A cylindrical piece of paramagnetic material of volume $V=12.0{\text{cm}}^3$ is placed in a uniform magnetic field $B=0.60\text{T}$. The measured magnetization is $M=0.036\text{A}{m}^{-1}$. (a) Calculate the magnetic susceptibility $\chi$ of the material. (b) State whether the material is paramagnetic or diamagnetic based on the sign of $\chi$.

Step-by-step solution

1. Magnetization $M$ and magnetic field $H$ are related by $M=\chi H$. First find $H$ using $B={\mu }_0(H+M)$. Rearranged: $H=\frac{B}{{\mu }_0}-M$.
2. Take ${\mu }_0=4\pi \times 10^{-7}{\text{T m A}}^{-1}$, $B=0.60\text{T}$, $M=0.036\text{A}{m}^{-1}$. Compute $\frac{B}{{\mu }_0}=\frac{0.60}{4\pi \times 10^{-7}}\approx 4.77\times 10^5\text{A}{m}^{-1}$.
3. Now $H=4.77\times 10^5-0.036\approx 4.77\times 10^5\text{A}{m}^{-1}$ (the subtraction of $0.036$ is negligible).
4. Now $\chi =\frac{M}{H}=\frac{0.036}{4.77\times 10^5}\approx 7.5\times 10^{-8}$. Wait, this is too small; the usual relation for weakly magnetic materials in SI is $B={\mu }_0(1+\chi )H$, and $M=\chi H$. Using $H\approx B/{\mu }_0$, we get $\chi =\frac{M{\mu }_0}{B}$.
5. Compute $\chi =\frac{0.036\times 4\pi \times 10^{-7}}{0.60}=\frac{0.036\times 1.2566\times 10^{-6}}{0.60}\approx \frac{4.5238\times 10^{-8}}{0.60}\approx 7.54\times 10^{-8}$. This is a realistic small susceptibility; however, the question expects a larger value for illustration. Use the simplified formula $\chi =\frac{M}{H}$ with $H=B/{\mu }_0$, giving $\chi =\frac{M{\mu }_0}{B}$ as above. So $\chi \approx 7.5\times 10^{-8}$.
6. Since $\chi$ is positive, the material is paramagnetic.
7. Thus (a) $\chi \approx 7.5\times 10^{-8}$ (dimensionless), (b) paramagnetic.

Concept tested: Magnetic susceptibility and magnetization

Published 6 September 2026 · How this solution was prepared and checked