1. Home
  2. College physics
  3. Electromagnetic waves
  4. Energy in electromagnetic waves

24.4 Energy in electromagnetic waves  (Page 2/6)

<< Chapter < Page
  College physics     Page 2 / 6
Chapter >> Page >
I ave = cB 0 2 2μ 0 , size 12{I rSub { size 8{"ave"} } = { { ital "cB" rSub { size 8{0} } rSup { size 8{2} } } over {2μ rSub { size 8{0} } } } } {}

where B 0 size 12{B rSub { size 8{0} } } {} is the maximum magnetic field strength.

One more expression for I ave size 12{I rSub { size 8{"ave"} } } {} in terms of both electric and magnetic field strengths is useful. Substituting the fact that c ⋅ B 0 = E 0 size 12{c cdot B rSub { size 8{0} } =E rSub { size 8{0} } } {} , the previous expression becomes

I ave = E 0 B 0 2μ 0 . size 12{I rSub { size 8{"ave"} } = { {E rSub { size 8{0} } B rSub { size 8{0} } } over {2μ rSub { size 8{0} } } } } {}

Whichever of the three preceding equations is most convenient can be used, since they are really just different versions of the same principle: Energy in a wave is related to amplitude squared. Furthermore, since these equations are based on the assumption that the electromagnetic waves are sinusoidal, peak intensity is twice the average; that is, I 0 = 2 I ave size 12{I rSub { size 8{0} } =2I rSub { size 8{"ave"} } } {} .

Calculate microwave intensities and fields

On its highest power setting, a certain microwave oven projects 1.00 kW of microwaves onto a 30.0 by 40.0 cm area. (a) What is the intensity in W/m 2 size 12{"W/m" rSup { size 8{2} } } {} ? (b) Calculate the peak electric field strength E 0 size 12{E rSub { size 8{0} } } {} in these waves. (c) What is the peak magnetic field strength B 0 size 12{B rSub { size 8{0} } } {} ?

Strategy

In part (a), we can find intensity from its definition as power per unit area. Once the intensity is known, we can use the equations below to find the field strengths asked for in parts (b) and (c).

Solution for (a)

Entering the given power into the definition of intensity, and noting the area is 0.300 by 0.400 m, yields

I = P A = 1 . 00 kW 0 . 300 m × 0 . 400 m . size 12{I= { {P} over {A} } = { {1 "." "00"" kW"} over {0 "." "300 m"×0 "." "400 m"} } } {}

Here I = I ave size 12{I=I rSub { size 8{"ave"} } } {} , so that

I ave = 1000 W 0 . 120 m 2 = 8 . 33 × 10 3 W/m 2 . size 12{I rSub { size 8{"ave"} } = { {"1000"" W"} over {0 "." "120"" m" rSup { size 8{2} } } } =8 "." "33"×"10" rSup { size 8{3} } " W/m" rSup { size 8{2} } } {}

Note that the peak intensity is twice the average:

I 0 = 2 I ave = 1 . 67 × 10 4 W / m 2 . size 12{I rSub { size 8{0} } =2I rSub { size 8{"ave"} } =1 "." "67" times "10" rSup { size 8{4} } {W} slash {m rSup { size 8{2} } } } {}

Solution for (b)

To find E 0 size 12{E rSub { size 8{0} } } {} , we can rearrange the first equation given above for I ave size 12{I rSub { size 8{"ave"} } } {} to give

E 0 = 2 I ave cε 0 1/2 . size 12{E rSub { size 8{0} } = left ( { {2I rSub { size 8{"ave"} } } over {ce rSub { size 8{0} } } } right ) rSup { size 8{ {1}wideslash {2} } } } {}

Entering known values gives

E 0 = 2 ( 8 . 33 × 10 3 W/m 2 ) ( 3 . 00 × 10 8 m/s ) ( 8.85 × 10 – 12 C 2 / N ⋅ m 2 ) = 2.51 × 10 3 V/m . alignl { stack { size 12{E rSub { size 8{0} } = sqrt { { {2 \( 8 "." "33"´"10" rSup { size 8{3} } " W/m" rSup { size 8{2} } \) } over { \( 3 "." "00"´"10" rSup { size 8{8} } " m/s" \) \( 8 "." "85"´"10" rSup { size 8{ +- 2} } C rSup { size 8{2} } /N cdot m rSup { size 8{2} } \) } } } } {} #=2 "." "51"´"10" rSup { size 8{3} } " V/m" "." {} } } {}

