125 nm 125 x 10 10 m Therefore inasmuch as the ca ratio for Co is 1623 then V C

# 125 nm 125 x 10 10 m therefore inasmuch as the ca

• Notes
• 554
• 92% (12) 11 out of 12 people found this document helpful

This preview shows page 539 - 542 out of 554 pages.

Subscribe to view the full document.

Subscribe to view the full document.

Unformatted text preview: 125 nm (1.25 x 10-10 m). Therefore, inasmuch as the c/a ratio for Co is 1.623, then V C = 1.623 ( ) 12 3 ( ) 1.25 x 10 − 10 m ( ) 3 = 6.59 x 10-29 m 3 And, now solving for n B yields n B = 1.3 x 10 6 A/m ( ) 6.59 x 10 − 29 m 3 /unit cell ( ) 9.27 x 10 − 24 A - m 2 Bohr magneton = 9.24 Bohr magneton unit cell Inasmuch as there are 1.72 and 0.60 Bohr magnetons for each of Co and Ni, and, for HCP, there are 6 equivalent atoms per unit cell, and if we represent the fraction of Ni atoms by x , then 539 n B = 9.24 Bohr magnetons/unit cell = 0.6 Bohr magnetons Ni atom ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 6x Ni atoms unit cell ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + 1.72 Bohr magnetons Co atom ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (6)(1 − x) Co atoms unit cell ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ And solving for x , the fraction of Ni atoms , x = 0.161, of 16.1 at% Ni. In order to convert this composition to weight percent, we employ Equation (4.7) as C Ni = C Ni ' A Ni C Ni ' A Ni + C Co ' A Co × 100 = (16.1 at%)(58.69 g/mol) (16.1 at%)(58.69 g/mol) + (83.9 at%)(58.93 g/mol) × 100 = 16.0 wt% 20.D2 This problem asks that we design a cubic mixed-ferrite magnetic material that has a saturation magnetization of 4.25 x 10 5 A/m. According to Example Problem 20.2 the saturation magnetization for Fe 3 O 4 is 5.0 x 10 5 A/m. In order to decrease the magnitude of M s it is necessary to replace some fraction of the Fe 2+ with a divalent metal ion that has a smaller magnetic moment. From Table 20.4 it may be noted that Co 2+ , Ni 2+ , and Cu 2+ , with 3, 2, and 1 Bohr magnetons per ion, respectively, have fewer than the 4 Bohr magnetons/Fe 2+ ion. Let us first consider Ni 2+ (with 2 Bohr magnetons per ion) and employ Equation (20.11) to compute the number of Bohr magnetons per unit cell ( n B ), assuming that the Ni 2+ addition does not change the unit cell edge length (0.839 nm, Example Problem 20.2). Thus, n B = M s a 3 µ B = 4.25 x 10 5 A /m ( ) 0.839 x 10 − 9 m ( ) 3 /unit cell 9.27 x 10 − 24 A - m 2 /Bohr magneton = 27.08 Bohr magnetons/unit cell 540 If we let x represent the fraction of Ni 2+ that have substituted for Fe 2+ , then the remaining unsubstituted Fe 2+ fraction is equal to 1 - x . Furthermore, inasmuch as there are 8 divalent ions per unit cell, we may write the following expression: n B = 8 2x + 4(1 − x) [ ] = 27.08 which leads to x = 0.308. Thus, if 30.8 at% of the Fe 2+ in Fe 3 O 4 are replaced with Ni 2+ , the saturation magnetization will be decreased to 4.25 x 10 5 A/m. Upon going through this same procedure for Co and Cu, we find that x Co = 0.615 (or 61.5 at%) and x Cu = 0.205 (20.5 at%) will yield the 4.25 x 10 5 A/m saturation magnetization. 20.D3 (a) The mechanism by which the VCR head records and plays back audio/video signals is essentially the same as the manner by which the head on a computer storage device reads and writes, as described in Section 20.10....
View Full Document