Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons (1 Bohr magneton equals $9.27$ $\times 10^{-24} \mathrm{~A} \mathrm{~m}^{2}$ ). The density of atoms in iron is $8.52 \times 10^{28}$ atoms $\mathrm{m}^{-3}$.
(a) Find the maximum magnetisation $/$ in a long cylinder of iron
(b) Find the maximum magnetic field $B$ on the axis inside the cylinder.
Given that in problem, No of atoms per unit volume is $f=8.52 \times 10^{28}$ atoms $/ \mathrm{m}^{3}$ Magnetisation per atom is $M=2 \times 9.27 \times 10^{-}$ ${ }^{24} \mathrm{~A}-\mathrm{m}^{2}$ (a) Intensity of magnetisation, $I=\mathrm{M}, \mathrm{V} \Rightarrow l=2 \times 9.27 \times 10^{-24} \times 8.52 \times 10^{28} \Rightarrow 1=1.58 \times 10^{6} \mathrm{~A} / \mathrm{m}$.
(b) For maximum magnetisation, the magnetising field will be equal to the intensity of magnetisation. So, $I=H$ Magnetic field (B) will be, $B=4 \pi$ ता $\times 10^{-7} \times 1.58 \times 10^{6} \Rightarrow B \approx 19.8 \times 10^{-1}=2.0 \mathrm{~T}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.