Basically you don't need to consume power to set-up an E-field unless you are taking power from that E-field or, the thing that generates the E-field is also generating a H-field i.e. it is a regular antenna. The first line in your question mentions a dipole antenna and this will transmit a regular E-M wave so the 1mW will disperse to the far corners of the universe BUT, this needn't be the case in the magnetic equivalent, a coil of wire.
The term "reference antenna" is mentioned because it is important to realize that a dipole antenna has directivity (also called gain) - different antennas have different gains. For instance, the dipole has a gain (compared to the theoretical isotropic antenna) of about 1.7 dB - hence it is appropriate to mention what the reference antenna is in the production of E-fields using a dipole antenna.
For the generation of a distant H-field, a coil is used with current flowing in the coil. See this diagram below: -
See also this excellent website for greater detail and in particular, how the above formula is derived from the Biot-Savart law.
Plugging in some numbers to the formula above will get you the flux density At the plane of the receive coil. Given that B = \$\mu H\$ you can calculate the H field intensity.
EXTRA INFORMATION
How much drive current (uA) is needed to achieving filed strength of 3
dΒμΑ/m at 10 metres at 1MHz?
Theoretically, the drive current supplied to the transmit coil can be much less than what is actually flowing in the transmit coil if you parallel resonate that coil with capacitors - a parallel resonant LC circuit has theoretically infinite impedance so, if you only wish to set up a constant 1MHz field and this field, in the absence of the receive coil, is not supplying power to structural stuff around the antenna (such as anything conducting or partially conducting), the power into the transmit coil needs only to be sufficient to overcome the resistive losses in the coil.
In reality, a few mW will be needed to pump several amps of current round the coil. Designing a circuit that can deliver the sinewave impetus to the coil can be a little tricky but not impossible. Number of turns on the transmit coil ought not to be greater than 10 (call it experience) and probably more like 3 or 4 (tightly wound to increase the inductance thus making the tuning capacitor smaller and Q higher).
So if you have 10 amps flowing around 4 turns you have NI = 40. If radius is 1m, the flux density at a distance of 10m is: -
\$\dfrac{4\pi\times 10^{-7}\times 40}{2(101)^{1.5}}\$ = 25 nano teslas
To convert this to H-field divide by \$4\pi\times 10^{-7}\$ to get 0.02 amps/metre.
I did this quickly on my calculator so forgive if I got it wrong! To meet your requirement for 1.4142 uA/m should be fairly easily achieved.
Should you be interested, the receive coil can also be parallel tuned with a capacitor(s) to increase the voltage output by anything up to ~20 times (again, experience designing food and pharmaceutical metal detectors is my background).
