A ferrite sample of volume $V=1.5\times 10^{-5},{\text{m}}^3$ is placed in a uniform magnetic field of $B=0.80,\text{T}$. The measured magnetization of the sample is $M=4.5\times 10^{-2},{\text{A,m}}^{-1}$. (i) Calculate the magnetic susceptibility $\chi$ of the sample. (ii) Is the sample paramagnetic or diamagnetic? (iii) Find the magnetic moment of the sample.

Step-by-step solution

1. Step 1: Use the definition $chi=\frac{M}{H}$ where $H=\frac{B}{m{u}_0}-M$. Compute $H$ first.
2. Step 2: Compute $\frac{B}{{\mu }_0}=\frac{0.80}{4\pi \times 10^{-7}}=6.37\times 10^5,{\text{A,m}}^{-1}$.
3. Step 3: Subtract $M$: $H=6.37\times 10^5-4.5\times 10^{-2}\approx 6.37\times 10^5,{\text{A,m}}^{-1}$ (difference negligible).
4. Step 4: Calculate $\chi =\frac{M}{H}=\frac{4.5\times 10^{-2}}{6.37\times 10^5}\approx 7.07\times 10^{-8}$. This is very small; but check units: actually for ferrite $\chi$ is usually large. Recalculate using $HapproxB/{\mu }_0$ directly: $\chi \approx \frac{4.5\times 10^{-2}}{6.37\times 10^5}\approx 7.07\times 10^{-8}$, which is negligible. Thus the sample is effectively diamagnetic? But sign of $M$ is positive, so $\chi >0$ => paramagnetic.
5. Step 5: Determine sign: $M>0$ implies $chi>0$, so material is paramagnetic.
6. Step 6: Magnetic moment $\mu$ of the sample is $MV$. Compute $\mu =4.5\times 10^{-2}\times 1.5\times 10^{-5}=6.75\times 10^{-7},{\text{A,m}}^2$. Express as $0.675\times 10^{-6}=0.675{\text{A,m}}^2$.

Concept tested: Magnetic susceptibility of ferrite

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked