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Why is a changing magnetic field and changing electric field perpendicular?

  1. Dec 31, 2009 #1
    Hi,

    My question is as per the title. Why 90 deg and no other angle else? I did a thorough Net Search and could find no satisfactory answer. Can it be explained using Maxwell's Equation? (Sorry, i dont really understand Maxwell's equations)

    I am a teacher who is apprehensive about teaching this topic due to its abstractness (to me anyway).

    Happy New Year to all!
     
  2. jcsd
  3. Dec 31, 2009 #2

    Vanadium 50

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    I am confused about the question. I can have an electric and a magnetic field pointing in any two arbitrary directions, and I can change either or both of them.
     
  4. Dec 31, 2009 #3
    Oops, sorry if i had not made myself clear.

    I can rephrase this question as "why is it that a changing magnetic field produces a changing electric field which is perpendicular to the changing magnetic field?"

    I hope i had cleared up some confusion. Sorry, i was too caught up in my own world using my own lingo...
     
  5. Dec 31, 2009 #4

    cgw

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    You are correct about Maxwell's equations. I was just reading about this the other night in The Feyman Lectures on Physics. If you have access to this book (Vol II) it is covered pretty well.
    Maybe someone here can put it in a nutshell though.
     
  6. Dec 31, 2009 #5
    This is basically the explanation of Lorentz force. A good way to visualize it is the right hand rule. You make a thumbs up position with your right hand and grab a (hopefully insulated) wire carrying current. If your thumb is pointing in the direction of current flow, your other 4 fingers represent the direction of the magnetic field around the wire. As to why this is, I'm not sure but is anyone really? I take it as a given and have never questioned why it's like that, not that you shouldn't.

    Another source of confusion is that engineers often refer to this as the left hand rule, because it has become a staple of print reading tradesmen to envision current flowing from pos to neg, even though this is not the case in electron theory. This was taught to me exactly as I described in the first paragraph but as the left hand rule in a motor theory class. (And I'm not talking about the vector type rule with motor torque to all you smarty pants out there!)
     
  7. Dec 31, 2009 #6

    diazona

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    To some extent it's a historical artifact of the way magnetic fields were defined. Originally, they were used to describe bar magnets attracting each other, as well as compasses responding to the Earth's magnetic field, and people just assumed that the field lines pointed out of the north pole of a magnet and into the south pole. Only later was it discovered that the currents in magnets ran perpendicular to the axis of the poles.

    Although now that I think about it, I can't really imagine a way of developing magnetism where the field would be parallel to the current. For instance, cyclotron motion shows that the magnetic force is always perpendicular to the velocity and to some other axis, not like the electric force which is always in one particular direction for a given uniform field.

    As to why it's the right hand rule rather than the left hand rule: that's purely convention. Probably because most people are right handed, I guess.
     
  8. Dec 31, 2009 #7

    diazona

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    And there I go drifting off topic again :blushing:
    There is a derivation of the mathematical description of electromagnetic waves on Wikipedia:
    http://en.wikipedia.org/wiki/Electromagnetic_waves
    Part of the derivation shows why the electric and magnetic fields have to be perpendicular to the direction of propagation. Basically, by putting together two of Maxwell's equations, you obtain the wave equation
    [tex]\biggl[\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\biggr]f(x, y, z, t) = 0[/tex]
    where f is any component of either an electric or magnetic field. By solving this equation, you find that the solutions have to be proportional to
    [tex]f(k_x x + k_y y + k_z z + \omega t)[/tex]
    where [itex]k_x^2 + k_y^2 + k_z^2 = \omega^2/c^2[/itex]. (Any function that depends on x, y, z, and t in that particular combination, and only that particular combination, satisfies the wave equation.) The vector with components [itex]k_x,k_y,k_z[/itex], the wavevector, points in the direction of propagation.

