Electronic – Measuring change in inductance

Arguably, the easiest way is to just measure it:

      +--------+          
      |        | 
      |      [50R]
      |        |
      |       [C]
    [GEN]      |
      |        +------+
      |        |      |
      |       [L]  [SCOPE]
      |        |      |
      +--------+------+ 

With \$C\$ known, adjust the generator's output frequency for a peak across the scope, then determine the reactance of the capacitor with:

$$X_c=\frac1{2\pi fC}$$

Next, since the reactance of the capacitor and the inductor will be equal at resonance, (the frequency at the maximum amplitude of the peak on the scope) solve for the inductance with:

$$L=\frac{X_c}{2\pi f}$$

Both equations can be combined:

$$L=\frac{1}{(2\pi f)² C}$$

You may need to adjust the value of the \$R\$ in order to get a nice peak.

Place the known and unknow inductors in series and apply a known AC voltage across the pair of them. Best to use an o-scope for this but it's not beyond impossible to use the multimeter on AC measurement.

The sum of the voltages across each inductor will add up to the voltage applied across them both. If the two inductors are equal then the voltage across either will be the same. The smaller inductance will have a lower AC voltage.

Use the voltage ratio to determine the ratio of the inductors. For instance a 2H coil in series with a 1H coil will have twice the voltage across it compared to the 1H coil.