A cylindrical paramagnetic sample has radius $2.0\text{cm}$ and length $5.0\text{cm}$. It is placed in a uniform magnetic field of $B=0.75\text{T}$. The measured magnetization of the sample is $M=5.4\times 10^{-2}{\text{A⋅m}}^{-1}$.\n(a) Calculate the magnetic susceptibility $\chi$ of the material.\n(b) State whether the material is paramagnetic or diamagnetic.\n(c) Find the total magnetic moment $\mu$ of the sample.

Step-by-step solution

1. Compute the volume of the cylinder: $V=\pi r^{2}l=\pi (0.02\text{m}{)}^2(0.05\text{m})=6.28\times 10^{-5}{\text{m}}^3$.
2. Find the magnetic field intensity $H$ using $B={\mu }_0(H+M)$, so $H=\frac{B}{{\mu }_0}-M$.
3. Insert ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$: $\frac{B}{{\mu }_0}=\frac{0.75}{4\pi \times 10^{-7}}=5.96\times 10^5\text{A/m}$.
4. Subtract the magnetization: $H=5.96\times 10^5-5.4\times 10^{-2}\approx 5.96\times 10^5\text{A/m}$.
5. Calculate susceptibility: $\chi =\frac{M}{H}=\frac{5.4\times 10^{-2}}{5.96\times 10^5}=9.1\times 10^{-8}$.
6. Since $\chi$ is positive, the material is paramagnetic.
7. The magnetic moment of the sample is $\mu =MV=(5.4\times 10^{-2})(6.28\times 10^{-5})=3.4\times 10^{-6}{\text{A⋅m}}^2$.

Concept tested: Magnetization, susceptibility and magnetic moment of matter

Published 17 September 2026 · How this solution was prepared and checked