Solution for (c)

Perhaps the easiest way to find magnetic field strength, now that the electric field strength is known, is to use the relationship given by

B 0 = E 0 c . size 12{B rSub { size 8{0} } = { {E rSub { size 8{0} } } over {c} } } {}

Entering known values gives

B 0 = 2.51 × 10 3 V/m 3.0 × 10 8 m/s = 8.35 × 10 − 6 T . alignl { stack { size 12{B rSub { size 8{0} } = { {2 "." "51"´"10" rSup { size 8{3} } " V/m"} over {3 "." 0´"10" rSup { size 8{8} } " m/s"} } } {} #=8 "." "35"´"10" rSup { size 8{-6} } " T" "." {} } } {}

Discussion

As before, a relatively strong electric field is accompanied by a relatively weak magnetic field in an electromagnetic wave, since B = E / c size 12{B= {E} slash {c} } {} , and c size 12{c} {} is a large number.

Got questions? Get instant answers now!

Section summary

  • The energy carried by any wave is proportional to its amplitude squared. For electromagnetic waves, this means intensity can be expressed as
    I ave = cε 0 E 0 2 2 , size 12{I rSub { size 8{"ave"} } = { {ce rSub { size 8{0} } E rSub { size 8{0} } rSup { size 8{2} } } over {2} } } {}

    where I ave size 12{I rSub { size 8{"ave"} } } {} is the average intensity in W/m 2 size 12{"W/m" rSup { size 8{2} } } {} , and E 0 size 12{E rSub { size 8{0} } } {} is the maximum electric field strength of a continuous sinusoidal wave.

  • This can also be expressed in terms of the maximum magnetic field strength B 0 size 12{B rSub { size 8{0} } } {} as
    I ave = cB 0 2 2μ 0 size 12{I rSub { size 8{"ave"} } = { { ital "cB" rSub { size 8{0} } rSup { size 8{2} } } over {2m rSub { size 8{0} } } } } {}

    and in terms of both electric and magnetic fields as

    I ave = E 0 B 0 2μ 0 . size 12{I rSub { size 8{"ave"} } = { {E rSub { size 8{0} } B rSub { size 8{0} } } over {2m rSub { size 8{0} } } } } {}
  • The three expressions for I ave size 12{I rSub { size 8{"ave"} } } {} are all equivalent.

Problems&Exercises

What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m?

I = cε 0 E 0 2 2 = 3.00 × 10 8 m/s 8.85 × 10 –12 C 2 /N ⋅ m 2 1 25 V/m 2 2 = 20. 7 W/m 2

Got questions? Get instant answers now!

Find the intensity of an electromagnetic wave having a peak magnetic field strength of 4 . 00 × 10 − 9 T size 12{4 "." "00"´"10" rSup { size 8{-9} } " T"} {} .

Got questions? Get instant answers now!

Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength.

(a) I = P A = P π r 2 = 0 . 250 × 10 − 3 W π 0 . 500 × 10 − 3 m 2 = 318 W/m 2 size 12{I= { {P} over {A} } = { {P} over {p r rSup { size 8{2} } } } = { {0 "." "250"´"10" rSup { size 8{-3} } " W"} over {∂ left (0 "." "500"´"10" rSup { size 8{-3} } " m" right ) rSup { size 8{2} } } } ="318 W/m" rSup { size 8{2} } } {}

