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I know that for EM waves (i.e sunlight) for any E wave in any direction, there is B field perpendicular to it. However, when we pass that EM wave through the linear polarization filter, what actually passes through the filter? Is it any component of E or B that is parallel to the filtering lines? If that is true then it means that any E or B field component that is filtered through is going to be parallel to each other, right? Are these filtered E and B, kind of independent fields or what kind of interaction do they hold with each other. I am confused because, I know that for any E(t) time-varying field there is always perpendicular B field associated with it, which is proven wrong by my assumption of the linear polarization. Can someone explain what I am assuming wrong here and the correct explanation behind it.

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There is an asymmetry in matter between E and B fields. This is due basically to the fact that all matter is composed out of electric monopoles, charged particles, whereas magnetic monopoles do not exist to take a part. Thus it is the spill over electric fields between atoms that are strong enough to filter and polarize EM radiation. The magnetic fields due to magnetic moments of atoms and molecules are very much weaker.

Filters of course are chosen because they are efficient in polarizing light.

emradiation

Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This 3D animation shows a plane linearly polarized wave propagating from left to right. Note that the electric and magnetic fields in such a wave are in-phase with each other, reaching minima and maxima together

It is the E field that will be allowed through the filter fields and the perpendicular B field is not affected in a good filter.

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  • $\begingroup$ what do you mean by B field is not affected in a good filter? does that mean, the B field associated with that filtered E field also passes along with filtered E field and doesn't get isolated? $\endgroup$ – dr3patel Feb 25 '16 at 4:44
  • $\begingroup$ It passes through filter matter without being affected , so it is the E, which is affected, that defines which space direction the coupled fields will have. $\endgroup$ – anna v Feb 25 '16 at 4:49
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When an un-polarized EM wave passes through a filter ( We can imagine a filter consisting of 'regular array of parallel metallic wires ) the component of $\vec E $ which is parallel to wire can interact with the electrons in the wire. This essentially is like an EM wave getting reflected from a metal surface. This reflected wave is $180^{\circ}$ out of phase with the incident wave and hence they cancel each and left with only the perpendicular component. The perpendicular component's interaction with the wire is minimal because the electrons movement is constrained in that direction ( we can assume the wire to be very thin). So the component which passes through the filter is perpendicular to the filtering lines.

The Magnetic field and the Electric field are not at all independent, they are intimately related to each other because they together have to satisfy the maxwell's relations. The magnetic field exist because there is a time varying electric field. Or you can very well say a similar story from the prescriptive of magnetic field and reach the same conclusion : Electric field exist because time varying magnetic field exist.

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  • $\begingroup$ so, E parallel (to wires) is reflected, E perpendicular (to the wires) is filtered through and also, the B perpendicular (to passed E) and parallel (to wires) is also passed through? $\endgroup$ – dr3patel Feb 25 '16 at 15:05
  • $\begingroup$ @dr3patel When we say that the E (parallel) got reflected , we mean the EM wave composed of both E(parallel) and corresponding Magnetic field (perpendicular to E(parallel) and hence to the wires) is getting reflected . So yes Magnetic field component parallel to the wires 'passes' through and not the other one. $\endgroup$ – varghese Feb 25 '16 at 15:08
  • $\begingroup$ You should not think of magnetic field and electric field as two different things. Think about what will happen to E(or B) and see what will be the other from maxwells relations. $\endgroup$ – varghese Feb 25 '16 at 15:14

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