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

We need to find the percentage increase in the magnetic field inside a toroid when the free space is filled with a material of magnetic susceptibility $$\chi = 2 \times 10^{-2}$$.

For a toroid with $$n$$ turns per unit length carrying current $$I$$, the magnetic field in free space is $$B_0 = \mu_0 n I$$.

When the interior is filled with a magnetic material, the relative permeability becomes $$\mu_r = 1 + \chi$$, so the total magnetic field is:

$$B = \mu_0(1 + \chi) n I = B_0(1 + \chi)$$

(This is because the magnetizing field $$H = nI$$ remains the same, being determined by the free current, and $$B = \mu H = \mu_0(1 + \chi)H$$.)

Now, the percentage increase in the magnetic field is:

$$\frac{\Delta B}{B_0} \times 100 = \frac{B_0 \chi}{B_0} \times 100 = \chi \times 100 = 2 \times 10^{-2} \times 100 = 2\%$$

Hence, the correct answer is Option 3.