:

1m0.10

2 !==

"

#k ⇒ m628.0=!

Examining the argument of the cosine function, we note that the angular frequency ω of the wave is:

19s1000.32!

"== f#$

Solving for f yields: MHz477

2

s1000.319

=!

=

"

#f

(c) Express the magnitude of S

!in

terms of E:

c

E

0

2

µ=S!

⇒ S

!cE0

µ=

Substitute numerical values and evaluate E:

( )( )( ) V/m1.194W/m100m/s10998.2N/A1042827=!!=

"#E

Because ( ) ( ) [ ] iS ˆcosW/m100,

22tkxtx !"=

!

and BES!!!

!=0

1

µ:

( ) ( ) [ ] jE ˆcosV/m194, tkxtx !"=!

where k = 10.0 rad/m and ω = 3.00 × 109 rad/s.

Use B = E/c to evaluate B:

nT 4.647m/s10998.2

V/m1.194

8=

!=B

Because BES!!!

!=0

1

µ, the direction

of B!

must be such that the cross product of E

! with B

!is in the

positive x direction:

( ) ( ) [ ]kB ˆcosnT 647, tkxtx !"=!

where k = 10.0 rad/m and ω = 3.00 × 109 rad/s.

5. T&M 30.P.046 An electromagnetic wave has a frequency of 100 MHz and is traveling in a vacuum. The magnetic field is given by B(z, t) = (1.00 x 10-8 T) cos (kz - ωt) (in the x direction). Solution: Picture the Problem We can use c = fλ to find the wavelength. Examination of the argument of the cosine function will reveal the direction of propagation of the wave. We can find the magnitude, wave number, and angular frequency of the electric vector from the given information and the result of (a) and use these results to obtain E

!(z, t). Finally, we can use its definition to find the Poynting

vector. (a) Relate the wavelength of the wave to its frequency and the speed of light:

f

c=!

Substitute numerical values and evaluate λ: m00.3

MHz100

m/s10998.28

=!

="

From the sign of the argument of the cosine function and the spatial dependence on z, we can conclude that the wave propagates in the +z direction.

(b) Express the amplitude of E

!: ( )( )

V/m00.3

T10m/s10998.288

=

!=="

cBE

Find the angular frequency and wave number of the wave:

( ) 18s1028.6MHz10022!

"=== ##$ f

and 1

m09.2m00.3

22 !===

"

#

"k

Because S

!is in the positive z direction, E

!must be in the negative y direction in

order to satisfy the Poynting vector expression:

( ) ( ) ( ) ( )[ ]jtztzE ˆs1028.6m09.2cosV/m00.3,181 !!

"!!=!

(c) Use its definition to express and evaluate the Poynting vector:

( )( )( ) ( ) ( )[ ]( )ijtzBEtzS ˆˆs1028.6m09.2cos

N/A104

T10V/m00.31,

1812

27

8

0

!!"!

"=!= ""

"

"

#µ

!!!

or ( ) ( ) ( ) ( )[ ]ktztzS ˆs1028.6m09.2cosmW/m9.23,

18122 !!"!=

!

The intensity of the wave is the average magnitude of the Poynting vector. The average value of the square of the cosine function is 1/2:

( )2

2

2

1

mW/m9.11

mW/m9.23

=

== S!

I

6. T&M 30.P.047 A circular loop of wire can be used to detect electromagnetic waves. Suppose a 104 MHz FM station radiates 40 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? Solution: The intensity of the radiation is given by the power divided by

the area: I =P

4!d2. This is related to B0, the amplitude of the

magnetic field, as follows: B0=

2µ0I

c. The induced rms voltage is

!rms

="B

0#r

2

2, such that

!rms =(2" f )"r

2

2

Pµ0

2"d2c=fr2

d

P"3µ0

c=(104 #10

6)(.3

2)

105

(40 #103)"

3(4" #10

$7)

2.998 #108

= 6.75mV

7. T&M 31.P.051

What is the polarizing angle for light in air that is incident on the

following? (a) water (n = 1.33) (a) glass (n = 1.50)

Solution: Picture the Problem The polarizing angle is given by Brewster’s law:

12ptan nn=! where n1 and n2 are the indices of refraction on the near and far sides of the interface, respectively.

Use Brewster’s law to obtain: !!

"

#$$%

&=

'

1

21

p tann

n(

(a) For n1 = 1 and n2 = 1.33:

°=!"#

$%&

='

1.5300.1

33.1tan

1

p(

(b) For n1 = 1 and n2 = 1.50:

°=!"#

$%&

='

3.5600.1

50.1tan

1

p(

8. T&M 31.P.053

Two polarizing sheets have their transmission axes crossed so

that no light gets through. A third sheet is inserted between the

first two such that its transmission axis makes an angle θ with

that of the first sheet. Unpolarized light of intensity I0 is incident

on the first sheet. Find the intensity of the light transmitted

through all three sheets for the following values of θ. (a) θ = 45° and (b) θ = 30°

Solution: Picture the Problem Let In be the intensity after the nth polarizing sheet and use

!2

0cosII = to find the intensity of the light transmitted through all three sheets

for θ = 45° and θ = 30°. (a) The intensity of the light between 02

1

1II =

the first and second sheets is: The intensity of the light between the second and third sheets is:

0412

021

2,1

2

1245coscos IIII =°== !

The intensity of the light that has passed through the third sheet is:

0812

041

3,2

2

2345coscos IIII =°== !

(b) The intensity of the light between the first and second sheets is:

02

1

1II =

The intensity of the light between the second and third sheets is:

0832

021

2,1

2

1230coscos IIII =°== !

The intensity of the light that has passed through the third sheet is:

03232

083

3,2

2

2360coscos IIII =°== !