What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?

As the frequency is cut in half, the current will double and the inductive reactance will be cut in half.

As we know inductance current formula is $\mathrm{i}=\frac{{\mathrm{V}}_0}{{\mathrm{X}}_{\mathrm{L}}}$

Where $\mathrm{\omega }$ is the angular frequency in the equation above, the inductive reactance in a purely inductive circuit is represented as ${\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L}$. L denotes inductance, f denotes frequency, and ${\mathrm{X}}_{\mathrm{L}}$ denotes inductance reactance.

$\mathrm{i}=\frac{{\mathrm{V}}_0}{{\mathrm{X}}_{\mathrm{L}}}$

So,

$f=f/2$

$\mathrm{i}=2\left(\frac{{\mathrm{V}}_0}{2\mathrm{\pi fL}}\right)$

Therefore,  the inductive reactance will be halved and current will be doubled.