User:Wikieditoroftoday/Solutions To Physics Textbooks/Physical Properties of Crystals (9780198511656)/Chapter 3

Chapter 3 - Paramagnetic and Diamagnetic Susceptibility

Exercise 3.1

p. 60

Problem: A magnetic field of strength $H_{1}$ is applied to a crystal along the $Ox_{1}$ axis and then a magnetic field of strength $H_{2}$ is added to the same crystal along the $Ox_{2}$ axis. Calculate the work (using (7) and (12)). Repeat the same calculation with $H_{2}$ applied first and $H_{1}$ applied later. Show that the work done is only equal for $\mu _{12}=\mu _{21}$.

Solution: With $\mathrm {d} \Psi =vH_{i}\mathrm {d} B_{i}$ (12) on p.59 and $B_{i}=\mu _{ij}H_{j}$ (7) on p.55 we get $\mathrm {d} \Psi =v\mu _{ij}H_{i}\mathrm {d} H_{j}$ (13) on p.59.

With $H_{1}$ applied first the result is ${\frac {1}{v}}\mathrm {d} \Psi =\mu _{11}H_{1}\mathrm {d} H_{1}$. Integration gives ${\frac {1}{v}}\Psi ={\frac {1}{2}}\mu _{11}H_{1}^{2}$. Applying the addition field of strength $H_{2}$ produces additional work ${\frac {1}{v}}\mathrm {d} \Psi =\mu _{12}H_{1}\mathrm {d} H_{2}+\mu _{22}H_{2}\mathrm {d} H_{2}$ and integration gives us ${\frac {1}{v}}\Psi =\mu _{12}H_{1}H_{2}+{\frac {1}{2}}\mu _{22}H_{2}^{2}$. The work done is ${\frac {1}{2}}\mu _{11}H_{1}^{2}+\mu _{12}H_{1}H_{2}+{\frac {1}{2}}\mu _{22}H_{2}^{2}$.

With $H_{2}$ applied first the result changes. First we get ${\frac {1}{v}}\mathrm {d} \Psi =\mu _{22}H_{2}\mathrm {d} H_{2}$ and after integration ${\frac {1}{v}}\Psi ={\frac {1}{2}}\mu _{22}H_{2}^{2}$. Applying $H_{1}$ results in ${\frac {1}{v}}\mathrm {d} \Psi =\mu _{11}H_{1}\mathrm {d} H_{1}+\mu _{21}H_{2}\mathrm {d} H_{1}$. The work done is thus ${\frac {1}{2}}\mu _{11}H_{1}^{2}+\mu _{21}H_{2}H_{1}+{\frac {1}{2}}\mu _{22}H_{2}^{2}$.

Both calculations are only equal if $\mu _{12}=\mu _{21}$.

Exercise 3.2

p. 64

Exercise 3.3

p. 64

Exercise 3.4

p. 65