The percentage change in the magnetic field (B) when the space within a current-carrying solenoid is filled with magnesium is calculated as follows:

We start by using the formula for percentage change in the magnetic field:

$ \% \text{ change in } B = \frac{B_{\text{new}} - B_{\text{old}}}{B_{\text{old}}} \times 100\% $

Substituting the expressions for the magnetic field, we have:

$ = \frac{\mu n i - \mu_0 n i}{\mu_0 n i} \times 100\% = \frac{\mu - \mu_0}{\mu_0} \times 100\% $

Where:

$\mu$ is the permeability of the material inserted (here, magnesium).

$\mu_0$ is the permeability of free space.

$n$ is the number of turns per unit length of the solenoid.

$i$ is the current through the solenoid.

Given that the relationship between the permeability of the medium ($\mu$) and relative permeability ($\mu_r$) is:

$ \mu = \mu_0 \mu_r $

We can simplify this to:

$ = \frac{\mu_0 \mu_r - \mu_0}{\mu_0} \times 100\% $

Which reduces to:

$ = (\mu_r - 1) \times 100\% $

The relative permeability $\mu_r$ is equal to $1 + \chi$, where $\chi$ is the magnetic susceptibility. Thus, we have:

$ = \chi \times 100\% $

When magnesium is used, with a magnetic susceptibility $\chi_{\text{Mg}} = 1.2 \times 10^{-5}$, the calculation becomes:

$ = 1.2 \times 10^{-5} \times 100\% = 1.2 \times 10^{-3}\% $

Therefore, the percentage increase in the magnetic field when the solenoid is filled with magnesium is $1.2 \times 10^{-3}\%$.