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

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 $$\mathrm{d}\Psi = v H_i \mathrm{d} B_i$$ (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