I have got some questions on this topic based on the best answers I get to these:

-1- If magnetic fields arise due to electric currents in conductors, we say they are not E-M waves in nature; it is only a field { in this case- magnetic}. How, then, can they be used for information transmission? source: http://www.sparkbangbuzz.com/els/magbdcst-el.htm

The link says a coil of wire is used, so we know there is no tank circuit or any of that sort to describe a radio transmitter.

-2- If infact, this works, why the war for bandwidth when we can start exploring magnetic field transmission? Plus, energy is stored and returned in each phase, so minimal power expended. {I know magnetic fields can interact with charges, and thus living cells, but still...}

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    \$\begingroup\$ The light from a star (an EM field) can travel billions of miles and is detectable, try measuring the magnetic field strength of the same star and differentiate it from all other magnetic fields. \$\endgroup\$ – JIm Dearden Jul 22 '15 at 9:26
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    \$\begingroup\$ sparkbangbuzz.com/els/magbdcst-el.htm - an excellent example of why should never believe what's written on the interweb. This article should never be used as a source reference for anything. \$\endgroup\$ – JIm Dearden Jul 22 '15 at 9:51

A transformer can certainly be used for data transmission between two points and it uses a magnetic field for transmission. Real power is transferred but only the H field is used. You can separate the two windings so that the coupling factor is very small and still transmit power but the demands on the coils increase to preserve decent efficiency.

RFID tags (a lot of them) only use magnetic fields for transfer of data and power for the hand-held device.

But, the main trouble with using a H field (or a E field) is that it reduces in amplitude with distance cubed. A proper EM wave reduces the H (or E field) with distance i.e. no inverse cube or square law. Note also that the power of an EM wave reduces with distance squared because the power is proportional to E multiplied by H and if H and E halve with distance, the power quarters.

On the other hand every H field has an associated E field (maxwell etc..) so why do we say the H field reduces as per distance cubed - the reason is this - the H and E fields have to be of the correct ratio to constitute a bona fide EM wave. The ratio has to be the impedance of free-space (approximately 377 ohms resistive) so E is bigger than H by a ratio 377:1: -

enter image description here

When you plug the numbers into the equation, the square root of the ratio of permeability to permittivity is ~377 ohms.

With a simple oscillator and loop antenna (coil) running at too low a frequency, the E field will be too small to meet the above criteria and only a very tiny EM wave is produced. But, the H field is still "usable" for transferring power and data (inverse cube law) but the associated E field is of no use. When a loop is used that is dimensionally close to the wavelength of the carrier frequency, the E field rises and a properly proportioned EM wave is produced. This is what we call a radio wave - it has electric and magnetic parts at a ratio of 377:1.

Regards the use of a H field to free-up bandwidth (Q2) - this will interfere with an EM transmission if at the same frequency - there is no magic bullet on this - interfere with the E or the H fields from another source and you'll disrupt the ability of a receiver to detect the original EM wave. The energy returned is not identical to the energy sent/stored if energy is removed. Energy has to be removed by a receiver or it receives nothing.

The article linked is profoundly incorrect when it states this: -

It is my belief that radio waves are nothing more than a simple magnetic field

The article also said this: -

A complete circuit is just plain not necessary

Of course a circuit is necessary and although a whip antenna may look incomplete the capacitance to to return wire (earth in the case of a quarter wave antenna) completes the circuit. No complete circuit = no current flow = nothing.

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Your question is too broad, but let me take a stab at it anyway. A field is just the distribution of something in space. A wave is the result of a time varying field. What Maxwell's laws tell us is that E and H fields (elec. and mag. resp.) cannot be changed independently, so a changing magnetic field is associated with a changing E-field. To transmit any information, you need to trasfer energy. A field cannot transmit energy - only a wave can.

Only, when you try to change the a H-field, the E-field will also change. So you can only have E-M waves and never just E-wave, and neither just H-waves.

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    \$\begingroup\$ A field certainly can transfer energy. How do you think a transformer works? The other problem in your answer is that you are missing the whole aspect of the ratio of E to H fields. When the ratio is correct i.e. 377:1 you get a radio wave that is optimized for transmission thru space. \$\endgroup\$ – Andy aka Jul 22 '15 at 10:51
  • \$\begingroup\$ Transformers have varying fields. You can't connect DC to a transformer. Due to material property (of the core) the change in H field is entirely* in the core. This results in changing E-fields which drive currents not only in the coils, but in the core itself (eddy currents). *See Leakage inductance. \$\endgroup\$ – kabZX Jul 22 '15 at 11:38
  • \$\begingroup\$ I fail to see why leakage inductance is in any part of your comment on transformers. I think you mean magnetization inductance not leakage. \$\endgroup\$ – Andy aka Jul 22 '15 at 11:47
  • \$\begingroup\$ @Andyaka No I meant leakage. I was just pointing out that the changing field in a transformer is not entirely in the core. \$\endgroup\$ – kabZX Jul 22 '15 at 11:56
  • \$\begingroup\$ You're right that any changing E creates a changing H and vice versa. But in the near field of a capacitor or a coil, the E or H dominates completely. So if you're less than say 1/6 wavelengths from the source, you can ignore the EM type wave and concentrate on the fields. This is the principal of inductive coupling, for example a contactless credit card. If you're a wavelength or two away, then the EM wave dominates. Radio waves etc. \$\endgroup\$ – tomnexus Jul 22 '15 at 21:47

To tranfer any kind of useful information with E-M fields requires a transfer of energy to "excite" the receiver (usually an antenna and RF receiver combined) and a modulation scheme. Thus the need for a "carrier" to modulate and also a proper frequency that will, in fact, radiate, causing E-M waves to be generated. It is impossible to separate the E from the M. They constantly interact and are generateed at the same time by their very nature. They create waves if used in this manner and those waves radiate (travel to some distance). Without the waves and the radiation there can be no information sent or received. Even CW uses a modulatin scheme (carrier on/off).

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