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

Why is b in the answer
Dahsolar Reply
how do you work it out?
Brad Reply
answer
Ernest
heheheehe
Nitin
(Pcos∅+qsin∅)/(pcos∅-psin∅)
John Reply
how to do that?
Rosemary Reply
what is it about?
Amoah
how to answer the activity
Chabelita Reply
how to solve the activity
Chabelita
solve for X,,4^X-6(2^)-16=0
Alieu Reply
x4xminus 2
Lominate
sobhan Singh jina uniwarcity tignomatry ka long answers tile questions
harish Reply
t he silly nut company makes two mixtures of nuts: mixture a and mixture b. a pound of mixture a contains 12 oz of peanuts, 3 oz of almonds and 1 oz of cashews and sells for $4. a pound of mixture b contains 12 oz of peanuts, 2 oz of almonds and 2 oz of cashews and sells for $5. the company has 1080
ZAHRO Reply
If  , , are the roots of the equation 3 2 0, x px qx r     Find the value of 1  .
Swetha Reply
Parts of a pole were painted red, blue and yellow. 3/5 of the pole was red and 7/8 was painted blue. What part was painted yellow?
Patrick Reply
Parts of the pole was painted red, blue and yellow. 3 /5 of the pole was red and 7 /8 was painted blue. What part was painted yellow?
Patrick
how I can simplify algebraic expressions
Katleho Reply
Lairene and Mae are joking that their combined ages equal Sam’s age. If Lairene is twice Mae’s age and Sam is 69 yrs old, what are Lairene’s and Mae’s ages?
Mary Reply
23yrs
Yeboah
lairenea's age is 23yrs
ACKA
hy
Katleho
Ello everyone
Katleho
Laurene is 46 yrs and Mae is 23 is
Solomon
hey people
christopher
age does not matter
christopher
solve for X, 4^x-6(2*)-16=0
Alieu
prove`x^3-3x-2cosA=0 (-π<A<=π
Mayank Reply
create a lesson plan about this lesson
Rose Reply
Excusme but what are you wrot?
Reply
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