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Question #84060

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The Question

A long isotropic ferromagnetic rod of intrinsic Young's modulus YY and Poisson's ratio ν\nu is placed in a uniform external magnetic field H0H_0 parallel to its axis. Due to the magnetostriction effect, the magnetic susceptibility χ\chi of the material depends on the volumetric strain θ=ΔVV0\theta = \frac{\Delta V}{V_0} as χ(θ)=χ0(1+λθ2)\chi(\theta) = \chi_0 (1 + \lambda \theta^2), where χ0\chi_0 and λ\lambda are positive dimensionless constants. The system is maintained at a constant temperature. Which of the following expressions correctly represents the effective Bulk Modulus KeffK_{eff} of the rod for small volumetric strains, and the critical external magnetic field HcH_c at which the rod becomes mechanically unstable?

Options

  • A.

    Keff=Y3(1−2ν)−μ0χ0λH02;Hc=Y3μ0χ0λ(1−2ν)K_{eff} = \frac{Y}{3(1-2\nu)} - \mu_0 \chi_0 \lambda H_0^2 \quad ; \quad H_c = \sqrt{\frac{Y}{3\mu_0 \chi_0 \lambda (1-2\nu)}}

  • B.

    Keff=Y3(1−2ν)+μ0χ0λH02;No instability occursK_{eff} = \frac{Y}{3(1-2\nu)} + \mu_0 \chi_0 \lambda H_0^2 \quad ; \quad \text{No instability occurs}

  • C.

    Keff=Y1−2ν−μ0χ0λH02;Hc=Yμ0χ0λ(1−2ν)K_{eff} = \frac{Y}{1-2\nu} - \mu_0 \chi_0 \lambda H_0^2 \quad ; \quad H_c = \sqrt{\frac{Y}{\mu_0 \chi_0 \lambda (1-2\nu)}}

  • D.

    Keff=Y3(1−2ν)−12μ0χ0λH02;Hc=2Y3μ0χ0λ(1−2ν)K_{eff} = \frac{Y}{3(1-2\nu)} - \frac{1}{2}\mu_0 \chi_0 \lambda H_0^2 \quad ; \quad H_c = \sqrt{\frac{2Y}{3\mu_0 \chi_0 \lambda (1-2\nu)}}

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How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Poisson's Ratio Limits|Magnetic Permeability and Susceptibility

Why This Question Is Tricky

Examiner Levers

Visualization Stress

Requires mental rotation or 3D mapping (Optics/Geometry/Vectors).

Difficulty Boost : +4

Common Mistakes

This question tests the thermodynamic coupling between elasticity and magnetism (magnetostriction).

A common trap is using the positive magnetic energy density um=+12μH2u_m = +\frac{1}{2}\mu H^2 instead of the magnetic Gibbs free energy gm=−12μH2g_m = -\frac{1}{2}\mu H^2 for a system at constant external field H0H_0.

Using the positive sign leads to the false conclusion that the magnetic field stiffens the material (Option B).

Furthermore, it tests the fundamental relation between Young's modulus, Poisson's ratio, and Bulk Modulus, requiring students to recognize how the intrinsic resistance to volume change is counteracted by the magnetic pressure, ultimately leading to mechanical instability.

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Question Information

SubjectPhysics
ChapterMechanical Properties of Solids - Class 11 Part 2
ConceptPoisson's Ratio Limits
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree