You could resonate it with a parallel or series capacitor and use a signal generator and o-scope for finding the resonant-frequency. You need to have a capacitor of at least 50 times it's likely self capacitance but, that can also be measured with a frequency generator and an o-scope. Then add a known capacitor (say 10nF) and you should see the resonant frequency drop at least ten if not 100 times. Use this formula: -
\$f_R = \dfrac{1}{2 \pi \sqrt{LC}}\$
I regularly build coils for transmitting power from fixed units to rotating electronics and the parallel resonance way is the most reliable for accuracy.
You should also note that depending on the material of the ferrite the inductance may change quite significantly with current passed through it - this is due to the onset of saturation but some ferrites are designed to be like this so, if possible try and run the test with an oscillator delivering enough voltage to impart the right amount of current into the coil.
1.
I made an inductor tester when I was designing large coils for an arc welder PFC stage. It consisted of a large capacitor=C1 that charges slowly through a resistor R1.
A large IGBT (shown as a switch, SW1) was connected in such a way to switch ON when a button was pressed. This IGBT would connect the capacitor to the unknown inductor.
A current transformer (AM2) was used to measure the current through the inductor and this current was plotted on an oscilloscope.
As we know, the equation for an inductor is V = L*dI/dt
Where L is the value of the inductor, V is the voltage across the inductor, and dI/dt is the rate of change of the current through the inductor. (Ignoring resistance.)
simulate this circuit – Schematic created using CircuitLab
In this tester the voltage is essentially constant -- the capacitor is large enough so that it does not appreciably discharge during the test.
This means that the product of L*dI/dt should be a constant. But as you know, the inductance will decrease when the current rises in an inductor.
This is observed on the oscilloscope as a linear ramp L*(dI/dt) when the test begins. As the inductance goes DOWN, the slope of the current (dI/dt) increases.
The point where the slope of the line goes non-linear is the point where the inductor begins to saturate.
You can measure the inductor value by picking two linear values and inserting it into the equation, L = V*dt/di.
2.
I believe you are defining core saturation. The core will definitely be saturated. It's not the voltage that causes the core to saturate. It's the level of current in the inductor, analogously the intensity of magnetic flux.
Best Answer
There are two quantities B and H, both often confusingly called 'magnetic field', so when we get down to discussing the detail, most people prefer to use the terms B-field and H-field for clarity. I like to think of H as magnetic induction or driving force, and B as magnetic flux density. One definition of B is that it's the change in the surface integral of B across the area of a linked conductor that generates a voltage in that conductor.
The relation between them is \$B=\mu_0\mu_rH\$, where \$\mu_0\$ is the permeability of free space, and \$\mu_r\$ is the relative permeability of the material with respect to a vacuum. \$\mu_r\$ has a value close to 1.000 for most ordinary materials, and up to many thousands for ferromagnetic materials.
The units of H are A/m, amps per metre. The H-field can be calculated over some path by counting the number of amps flowing through the area bounded by the path, divided by the length of the path. In the case of a straight wire, then for a circle centred on the wire with radius d, the path length is the circumference, 2*pi*d. The current is whatever is flowing through the wire, so the H-field is \$\frac{I}{2\pi d}\$. When we multiply by \$\mu_0\mu_r\$ to get to B-field, we have your first expression for B-field.
The geometry of the first situation was easy, even I can work out the length of a circumference of a circle, and the cylindrical symmetry kept it easy for relating the fields to each other.
The field in the centre of a loop is more tricky, as the effective path lengths have to be integrated over the range of paths that hug the wires out to paths that go off to infinity. The integrals are beyond me (mathematicians welcome to post a derivation in another answer), so I have to resort to that dreaded phrase 'it can be shown that' the effective path length is 0.5R, where R is the radius of the loop. The effective current is NI, where the loop has N turns, and so you get your second expression, for the B-field at the centre of the loop.
With a 'long' solenoid, the same sort of arguments and integration give your third expression.