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What is the maximum magnetic field strength?

  1. Apr 24, 2007 #1
    1. The problem statement, all variables and given/known data
    The figure shows a vertically polarized radio wave of frequency 1.0 times 10^6 traveling into the page. The maximum electric field strength is 1000V/m.
    What is the maximum magnetic field strength?
    What is the magnetic field strength at a point where E=500V/m down
    What is the smallest distance between a point on the wave having the magnetic field of part b and a point where the magnetic field is at maximum strength?


    2. Relevant equations
    Not sure
    only equation it gives in book ins I=Io*cos^2(theta)

    3. The attempt at a solution
    I really have no idea how to even start this problem our physics book doesn't give any relationships that I can find to calculate this problem. If someone could provide with me the equations that can help me or a source that explains this kind of situation it would be helpful.
     
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Apr 24, 2007 #2

    mezarashi

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    Homework Helper

    The equation you list is useful if you would like to calculate the output power of unpolarized light going through a polaroid filter.

    Therefore I'm not sure what kind of EM you're studying at the moment. There is an equation that relates the electric and magnetic fields of an EM wave. It is based on the known fact that the energy is distributed evenly among the two.

    [tex] E = \frac{B}{\sqrt{\epsilon_0 \mu_0}}[/tex]
     
  4. Nov 24, 2008 #3
    Re: Polarization

    Obviously I'm seeing this a bit late to help the person who originally posted, but in case anybody else has issues...

    The post before this is correct - the easiest way to calculate an EM wave's magnetic field from its electric field, or vice versa, is to remember that energy is shared equally between the fields.

    [tex]E=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}B[/tex]

    [tex]\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}[/tex] is the speed of light, c, which is equal to [tex]3.00\times10^{8} m/s[/tex].

    [tex]E=cB[/tex]

    Answering the first two questions should be simple here. Since it doesn't appear that you are being asked for a direction/phase sign, you shouldn't need to use the right-hand rule.

    For the last question, consider the amplitude of E. You are told that the maximum E-field is equal to 1000 V/m. This corresponds to a maximum of a sinusoidal wave with an amplitude of 1000 units. You are asked to find the shortest distance on the wave between this point and a point where the E-field is 500 V/m, or half the amplitude. Since E-fields and B-fields are not measured in meters, you are really being asked to find this distance on the axis of propagation. If we take the simple unit circle of circumference [tex]2\pi[/tex], and find the distance on the circle between the point where the y-value is 1 (the maximum) and a point where the y-value is 1/2 (half of the maximum), we calculate the length of that section of the circle is

    [tex]\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}[/tex]

    which is 1/6 of the total circumference. Since one turn of the circle represents one complete wavelength, and we know that the frequency of the wave is [tex]1.0\times10^{6}s^{-1}[/tex], we can calculate

    [tex]\lambda=\frac{c}{f}=\frac{3.0\times10^{8}ms^{-1}}{1.0\times10^{6}s^{-1}}=300m[/tex]

    1/6 of the total wavelength is 50m.

    Sorry for the wordiness - a picture would have helped for the last question.

    Matt Kramer Madison, WI mfkramer@wisc.edu
     
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