# E and H fields created by fiber optics?

When an EM wave travels down a conductor, it creates and electric and magnetic field around (H) the wire and normal to (E) the wire.

My question is, when light travels down an optical material such as fiber optics, is there an similar magnetic and electric field created around the fiber?

My gut tells me there should be, but I cant reason it out. I'm looking for some help with this thought experiment.

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If we think about it for a moment, and if we know that light is an E and H field, then your gut feeling is right -- there must be one. However, if you look at the end of an optical fiber with light in it (taking appropriate safety precautions!) you see the light in the fiber, not around it. So the field must be inside, not outside the fiber.

To see why, we will turn to ray optics. Imagine an optical fiber stretched straight. A ray of light entering the fiber at parallel incidence simply goes down the fiber and emerges from the other end. If a ray enters at an angle, however, then it might escape through the wall of the fiber. But if the angle is small enough, it will bounce off the inside wall of the fiber due to total internal reflection.

This is perhaps best illustrated by this image courtesy of Wikipedia.

So the light, and therefore the E and H fields, are confined inside the fiber.

If you know a little about total internal reflection, you might realize this is not entirely true -- there are evanescent fields on the outside of the fiber. "Evanescent" means they don't propagate away from the fiber, but just decay exponentially, so this is entirely dissimilar from the fields around a current traveling through a conductor.

Also, this description doesn't hold for a single-mode fiber, which can't be analyzed with ray optics -- you need to consider modes instead. Nonetheless, this is a good intuitive way of looking at it.

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Thankyou. Very helpful. Almost feels like a "see the forest for the trees" conundrum. Light is an EM wave, which is the propagation of electromagnetic energy, which kind of makes the whole question moot. Thanks again. – Michael Sep 21 '12 at 2:32