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E field - B field interaction

  1. Aug 19, 2012 #1
    Ok so I was wondering, lets say I have a device (eg., shaped like a toothpaste box with the two square ends cut off). I make 2 of the opposite faces from copper plate and the other 2 from iron plate (for example). I then generate an electric field in the box with the copper plates, and a magnetic field from the iron plates. If I pulse both the E and B fields at visible light frequency, will the result be visible light in the box?
     
  2. jcsd
  3. Aug 19, 2012 #2
    What you are basically constructing is an antenna. I don't believe there are any antennas capable of generating/receiving frequencies as high as visible light. Your setup would probably have VERY high impedance for frequencies that high, so it wouldn't work.

    The 'fastest' antenas we have are 'nanotube' antennas, which are effective in the terahertz (10^12 Hz), which is still much lower than visible light (10^15 Hz).
     
  4. Aug 19, 2012 #3
    its a thought experiment, assume we get past technical issues.
     
  5. Aug 19, 2012 #4
    Okay, but you still mention copper/iron, which wouldn't react fast enough to an applied voltage to achieve these frequencies. :uhh:

    But, if you could construct some antenna which can achieve frequencies high enough, you would see light.
     
  6. Aug 19, 2012 #5
    thats awesome, it means if we ever get there we can have light sources with incredible bandwidth and resolution. photochemistry would never be the same.
     
  7. Aug 19, 2012 #6
    You don't need a separate magnetic field source. A variable electric field induces a magnetic field, according to Ampere-Maxwell's Law, which in integral form is:
    [tex]
    \oint_{C}{\vec{B} \cdot d\vec{\mathcal{l}}} = \frac{1}{c^2} \, \frac{\partial}{\partial t} \left( \, \iint_{S}{(\vec{E} \cdot \hat{n}) \, da} \right) + \mu_0 \, I
    [/tex]
    where C is a closed contour with a line element [itex]d\vec{\mathcal{l}}[/itex], and S is any surface subtended on the contour C with a unit normal [itex]\hat{n}[/itex]. I is the total (convective) electric current passing through the contour C.

    In vacuum, the convective current I is zero, but the displacement current (the first term on the r.h.s. containing a time derivative of the electric flux) is usually large due to a rapidly oscillating field.

    This (osillating) magnetic field, in turn, generates additional electric fields, according to Faraday's Law:
    [tex]
    \oint_{C}{\vec{E} \cdot d\vec{\mathcal{l}}} = -\frac{\partial}{\partial t} \left( \iint_{S}{(\vec{B} \cdot \hat{n}) \, da}\right)
    [/tex]

    This is essentially the mechanism for emitting electromagnetic waves that propagate freely in space away from the initial source of the oscillating electric field.

    However, as mentioned by other posters, the frequency of electromagnetic waves generated by this mechanism is limited to frequencies that correspond to a wavelength comparable to the linear dimensions of the circuit elements.
     
  8. Aug 19, 2012 #7
    indeed, the only reason i suggest both is to have increased control over the photon's EM vector
     
  9. Aug 19, 2012 #8
    But, you will not have because the two fields will be incoherent, thus forming independent electromagnetic waves.
     
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