<< Chapter < Page Chapter >> Page >
I ave = cB 0 2 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 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 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 = 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 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 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 = 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 0 B 0 = 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

A body travelling at a velocity of 30ms^-1 in a straight line is brought to rest by application of brakes. if it covers a distance of 100m during this period, find the retardation.
Pamilerin Reply
what is distribution of trade
Grace Reply
what's acceleration
Joshua Reply
The change in position of an object with respect to time
Mfizi
Acceleration is velocity all over time
Pamilerin
hi
Stephen
It's not It's the change of velocity relative to time
Laura
Velocity is the change of position relative to time
Laura
acceleration it is the rate of change in velocity with time
Stephen
acceleration is change in velocity per rate of time
Noara
what is ohm's law
Stephen
Ohm's law is related to resistance by which volatge is the multiplication of current and resistance ( U=RI)
Laura
how i don understand
Willam Reply
how do I access the Multiple Choice Questions? the button never works and the essay one doesn't either
Savannah Reply
How do you determine the magnitude of force
Peace Reply
mass × acceleration OR Work done ÷ distance
Seema
Which eye defect is corrected by a lens having different curvatures in two perpendicular directions?
Valentina Reply
acute astigmatism?
the difference between virtual work and virtual displacement
Noman Reply
How do you calculate uncertainties
Ancilla Reply
What is Elasticity
Salim Reply
using a micro-screw gauge,the thickness of a piece of a A4 white paper is measured to be 0.5+or-0.05 mm. If the length of the A4 paper is 26+or-0.2 cm, determine the volume of the A4 paper in: a). Cubic centimeters b). Cubic meters
Ancilla Reply
what is module
Alex Reply
why it is possible for an object(man) to stay on air without falling down?
akande Reply
its impossible, what do you mean exactly?
Ryan
Exactly
Emmanuella
it's impossible
Your
Why is it not possible to stand in air?
bikko
the air molecules are very light enough to oppose the gravitational pull of the earth on the man..... hence, freefall occurs
Arzail
because of gravitational forces
Pamilerin
this mostly occur in space
Stephen
what is physics
Joshua Reply
no life without physics ....that should tell you something
Exactly
Emmanuella
😎👍
E=MC^2
study of matter and energy and an inter-relation between them.
Minahil
that's how the mass and energy are related in stationery frame
Arzail
Ketucky tepung 10m
firdaus
Treeskin, 6m Cloud gam water 2m Cloud gam white 2m And buur
firdaus
Like dont have but have
firdaus
Two in one
firdaus
Okay
firdaus
DNA card
firdaus
hey am new over hear
Shiwani
War right? My impesilyty again. Don't have INSURAN for me
firdaus
PUSH
firdaus
I give
firdaus
0kay
firdaus
Hear from long
firdaus
Hehehe
firdaus
All physics... Hahahaha
firdaus
Tree skin and two cloud have tokside maybe
firdaus
Sold thing
firdaus
PUSH FIRST. HAHAHAAHA
firdaus
thanks
firdaus
Kinetic energy is the energy due to montion of waves,electrons,atoms, molecule,substances an object s.
Emmanuella
Opjective 0
firdaus
Atom nber 0
firdaus
SOME N
firdaus
10.000m permonth. U use momentom with me
firdaus
hi
Hilal
plz anyone can tell what is meteor and why meteor fall in night? can meteor fall in the day
Hilal
meteor are the glowy (i.e. heated when the enter into our atmosphere) parts of meteoroids. now, meteoroids are the debris resulting from the collision of asteroids or comets. yes, it occurs in daytime too, but due to the daylight, we cant observe it as clearly as in night
Arzail
thank's
Hilal
hello guys
Waka
wich method we use to find the potential on a grounded sphere
Noman
hello
Pamilerin
Physics is the science that studies everything around us from the smallest things like quarks to the biggest things like galaxies. It's simply everything.
Laura
Good day everyone
Divine
It talks mainly about matter with related topics such as forces energy gravity and time. It's amazing
Laura
Hi
Alpha
Physics generally is the study of everything around us.
Steven
physics is the branch of sceince
shafiu
physics is the branch of sceince that deal with motion
shafiu
physics is the branch of sceince that deal with motion &energy
shafiu
with out a physics the life is nothing to see
Yilma Reply
What do you want to talk about😋😋
Emmanuella
the study of all the natural events occuring around us..... this is Physics (until those events obey the laws of physics)
Arzail
Conservation of energy😰
Emmanuella
yeah, that too
Arzail
Energy, it always remains there in a physical system. it can only take the form either in motion (kinetic energy) or in rest (potential energy)
Arzail
In nature organisms feed on one another in an orderly way.
Emmanuella
that describes the food chain, in which we humans are at the top
Arzail
The energy that came initially from the sun 🌞is converted into a form in which it can be stored in green plant.
Emmanuella
Therefore, there is conservation of energy.
Emmanuella
DNA CARD
firdaus
"card"
firdaus
Darag
firdaus
Practice Key Terms 2

Get the best College physics course in your pocket!





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