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QPages 286-287: Q8: HC Verma: Concepts of Physics - Volume 2 Magnetic Properties of Matter Pages 286-287: Q8 Solution: Assume that each iron atom has a permanent magnetic moment…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse magnetization ($I$ or $M$) with the magnetic field ($B$) or forget to multiply by $mu_0$ when converting from magnetization to field. Remember that $I = n m$ gives the dipole moment per unit volume, while the internal field in a long solenoid-like cylinder is $B = mu_0 I$.

Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons 1 Bohr magneton equals 9.27×10−24 A m29.27 \times 10^{-24} \text{ A m}^2). The density of atoms in iron is 8.52×1028 atoms m−38.52 \times 10^{28} \text{ atoms m}^{-3}. a) Find the maximum magnetisation I in a long cylinder of iron b) Find the maximum magnetic field B on the axis inside the cylinder.

Step-by-Step Verified Solution

**Answer:** (a) Imax≈1.58×106 A m−1I_{max} \approx 1.58 \times 10^6 \text{ A m}^{-1} (b) Bmax≈1.99 TB_{max} \approx 1.99 \text{ T}. **Step-BY-Step Explanation:** Given: n=8.52×1028 atoms/m3n = 8.52 \times 10^{28} \text{ atoms/m}^3, magnetic moment per atom =2×9.27×10−24 A m2= 2 \times 9.27 \times 10^{-24} \text{ A m}^2. (a) Maximum magnetisation (all moments aligned): I=n×m=8.52×1028×2×9.27×10−24≈1.58×106 A/mI = n \times m = 8.52 \times 10^{28} \times 2 \times 9.27 \times 10^{-24} \approx 1.58 \times 10^6 \text{ A/m}. (b) Maximum field: B=μ0I=4π×10−7×1.58×106≈1.99 TB = \mu_0 I = 4\pi \times 10^{-7} \times 1.58 \times 10^6 \approx 1.99 \text{ T}.

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