Question #43870

A paramagnetic rod of length $L$, initial cross-sectional area $A_0$, and initial density $\rho_0$ at temperature $T_0$ is placed between two perfectly rigid, immovable walls at $x=0$ and $x=L$. The rod is free to expand laterally, and the Poisson's ratio of the material is $\nu = 0$. The coefficient of linear thermal expansion is $\alpha$, and Young's modulus is $Y$. The mass magnetic susceptibility of the rod obeys Curie's Law, $\chi_m = \frac{C}{T}$, where $C$ is a constant and $T$ is the absolute temperature. A non-uniform external magnetic field $B(x) = B_0 \frac{x}{L}$ is applied along the x-axis. The temperature of the rod is then uniformly increased to $T = T_0 + \Delta T$. Assuming the material is linearly elastic, $\chi_m \ll 1$, and the internal magnetic field is approximately the external field, find the volumetric magnetization $M$ of the rod at the end $x=L$.

Express your answer in terms of $M_0 = \frac{\rho_0 C B_0}{T \mu_0}$ and the dimensionless parameter $\beta = \frac{\rho_0 C B_0^2}{Y T \mu_0}$. (Assume $\alpha \Delta T \ll 1$ and $\beta \ll 1$).