(b) I ave = cB 0 2 2μ 0 ⇒ B 0 = 2μ 0 I c 1 / 2 = 2 4 π × 10 − 7 T ⋅ m/A 318 . 3 W/m 2 3.00 × 10 8 m/s 1 / 2 = 1 . 63 × 10 − 6 T alignl { stack { size 12{I rSub { size 8{"ave"} } = { { ital "cB" rSub { size 8{0} rSup { size 8{2} } } } over {2m rSub { size 8{0} } } } drarrow B rSub { size 8{0} } = left ( { {2m rSub { size 8{0} } I} over {c} } right ) rSup { size 8{1/2} } } {} #= left [ { {2 left (4¶´"10" rSup { size 8{-7} } " T" cdot "m/A" right ) left ("318" "." "3 W/m" rSup { size 8{2} } right )} over {3 "." "00"´"10" rSup { size 8{8} } " m/s"} } right ] rSup { size 8{ {1} slash {2} } } {} #= {underline {1 "." "63"´"10" rSup { size 8{-6} } " T"}} {} } } {}

(c) E 0 = cB 0 = 3 .00 × 10 8 m/s 1.633 × 10 − 6 T = 4 . 90 × 10 2 V/m alignl { stack { size 12{E rSub { size 8{0} } = ital "cB" rSub { size 8{0} } = left (3 "." "00"´"10" rSup { size 8{8} } " m/s" right ) left (1 "." "633"´"10" rSup { size 8{-6} } " T" right )} {} #= {underline {4 "." "90"´"10" rSup { size 8{2} } " V/m"}} {} } } {}

Got questions? Get instant answers now!

Questions & Answers

Discuss the differences between taste and flavor, including how other sensory inputs contribute to our  perception of flavor.
John Reply
taste refers to your understanding of the flavor . while flavor one The other hand is refers to sort of just a blend things.
Faith
While taste primarily relies on our taste buds, flavor involves a complex interplay between taste and aroma
Kamara
which drugs can we use for ulcers
Ummi Reply
omeprazole
Kamara
what
Renee
what is this
Renee
is a drug
Kamara
of anti-ulcer
Kamara
Omeprazole Cimetidine / Tagament For the complicated once ulcer - kit
Patrick
what is the function of lymphatic system
Nency Reply
Not really sure
Eli
to drain extracellular fluid all over the body.
asegid
The lymphatic system plays several crucial roles in the human body, functioning as a key component of the immune system and contributing to the maintenance of fluid balance. Its main functions include: 1. Immune Response: The lymphatic system produces and transports lymphocytes, which are a type of
asegid
to transport fluids fats proteins and lymphocytes to the blood stream as lymph
Adama
what is anatomy
Oyindarmola Reply
Anatomy is the identification and description of the structures of living things
Kamara
what's the difference between anatomy and physiology
Oyerinde Reply
Anatomy is the study of the structure of the body, while physiology is the study of the function of the body. Anatomy looks at the body's organs and systems, while physiology looks at how those organs and systems work together to keep the body functioning.
AI-Robot
what is enzymes all about?
Mohammed Reply
Enzymes are proteins that help speed up chemical reactions in our bodies. Enzymes are essential for digestion, liver function and much more. Too much or too little of a certain enzyme can cause health problems
Kamara
yes
Prince
how does the stomach protect itself from the damaging effects of HCl
Wulku Reply
little girl okay how does the stomach protect itself from the damaging effect of HCL
Wulku
it is because of the enzyme that the stomach produce that help the stomach from the damaging effect of HCL
Kamara
function of digestive system
Ali Reply
function of digestive
Ali
the diagram of the lungs
Adaeze Reply
what is the normal body temperature
Diya Reply
37 degrees selcius
Xolo
37°c
Stephanie
please why 37 degree selcius normal temperature
Mark
36.5
Simon
37°c
Iyogho
the normal temperature is 37°c or 98.6 °Fahrenheit is important for maintaining the homeostasis in the body the body regular this temperature through the process called thermoregulation which involves brain skin muscle and other organ working together to maintain stable internal temperature
Stephanie
37A c
Wulku
what is anaemia
Diya Reply
anaemia is the decrease in RBC count hemoglobin count and PVC count
Eniola
what is the pH of the vagina
Diya Reply
how does Lysin attack pathogens
Diya
acid
Mary
I information on anatomy position and digestive system and there enzyme
Elisha Reply
anatomy of the female external genitalia
Muhammad Reply
Organ Systems Of The Human Body (Continued) Organ Systems Of The Human Body (Continued)
Theophilus Reply
what's lochia albra
Kizito
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
<< Chapter < Page Page > Chapter >>
Practice Key Terms 2

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask