The free space inside a current carrying toroid is filled with a material of susceptibility $2 \times 10^{-2}$. The percentage increase in the value of magnetic field inside the toroid will be

The magnetic susceptibility ($\chi_m$) of a material is the measure of how much the material becomes magnetized in response to an external magnetic field. When the material is placed in a magnetic field, the net magnetic field ($B$) inside the material is the sum of the external magnetic field ($B_0$) and the field produced by the material itself ($B_{material}$). This relationship can be expressed as : <br/><br/> $B = B_0 + B_{material}$ <br/><br/> Magnetic susceptibility is related to the relative permeability ($\mu_r$) of the material : <br/><br/> $\mu_r = 1 + \chi_m$ <br/><br/> The magnetic field inside the toroid can be calculated using the following formula : <br/><br/> $B = \mu_0 \mu_r H$ <br/><br/> where $\mu_0$ is the permeability of free space, $\mu_r$ is the relative permeability of the material, and $H$ is the magnetic field strength. <br/><br/> Now, let's consider the percentage increase in the magnetic field when the material is placed inside the toroid. The initial magnetic field ($B_0$) is given by : <br/><br/> $B_0 = \mu_0 H$ <br/><br/> After the material is placed inside the toroid, the magnetic field becomes : <br/><br/> $B = \mu_0 \mu_r H = \mu_0 (1 + \chi_m) H$ <br/><br/> The percentage increase in the magnetic field can be calculated as : <br/><br/> $$\frac{B - B_0}{B_0} \times 100\% = \frac{\mu_0 (1 + \chi_m) H - \mu_0 H}{\mu_0 H} \times 100\%$$ <br/><br/> Substitute the value of $\chi_m = 2 \times 10^{-2}$ : <br/><br/> $$\frac{B - B_0}{B_0} \times 100\% = \frac{(1 + 2 \times 10^{-2}) - 1}{1} \times 100\% = 2\%$$ <br/><br/> Thus, the percentage increase in the value of the magnetic field inside the toroid is 2%.