1. A magnetic what?
If you mean an output transformer, to provide a ratio and possibly isolation as well, then you must know the compliance (voltage) range of that AC current source. You're also worsening the current-sourcing ability (i.e., the high impedance being made lower) by connecting a transformer's magnetizing impedance in parallel with it. (If this impedance is largely inductive, and the frequency is fixed, you can resonate it out with a capacitor.)
If the output compliance range is ~10mV and the current is ~100mA, then the load impedance is around 0.1 ohms, and a generous ratio would indeed interface more closely with ordinary circuitry (voltages in the 5-15V range). You then need to decide how you will maintain the output impedance, whether it really needs to be a current source at all, or if it's adequate to measure the current, or if you need a more optimized transformer to do everything.
Note that, you might measure the current with a current transformer, which itself is applied with a very similar circuit. However, its load impedance is usually much smaller than its magnetizing impedance, so the latter can be ignored, without needing a special design.
The frequency range of a ferrite material relates to the application, its saturation flux density, and losses. Losses in turn are more-or-less related to relative permeability.
Power materials are concerned with power density, or cost. They want more flux density with lower losses, at a given frequency. The permeability tends to be modest (500-3k), and so the magnetizing impedance is also modest.
It sounds like what you are after, is a pulse transformer. A transformer that more closely resembles an ideal transformer with infinite magnetizing impedance. For this, you choose a high permeability. Typical applications are pulse transformers and common mode chokes. The frequency range tends to be low (mu' rolling off around 100kHz or so), but this does not impact the losses very much, because the magnetizing impedance is so high to begin with -- if it becomes all resistive at some frequency, it's still a high resistance.
In practice, the magnetizing impedance goes up with frequency (inductive at low frequencies), slows down at higher frequencies (20-200kHz), then flattens out at still-higher frequencies (1-10MHz). It may fall down in the 100MHz range. That's only visible on single-turn ferrite beads; more likely, the self-capacitance of your winding will take over in the MHz range. High-frequency characteristics of pulse transformers are almost exclusively due to the winding design, with little or no impact from the core.
A pulse transformer needs to handle some amount of voltage, at some frequency or pulse width, before saturating; this is usually specified on datasheets for finished parts. For new design, use:
N = V_pk / (4 A_e B_pk F_sw)
N = turns
V_pk = peak square wave voltage (bipolar, no DC)
A_e = effective core area (see core datasheet)
B_pk = desired peak flux density (typically 0.2-0.3 T (just below saturation) at low frequencies; derate at high frequencies due to core losses)
F_sw = switching frequency
For unipolar waves (e.g., flyback converter), change the 4 to a 2.
For (bipolar) sine waves, change V_pk to V_rms, and change the 4 to 4.44.
Or for general waves, take the peak flux: the integral of voltage with respect to time. For square pulses, this is width * height. Divide that by A_e to get flux density-turns, and divide by B_pk to get turns.
A useful lesson here: dimensional analysis works and none of these magnetic relations have constants you need to be weary of. (Well, there's the 4 in the above equation, but that's from the waveform, which is obviously a special case.) Flux density B is units of volt-seconds per square meter; multiply it by an area and you get a flux, which is the area under a [voltage] curve. If you know the waveform and its timing, you can find the voltage.
I should cover permeability, since I mentioned that above. That's used in the inductance formula:
A_L = mu_0 mu_r A_e / l_e
Inductance is just the ratio of flux to current. B and H are the flux density and magnetization, the field versions of flux and current. There's always the duality between applied flux or applied current, and mu is the conversion factor between them: B = mu H. To get in-circuit inductance, we need to know the winding, and the core dimensions, to convert from fields to in-circuit parameters. So the above equation uses mu, area and length to do that.
(You of course need L = N^2 A_L to get actual in-circuit inductance.)
Note that air gap effectively reduces mu_r, or alternately, you can think of the core as being an air gap of l_e / mu_r. In that case, adding an air gap yields the apparent result:
A_L = mu_0 A_e / (l_e / mu_r + l_g)
Tim
P.S. Meh, I should probably format these formulas with MathJax...