The percentage increase in magnetic field (B) when space within a current carrying solenoid is filled with magnesium (magnetic susceptibility $\chi_{\mathrm{Mg}}=1.2 \times 10^{-5}$ ) is :

<p>The percentage change in the magnetic field (B) when the space within a current-carrying solenoid is filled with magnesium is calculated as follows:</p> <p>We start by using the formula for percentage change in the magnetic field:</p> <p>$ \% \text{ change in } B = \frac{B_{\text{new}} - B_{\text{old}}}{B_{\text{old}}} \times 100\% $</p> <p>Substituting the expressions for the magnetic field, we have:</p> <p>$ = \frac{\mu n i - \mu_0 n i}{\mu_0 n i} \times 100\% = \frac{\mu - \mu_0}{\mu_0} \times 100\% $</p> <p>Where:</p> <p><p>$\mu$ is the permeability of the material inserted (here, magnesium).</p></p> <p><p>$\mu_0$ is the permeability of free space.</p></p> <p><p>$n$ is the number of turns per unit length of the solenoid.</p></p> <p><p>$i$ is the current through the solenoid.</p></p> <p>Given that the relationship between the permeability of the medium ($\mu$) and relative permeability ($\mu_r$) is:</p> <p>$ \mu = \mu_0 \mu_r $</p> <p>We can simplify this to:</p> <p>$ = \frac{\mu_0 \mu_r - \mu_0}{\mu_0} \times 100\% $</p> <p>Which reduces to:</p> <p>$ = (\mu_r - 1) \times 100\% $</p> <p>The relative permeability $\mu_r$ is equal to $1 + \chi$, where $\chi$ is the magnetic susceptibility. Thus, we have:</p> <p>$ = \chi \times 100\% $</p> <p>When magnesium is used, with a magnetic susceptibility $\chi_{\text{Mg}} = 1.2 \times 10^{-5}$, the calculation becomes:</p> <p>$ = 1.2 \times 10^{-5} \times 100\% = 1.2 \times 10^{-3}\% $</p> <p>Therefore, the percentage increase in the magnetic field when the solenoid is filled with magnesium is $1.2 \times 10^{-3}\%$.</p>