The value of aluminium susceptibility is $$2.2 \times 10^{-5}$$. The percentage increase in the magnetic field if space within a current carrying toroid is filled with aluminium is $$\frac{x}{10^4}$$. Then the value of $$x$$ is ___.

First recall the expression for the magnetic field in the core of a toroid. In an empty (air-filled) toroid the magnetic permeability is simply $$\mu_0$$, so the field is

$$B_0 = \mu_0\,nI,$$

where $$n = \dfrac{N}{\ell}$$ represents the number of turns per unit length and $$I$$ is the current.

When we fill the same space with a magnetic material of susceptibility $$\chi$$, its relative permeability becomes

$$\mu_r = 1 + \chi.$$

The actual permeability inside the material is therefore

$$\mu = \mu_0\,\mu_r = \mu_0(1+\chi).$$

Substituting this new permeability into the toroid formula, the magnetic field now becomes

$$B = \mu\,nI = \mu_0(1+\chi)\,nI.$$

The increase in the field strength is

$$\Delta B = B - B_0 = \mu_0(1+\chi)\,nI - \mu_0\,nI = \mu_0 nI\,\chi.$$

The fractional (or relative) increase is then

$$\frac{\Delta B}{B_0} = \frac{\mu_0 nI\,\chi}{\mu_0 nI} = \chi.$$

To turn this into a percentage increase, we multiply by 100 %:

$$\text{Percentage increase} = \chi \times 100\%.$$

For aluminium we are given

$$\chi = 2.2 \times 10^{-5}.$$

Hence

$$\text{Percentage increase} = 2.2 \times 10^{-5} \times 100 = 2.2 \times 10^{-3}.$$

According to the problem statement, this percentage is expressed as $$\dfrac{x}{10^{4}}$$. Setting the two expressions equal gives

$$\frac{x}{10^{4}} = 2.2 \times 10^{-3}.$$

Now we solve for $$x$$:

$$x = 2.2 \times 10^{-3} \times 10^{4} = 2.2 \times 10^{1} = 22.$$

So, the answer is $$22$$.