Chapter 16.13, Problem 66AAP

To determine

The saturation induction for nickel.

Answer to Problem 66AAP

The saturation induction for nickel is $0.65\text{T}$.

Explanation of Solution

Write the expression to calculate saturation magnetization for nickel $\left(M_s\right)$.

  $M_s=\left(\text{Atomic density}\right)N{\mu }_B$                                                                                  (I)

Here, magnetic moment is $N$ and Bohr magneton is ${\mu }_B$.

Write the expression to calculate atomic density for nickel.

  $\begin{array}{c}\text{Atomic density}=\frac{n}{V}\\ =\frac{n}{a^3}\end{array}$                                                                                            (II)

Here, number of atoms per unit cell is $n$, volume of the unit cell is $V$ and lattice constant is a.

Write the expression to calculate saturation induction for nickel $\left(B_s\right)$.

  $B_s\approx {\mu }_0{M}_s$                                                                                                (III)

Here, permeability constant is ${\mu }_0$.

Conclusion:

The value for lattice constant $\left(a\right)$ for nickel is $0.352\text{nm}$.

Substitute $4\text{atoms}/\text{unit cell}$ for n and $0.352\text{nm}$ for $a$ in Equation (II).

 $\begin{array}{c}\text{Atomic density}=\frac{4\text{atoms}/\text{unit cell}}{{\left(0.352\text{nm}\right)}^3}\\ =\frac{4\text{atoms}/\text{unit cell}}{{\left(0.352\text{nm}\times \frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)}^3}\\ =9.171\times {10}^{28}\text{atoms}/{\text{m}}^3\end{array}$

The value of Bohr magneton $\left({\mu }_B\right)$ is taken to be $9.27\times {10}^{-24}\text{A}\cdot {\text{m}}^2$.

Substitute $9.171\times {10}^{28}\text{atoms}/{\text{m}}^3$ for $\text{Atomic density}$, $0.604\text{Bohr magnetons}/\text{atom}$ for $N$ and $9.27\times {10}^{-24}\text{A}\cdot {\text{m}}^2$ for ${\mu }_B$ in Equation (I).

 $\begin{array}{c}{M}_s=\left(9.171\times {10}^{28}\text{atoms}/{\text{m}}^3\right)\left(0.604\text{Bohr magnetons}/\text{atom}\right)\left(9.27\times {10}^{-24}\text{A}\cdot {\text{m}}^2\right)\\ =5.135\times {10}^5\text{A}/\text{m}\end{array}$

The value of permeability constant $\left({\mu }_0\right)$ is taken to be $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$.

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $5.135\times {10}^5\text{A}/\text{m}$ for $M_s$ in Equation (III).

 $\begin{array}{c}{B}_s=\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.135\times {10}^5\text{A}/\text{m}\right)\\ =0.65\text{T}\end{array}$

Thus, the saturation induction for nickel is $0.65\text{T}$.