    Now, the other two of Maxwell's equations say that the fields are divergence-free; that is,
    [tex]\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = 0[/tex]
    and similarly for the magnetic field. Remember that each of these components has to take a form like
    [tex]E_x = E_{(0)x}f(k_x x + k_y y + k_z z + \omega t)[/tex]
    where [itex]E_{(0)x}[/itex] is some amplitude. Anyway, if you put this type of function into the divergence-free condition from a couple lines above, you find that
    [tex]E_x k_x + E_y k_y + E_z k_z = 0[/tex]
    which is just saying that
    [tex]\vec{E}\cdot\vec{k} = 0[/tex]
    This tells you that the electric field is perpendicular to the direction of propagation.

    By the way, I commend you on taking the time to learn the material well before you teach it. It's disappointing how often people seem to not do that.
     
  9. Jan 1, 2010 #8
    Thanks Diazona for the detailed explanation. Even though i cant say fully i understood the equations, i felt better with others concurring with me on this i.e. we have to fall back on Maxwell's Eqns to explain, most of the time we take it as a given...

    Thanks Diazona also for the encouragement in my teaching. :!!)
     
  10. Jan 1, 2010 #9
    Hmm. I don't know as diazona'a results are generally true. The vacuum wave equation assumes an isotropic, homogeneus media having zero charge and current density.
     
  11. Jan 1, 2010 #10

    diazona

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    Yes, so? We're talking about electromagnetic waves in a vacuum (or at least that was my impression).
     
  12. Jan 1, 2010 #11
    AP_cycles may be talking about Maxwell's equation curl E = - ∂B/∂t, commonly known as the differential form of Faraday's Law. In the integral form,

    E·dl = - (d/dt)∫B·n dA

    where a changing magnetic field B through an area A creates an electric field around the perimeter ∫dl of the area.

    The most often application of this is the common electrical transformer.

    Bob S
     
  13. Jan 1, 2010 #12
    Read post 1 again. Neither is it clear that the OP knows there could be differences.
     
  14. Jan 1, 2010 #13

    diazona

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    Again, so? I don't see what you're getting at.

    For clarity, I will state that regardless of what else may be going on, my post #7 is talking about electromagnetic waves in a vacuum. Maybe I should have been more specific about that.
     
  15. Jan 1, 2010 #14
    I have always wondered whether this property is an inherently chiral one, or just an artifact of nomenclature. I ask this: would the laws of the universe be invariant if it simply were the left hand rule? Is there something real that actually takes that direction, or is it just what is called what?

    It is hard to phrase the question correctly. Sorry if it's confusing.
     
  16. Jan 2, 2010 #15

    ideasrule

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    It's just an artifact of nomenclature. An alien could envision magnetic fields as going from south to north; then the left-hand rule would apply.
     
  17. Jan 2, 2010 #16
    ideasrule has answered that question, but you might be interested in parity (and CP) symmetry violations (try wikipedia).
     
  18. Jan 2, 2010 #17
    This property is NOT a chiral one. I believe that fields due to electric, magnetic, and gravity, are non-chiral. They look the same in a mirror. We could use the left-hand rule as well, but the majority are right-handed. It is indeed arbitrary.

    The weak electric force, however, exhibits chirality. This was discovered in 1957 by Dr. Wu. Beta particles, i.e. electrons, expelled from an atom, have a predominant left-handed spin. In a mirror they exhibit a right-handed spin. As of now, the universe is asymmetrical.

    Claude
     
  19. Jan 3, 2010 #18
    That I understand. The question's sometimes the hardest part.

    It's more than convension--you will never see freely propagating, left-handed electromagnetic waves. However we do have to established what right-handed or left-handed means by some physical standard.

    It is the naming that is arbitrary. Right-handed could just as well be called left-handed, and with everyone agreeing on the name, radiation would still be chiral.

    Free electromagnetic waves are invariant under PT symmetry. P is parity or chirality. T is time. Replace your (x,y,z,t) coordinate system with one or all spatial dimensions, as well as the time coordinate inverted so that you get, say (-x,y,z,-t). PT(x,y,z,t)=(-x,-y,-z,-t).

    If there are no physically measurable difference for a system under this change, it is said to be PT invariant.

    There's another way to look at it. If you are looking at an electromagnetic wave reversed in time, you have to change its handedness so that it looks correct.
     
    Last edited: Jan 3, 2010
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