JEE Main 2023 — Magnetic Properties Of Matter Question with Solution

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 :

$$B=B_0+B_{material}$$

Magnetic susceptibility is related to the relative permeability ($${\mu }_r$$) of the material :

$${\mu }_r=1+{\chi }_m$$

The magnetic field inside the toroid can be calculated using the following formula :

$$B={\mu }_0{\mu }_{r}H$$

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.

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 :

$$B_0={\mu }_{0}H$$

After the material is placed inside the toroid, the magnetic field becomes :

$$B={\mu }_0{\mu }_{r}H={\mu }_0(1+{\chi }_m)H$$

The percentage increase in the magnetic field can be calculated as :

$$\frac{B-B_0}{B_0}\times 100\%=\frac{{\mu }_0(1+{\chi }_m)H-{\mu }_{0}H}{{\mu }_{0}H}\times 100\%$$

Substitute the value of $${\chi }_m=2\times 10^{-2}$$ :

$$\frac{B-B_0}{B_0}\times 100\%=\frac{(1+2\times 10^{-2})-1}{1}\times 100\%=2\%$$

Thus, the percentage increase in the value of the magnetic field inside the toroid is 2%.