A long solenoid has 1000 turns per metre and carries a current of 1 A. It has a soft iron core of µr = 1000. The core is heated beyond the Curie temperature, Tc. a. The H field in the solenoid is (nearly) unchanged but the B field decreases drastically. b. The H and B fields in the solenoid are nearly unchanged. c. The magnetisation in the core reverses direction. d. The magnetisation in the core diminishes by a factor of about 108.

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KollegeApply · Accepted Answer

a and d Explanation: The magnetic field intensity H = nl, where n = number of turns per metre of a solenoid and I = current and B = μ0μrI. Also, at normal temperature, a solenoid behaves as a ferromagnetic substance and at the temperature beyond the Curie temperature, it behaves as a paramagnetic substance. n = 1000 turns per metre, μr = 1000 H = nl = 1000 $\times$ 1 = 1000 Amp. So H is constant verifies the answer a B = μ0μr nl = (μ0nl)μr = Kμr (K = constant) So, B oo μ_r but there is a large decrease in the susceptibility of the core on heating it beyond critical temperature, hence magnetic field will decrease drastically. Now, for magnetisation in the core, when temperature of the iron core of a solenoid is raised beyond Curie temperature, then it behaves as a paramagnetic material, where and (chi_m)_(Fero) ≈ 10^3 (chi_m)_(Para) ≈ 10^-5 $\Rightarrow$ `((chi_m)_(Fero))/((chi_m)_(Para)) = 10^$\frac{3}{10}$^-5 = 10^8`

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