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Electromagnetic wave from an antenna

  1. Jun 29, 2011 #1
    1. The problem statement, all variables and given/known data
    An electromagnetic wave from a wire antenna travels (from the reader) toward the plane of the paper. At time t = 0.0 s it strikes the paper at normal incidence. At point O and t = 0.0 s, the magnetic field vector has its maximum value, 5.01E-8 T, pointing in the negative y-direction. The frequency of this wave is 2.86E6 Hz.

    (i.) What is the x-component of the associated electric field E at time t = 0.0 s? (Use the right-hand rule to determine the direction of E, and hence the sign of the x-component.)
    Answer is: 15 V/m

    (ii.) What is the magnitude of the Poynting vector of the wave at time t = 0.0 s?
    Answer is: .599 W/m^2

    (iii.) What is the y component of the magnetic field at point O at time 1.75×10-7 s?
    *HELP!


    2. Relevant equations
    B=B(max)*sin(kx-wt)


    3. The attempt at a solution
    Tried to kind x and k and w from equations like T=1/f
     
  2. jcsd
  3. Jun 29, 2011 #2
    Basically the question is saying at t = 0, B = B(max), thus sin(kx) = 1. So from this you can work out what x is at t = 0.

    k is known as the wave vector and is related to the wavelength of a wave by:

    [tex] k = 2 \pi /\lambda [/tex]

    where λ is the wavelength. Do you know how to get the wavelength of a wave from its frequency? Once you know that, you can work out k.

    Also it might help if you knew that w = 2πf, where f is the frequency.
     
  4. Jun 30, 2011 #3
    Well, so the angle whos sine is 1 would be pi/2, so at t=0, it starts at pi/2 ?
    And for the wavelength i have [itex]\lambda[/itex]= c/frequency

    and i have [itex]\omega[/itex]=2[itex]\pi[/itex]*frequency


    But, how do I figure out what x is at the time given? I heard something from students about finding the period and doing t/T ? I'm a little confused with that.
     
  5. Jun 30, 2011 #4
    If sin(kx) = 1 at t=0, then kx = pi/2 at t=0. You can calculate k from the equation I gave you in the last post. Once you've got k, you've got x.
     
  6. Jun 30, 2011 #5
    Okay, but that would just give x for t=0, right, don't I need it for the given time too ?
     
  7. Jul 1, 2011 #6
    It doesn't actually matter what x is at point O, only the value of kx matters. If you read the question it says to find the magnetic component at the same point O (thus same kx) at a different time.

    At t = 0, ωt was zero but at t = 1.75×10-7 s ωt is clearly not going to be zero. Find out what ωt is at this new time.
     
  8. Jul 5, 2011 #7
    Okay, so kx would equal pi/2 again so technically you don't have to find out what k and x is individually right? [itex]\omega[/itex] equals 17969909.97853 (2*pi*f). My final answer that I came out with is -1.3760938981E-9. Is this correct? I only have two more tries left and I don't want to waste it if it's not looking right. Should it be negative?
     
  9. Jul 7, 2011 #8
    Okay, I got that as a wrong answer.. I tried it again..
    I did B= B(max)*sin(kx-wt)

    k at t=0, k= (2*pi*f)/c= 0.0598997
    and since sin(kx)=1 [max], kx= pi/2 which makes x=26.22377622.

    k at t=1.75E-7, k= (2*pi) / (c*(t/T))= 4.18461E-8

    w at t= 1.75E-7 = (2*pi) / (t/T)= 12.5538168.

    [t/T=0.5005= # of periods passed in the amount of time given]

    so,

    B= 5.01E-8*sin[(1.09736E-6)-(12.538168*1.75E-7)]
    B=-9.61463001E-16 T

    Is this right??
     
  10. Jul 11, 2011 #9
    Can anyone tell me if this is right? I'm afraid to try it, only one try left !
     
  11. Jul 14, 2011 #10
    can anyone at least tell me if it definitely should be negative ?
     
  12. Aug 9, 2011 #11
    Would someone be able to check over my work for me and tell me if my answer should be right? I only have one more try ..
     
  13. Aug 10, 2011 #12
    nobody wants to tell me if im close to being right ..?
     
  14. Aug 10, 2011 #13

    ehild

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    B=Bmaxsin(kx-ωt). At t = 0, B=-Bmax, so kx =-pi/2.
    Recalculate B.


    ehild
     
  15. Aug 10, 2011 #14
    okay so my final answer came out to be -2.88055 E-15 would that be right?
     
  16. Aug 11, 2011 #15

    ehild

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    Show how you got that result.

    ehild
     
  17. Aug 11, 2011 #16
    Oh, I just did everything the same just a few changes.... :

    I did B= B(max)*sin(kx-wt)

    At t=0 sec:
    B=-B(max)
    sin(kx)= -1 [max] ==> kx= -pi/2
    x=-26.224 because k(t=0)= (2*pi*f)/c = 0.0598997

    At t= 1.75E-7 sec:
    k=(2*pi)/(c*(t/T)) = 4.18E-8
    w= (2*pi)/(t/T) = 12.55
    t/T = 0.5005

    so,
    B= 5.01E-8*sin[(-1.09736E-6)-(12.538168*1.75E-7)]
    B= -2.88055 E-15 T
     
  18. Aug 11, 2011 #17

    ehild

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    B=Bmaxsin(kx-wt)

    You do not need to calculate k and x separately as the formula contains kx (k multiplied by x). We already agreed that kx=-pi/2 at t=0 and at point O.

    It is given that the frequency is f=2.86 E6 Hz. You need to calculate ωt (ω multiplied by t) at time t=1.75 E-7 s.
    ω is the angular frequency. It is constant, characteristics of the wave, just like frequency. How is the angular frequency related to the frequency of any periodic motion? Check your notes or google "angular frequency".


    ehild
     
  19. Aug 12, 2011 #18
    thats what I thought but that other person was telling me to figure out what they were separately. And w= 2*pi*f , i know that. so just that times the given time right
     
  20. Aug 12, 2011 #19
    -2.38672E-08
    so thats my actual answer then?
     
  21. Aug 12, 2011 #20
    which was from:

    B=B(max) * sin[ (-pi/2)-(2*pi*2.86E6*1.75E-7) ]
     
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