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Question #43028

Difficulty: 8/10 | 🟢 Free

The Question

A cylindrical sample of a high-resistivity ferrimagnetic material of volume V=5.0×10−4 m3V = 5.0 \times 10^{-4} \text{ m}^3 is suspended vertically by a light torsional wire of torsional constant C=2.0π2 N⋅m/radC = \frac{2.0}{\pi^2} \text{ N}\cdot\text{m/rad}. A uniform horizontal magnetic field B⃗\vec{B} is applied. Initially, the wire is untwisted, and the sample is in equilibrium. An external agent slowly rotates the cylinder around the vertical axis by exactly one full turn (2π rad2\pi \text{ rad}) and holds it in this new position. During this 360∘360^\circ rotation, the magnetization of the sample undergoes exactly one complete hysteresis cycle. Assume the energy dissipated per unit volume during this rotation is equal to the area of the material's B−HB-H hysteresis loop, which is given as AH=4000 J/m3A_H = 4000 \text{ J/m}^3. Assuming eddy current losses are negligible, what is the total work done by the external agent?

Options

  • A.

    6.0 J6.0 \text{ J}

  • B.

    4.0 J4.0 \text{ J}

  • C.

    2.0 J2.0 \text{ J}

  • D.

    3.0 J3.0 \text{ J}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Magnetic Dipole Moment Orientation|Lenz’s Law Energy Conservation|Elastic Hysteresis Loop

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Work-Energy Theorem in the presence of non-conservative magnetic forces.

A common misconception among students is treating the magnetic potential energy (U=−m⃗⋅B⃗U = -\vec{m} \cdot \vec{B}) as purely conservative over a closed cycle, leading them to assume the net work done by the magnetic field is zero.

However, in ferromagnetic or ferrimagnetic materials, the reorientation of magnetic domains involves path-dependent energy dissipation (hysteresis).

Students must recognize that the external agent must supply energy not only to store mechanical potential energy in the twisted wire but also to overcome the non-conservative magnetic torque, which dissipates energy equal to the area of the B−HB-H loop multiplied by the material's volume.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptMagnetic Dipole Moment Orientation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree