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QPages 286-287: Q2: HC Verma: Concepts of Physics - Volume 2 Magnetic Properties of Matter Pages 286-287: Q2 Solution: A rod is inserted as the core in the current-carrying solenoid…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often substitute the magnetic flux density $B$ for the magnetic field intensity $H$, assuming the core material changes $H$; remember that $H$ is set only by the free current ($H=nI_{\text{free}}$ for a long solenoid) and is independent of the rod’s magnetization. Use the relation $\chi = M/H$ (with $M$ the magnetization) to find susceptibility, not $B/H$.

A rod is inserted as the core in the current-carrying solenoid of the previous problem. a) What is the magnetic intensity H at the centre? b) If the magnetization I of the core is found to be 0.12 A m−10.12 \text{ A m}^{-1}, find the susceptibility of the material of the rod. c) Is the material paramagnetic, diamagnetic or ferromagnetic?

Step-by-Step Verified Solution

**Answer:** (a) H=1500 A m−1H = 1500 \text{ A m}^{-1} (b) χ=8×10−5\chi = 8 \times 10^{-5} (c) Paramagnetic. **Step-by-Step Explanation:** (a) The magnetic intensity HH at the centre of a long solenoid depends only on the current and the number of turns per unit length, not on the core material. So H=1500 A m−1H = 1500 \text{ A m}^{-1}. (b) Given: I=0.12 A m−1I = 0.12 \text{ A m}^{-1}. Susceptibility χ=I/H=0.12/1500=8×10−5\chi = I/H = 0.12/1500 = 8 \times 10^{-5}. (c) Since χ\chi is small and positive (0<χ≪10 < \chi \ll 1), the material is paramagnetic.

All Questions in this Chapter

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