Buck Converter - Ripple Current
Assuming you know the relationship between Vin and Vout , the buck is in steady-state and continuous conduction.

During the charging phase, the current can be compute from:

$$V_{L}=V_{in}-V_{out}=L\frac{\mathrm{d}i }{\mathrm{d} t}$$

From our sequence commutation waveform analysis(drawing waveform for both on and off cycle), we can say:

$$ V_{in}-V_{out}=L \frac{I_{Lmax}-I_{Lmin}}{t_{on}} = L \frac{\Delta I_{L}}{t_{on}}$$

knowing:

$$ \alpha V_{in}= V_{out} $$

$$ \alpha=\frac{ t_{on}}{T} $$

We can manipulate the equation and isolate the current ripple

$$ V_{in}- \alpha V_{in}= L \frac{\Delta I_{L}}{\alpha T}$$

$$ \Delta I_{L} = \frac{V_{in}\alpha (1- \alpha)}{F L} $$