JEE Main 2025 — Magnetic Properties Of Matter Question with Solution

The percentage change in the magnetic field (B) when the space within a current-carrying solenoid is filled with magnesium is calculated as follows:

We start by using the formula for percentage change in the magnetic field:

$\%\text{ change in }B=\frac{B_{\text{new}}-B_{\text{old}}}{B_{\text{old}}}\times 100\%$

Substituting the expressions for the magnetic field, we have:

$=\frac{\mu ni-{\mu }_{0}ni}{{\mu }_{0}ni}\times 100\%=\frac{\mu -{\mu }_0}{{\mu }_0}\times 100\%$

Where:

$\mu$ is the permeability of the material inserted (here, magnesium).

${\mu }_0$ is the permeability of free space.

$n$ is the number of turns per unit length of the solenoid.

$i$ is the current through the solenoid.

Given that the relationship between the permeability of the medium ($\mu$) and relative permeability (${\mu }_r$) is:

$\mu ={\mu }_0{\mu }_r$

We can simplify this to:

$=\frac{{\mu }_0{\mu }_r-{\mu }_0}{{\mu }_0}\times 100\%$

Which reduces to:

$=({\mu }_r-1)\times 100\%$

The relative permeability ${\mu }_r$ is equal to $1+\chi$, where $\chi$ is the magnetic susceptibility. Thus, we have:

$=\chi \times 100\%$

When magnesium is used, with a magnetic susceptibility ${\chi }_{\text{Mg}}=1.2\times 10^{-5}$, the calculation becomes:

$=1.2\times 10^{-5}\times 100\%=1.2\times 10^{-3}\%$

Therefore, the percentage increase in the magnetic field when the solenoid is filled with magnesium is $1.2\times 10^{-3}\%$.