Magnetic Effects of Current and Magnetism Part 2

Given, number of turns, $n=1000 \mathrm{m}^{-1}$ Electric current, i = 0.75 A Relative permeability of 1st material $\mu_{1}=500$ Relative permeability of 2nd material $\mu_{2}=750$ we know that $\mu_{r}=1+\chi \Rightarrow \chi$ $=\mu_{r}-1$ For 1st and 2nd material, $\chi_{1}=\chi_{1}-1=500-1=499$ and $\chi_{2}=\mu_{2}-1=750-1=749$ Now, magnetic intensity, M = ni Let $M_{1}$ and $M_{2}$ be the magnetisation and $m_{1}$ and $m_{2}$ be the magnetic moments of material 1st and 2nd. Using, $\frac{I_{1}}{M}=\chi_{1}$ ...(i) and $\frac{I_{2}}{M}=\chi_{2}$ ...(ii) Dividing Eqs. (i) and (ii), $\frac{I_{1}}{I_{2}}=\frac{\chi_{1}}{\chi_{2}}$ ...(iii) We also know that, i.e $I=\frac{m}{V}$ $\frac{I_{1}}{I_{2}}=\frac{m_{1}}{m_{2}}$ ...(iv) From Eqs. (iii) and (iv), we get $\frac{m_{1}}{m_{2}}=\frac{\chi_{1}}{\chi_{2}}=\frac{499}{749}$ Hence, fraction change in magnetic moment, $\frac{\Delta m}{m_{1}}=\frac{m_{2}-m_{1}}{m_{1}}=\left(\frac{m_{2}}{m_{1}}-1\right)$ $=\left(\frac{749}{499}-1\right)=\frac{250}{499}$ Thus, value of χ is 250.