Question 20.7: Design/Materials Selection for a Ceramic Magnet Design a cub......
Design/Materials Selection for a Ceramic Magnet
Design a cubic ferrite magnet that has a total magnetic moment per cubic meter of 5.50 × 10^5 A/m.
We found in Example 20-6 that the magnetic moment per cubic meter for Fe_{3}O_{4} is 5.06 × 10^5 A/m. To obtain a higher saturation magnetization, we must replace Fe^{2+} ions with ions having more Bohr magnetons per atom. One such possibility (Table 20-7) is Mn^{2+}, which has five Bohr magnetons.
Assuming that the addition of Mn ions does not appreciably affect the size of the unit cell, we find from Example 20-6 that
V_{cell} = 5.864 × 10^{-22} cm³ = 5.864 3× 10^{-28} m³
Let x be the fraction of Mn^{2+} ions that have replaced the Fe^{2+} ions, which have now been reduced to 1 – x. Then, the total magnetic moment is
Total moment
= \frac{(8 \ \text{subcells}) \left[(x)(5 \ \text{magnetons}) \ + \ (1 \ – \ x)(4 \ \text{magnetons})\right](9.274 \ × \ 10^{-24} \ A ⋅ m²)}{5.864 \ × \ 10^{-28} \ m³}
= \frac{(8)(5x \ + \ 4 \ – \ 4x)(9.274 \ × \ 10^{-24} )}{5.864 \ × \ 10^{-28}}=5.50 \ × \ 10^{5}
x = 0.347
Therefore we need to replace 34.7 at% of the Fe^{2+} ions with Mn^{2+} ions to obtain the desired magnetization.
| Table 20-7 Magnetic moments for ions in the spinel structure | |||
| Ion | Bohr Magnetons | Ion | Bohr Magnetons |
| Fe^{3+} | 5 | Co^{2+} | 3 |
| Mn^{2+} | 5 | Ni^{2+} | 2 |
| Fe^{2+} | 4 | Cu^{2+} | 1 |
| Zn^{2+} | 0 | ||