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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.

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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

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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"} {} .

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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"}} {} } } {}

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Questions & Answers

What is physics?
Jeuloriz Reply
physics is a branch of science in which we are dealing with the knowledge of our physical things. macroscopic as well as microscopic. we are going look inside the univers with the help of physics. you can learn nature with the help of physics. so many branches of physics you have to learn physics.
vijay
What are quarks?
Breanna Reply
6 type of quarks
Neyaz
what is candela
Akani Reply
Candela is the unit for the measurement of light intensity.
Osei
any one can prove that 1hrpower= 746 watt
Neyaz Reply
Newton second is the unit of ...............?
Neyaz
Impulse and momentum
Fauzia
force×time and mass× velocity
vijay
Good
Neyaz
What is the simple harmonic motion?
Fauzia Reply
oscillatory motion under a retarding force proportional to the amount of displacement from an equilibrium position
Yuri
Straight out of google, you could do that to, I suppose.
Yuri
*too
Yuri
ok
Fauzia
Oscillatory motion under a regarding force proportional to the amount of displacement from an equilibrium position
Neyaz
examples of work done by load of gravity
Maureen Reply
What is ehrenfest theorem?
Fauzia Reply
You can look it up, faster and more reliable answer.
Yuri
That isn't a question to ask on a forum and I also have no idea what that is.
Yuri
what is the work done by gravity on the load 87kj,11.684m,mass xkg[g=19m/s
Maureen
What is law of mass action?
Fauzia Reply
rate of chemical reactions is proportional to concentration of reactants ...
muhammad
ok thanks
Fauzia
what is lenses
Ndobe Reply
lenses are two types
Fauzia
concave and convex
muhammad
right
Fauzia
speed of light in space
Vikash Reply
in vacuum speed of light is 3×10^8 m/s
vijay
ok
Vikash
2.99×10^8m/s
Umair
2.8820^8m/s
Muhammed
which is correct answer
Vikash
he is correct but we can round up in simple terms
vijay
3×10^8m/s
vijay
is it correct
Fauzia
I mean 3*10^8 m/s ok
vijay
299792458 meter per second
babar
3*10^8m/s
Neyaz
how many Maxwell relations in thermodynamics
vijay
how we can do prove them?
vijay
What is second law of thermodynamics?
Neyaz
please who has a detailed solution to the first two professional application questions under conservation of momentum
Kwaku Reply
I want to know more about pressure
Osei
I can help
Emeh
okay go on
True
I mean on pressure
Emeh
definition of Pressure
John
it is the force per unit area of a substance.S.I unit is Pascal 1pascal is defined as 1N acting on 1m² area i.e 1pa=1N/m²
Emeh
pls explain Doppler effect
Emmex
solve this an inverted differential manometer containing oil specific gravity 0.9 and manometer reading is 400mm find the difference of pressure
Abayomi Reply
Einstine claim that nothing can go with the speed of light even its half (50%) but in to make antimatter they they hit the sub atomic particals 99.9%the speed of light how is it possible
Salima Reply
nothing with physical properties. this doesn't include things like particles and gravitational waves
Mustafa
that particles are of very small mass.... near equals to massless
Aritra
but they exist
vijay
yes they exist but mass is too less
Aritra
ok
vijay
greet all
Abayomi
the unit of radioactivity is .....?
Neyaz
Great Sharukh ! Do you have question in physics?
Bibekbir Reply
book says that when wave enter from one medium to another its wavelenght changes but frequency not how ? and f is inversely related to wavelenth
Sharukh
yes but how comes
Sani
how are you?
Sharukh Reply
please help me
World
what's the problem
Aritra
I really don't know physics.. I need help,in solving
Amara
me too
Ewulum
hii
Cheeru
I really don't know physics.. I need help,in solving
Cheeru
me too
True
I can teach u if u are ready
latunde
yes I am ready
True
hi
Emeh
Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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