You can't tell by visual inspection, that's for sure because some of them are lacquered/painted and even those that aren't all tend to look dark-grey. What you are asking is really tricky to fathom because there are so many characteristics that look the same between two ferrites at one frequency but are vastly different at another. If you are still interested I'll try and say what I'd do (what I'd really do is throw all my unboxed/unmarked ferrites in the trash and buy some more).
I'd consider winding (say) 5 equally spaced turns and putting the coil in a circuit to see what its inductance was - maybe a colpitts oscillator with a few caps that can be switched in and out. Maybe even make a band-pass filter from it and see where it resonates if you have a signal generator.
First type of result this will tell you is the inductance of the wound core. Then using the squared relationship between turns and inductance you can deduce its "effective permeability". This should enable you to narrow down the type of core to a range of possibilities.
You need to be be avoiding "test frequencies" significantly above 100kHz and preferably more like 10kHz - this is to reduce parasitic capacitance giving you errors.
OK so far, you might have determined the approximate "effective permeability" of the core BUT there are plenty of suppliers toting vastly different materials that you'd have to read through to try and identify the part so I'd next consider seeing how the indctance varied with temperature.
You don't need to test over a vast range, maybe just 25ºC to 50ºC would give you a decent shot at trying to uncover the ferrite. Use the oscillator/filter idea mentioned earlier and a controlled temperature - almost certainly the inductance will rise with temperature although there are a small percentage that will stay stable or fall but this will give you another tell-tale characteristic of the ferrite.
So now you have effective permeability and some idea what its temperature characteristic looks like. Scanning through various supplier's websites might narrow down the ferrite to maybe five or ten types.
It's going to be a long process this way and you may never uncover what it is that is sitting in your junk box. I suppose if your effective permeability is low it's likely to be either very temperature stable (i.e. good for filters up to (say) 1MHz) or it could have very low losses up to over 50MHz. The temperature test that indicated hardly any change in inductance across 25ºC might tell you its a material like Ferroxcube's 3D3: -
Also shown is 3C90 for comparison. 3D3 has a flat curve of inductance/permeability against temperature; probably changing something like 5% in a 25ºC change around ambient. 3C90 probably changes about 20%. It also has a much higher permeabilty. I'd recognize these two ferrites from their characteristics!
I think I've definitely convinced myself to throw all unrecognizable ferrites in the bin.
Bottom line - if you have a target circuit try it.
EDIT Also, here's is a question/answer on EE stack exchange that might also be useful or provoke some other ideas.
This is a much simplified answer: -
There is this thing called a transformer - it has two coils that are closely magnetically connected. The two coils are called primary and secondary. If the primary has 1000 turns and the secondary has 100 turns it will step-down an AC voltage of 120V to 12V. There will be a slight power loss but usually transformers are better than 90% efficient. If we assumed near 100 % efficiency it follows that if we step down the voltage by ten times we also can step-up the current by ten times so, simplistically: -
\$V_{in} \cdot I_{in} = V_{out} \cdot I_{out}\$
The transformer is at the heart of an off-line switching power supply but instead of the transformer being fed with raw AC at 60Hz or 50Hz it is fed at a much higher frequency and this means a much smaller transformer can be used.
So the incoming 110V AC is rectified and smoothed (not to 110V DC) but to a voltage more like 154 volts DC. This feeds a power oscillator (that might run at 100kHz) and this drives the transformer primary.
On the secondary (the low voltage side) there is another rectifier and more smoothing capacitors. There may also be a feedback system that keeps the output closely regulated. Here's a typical block diagram: -
(source: philpem.me.uk)
So I should be able to get by with a smaller gauge wire going to the
input (at high voltage) of my switching power supply while having to
use a heavier gauge wire for the output (at lower voltage).
Correct!
Best Answer
P(t) = u(t) x i(t) is the total loss due to copper resistance, hysteresis and eddy-currents. The hysteresis loss must be calculating using the magnetization curve for the material, the current, number of turns and the material dimensions.
The magnetization curve required would be an AC magnetization curve or hysteresis loop. I believe the hysteresis loop for the specific operating frequency would be required. The material manufacturer would probably provide curves showing core loss per kilogram of material as a function of peak flux density at specific frequencies.
A core manufacturer would probably provide loss information for specific cores.
It may be very difficult to find information that separates hysteresis from eddy-current losses.