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Question #43284

Difficulty: 8/10 | 🟢 Free

The Question

A uniform rigid rod of mass M=0.6 kgM = 0.6 \text{ kg} and length L=1.0 mL = 1.0 \text{ m} is freely pivoted at its upper end. A small soft iron core of mass m=0.2 kgm = 0.2 \text{ kg} and volume V=5×10−4 m3V = 5 \times 10^{-4} \text{ m}^3 is attached to the lower end. The pendulum is released from rest in the horizontal position (θ=90∘\theta = 90^\circ). At the lowest point of its swing, the iron core passes through a narrow region of localized magnetic field. During this brief passage, the external magnetic field exerts a total retarding angular impulse of J=0.4 kg m2/sJ = 0.4 \text{ kg m}^2/\text{s} on the pendulum about the pivot. The soft iron core undergoes exactly one complete hysteresis cycle during the passage. The area of the hysteresis loop for the soft iron material is Ah=2000 J/m3A_h = 2000 \text{ J/m}^3. Assuming the rod is non-magnetic and non-conducting, what is the energy dissipated as heat due to eddy currents in the iron core? (Take g=10 m/s2g = 10 \text{ m/s}^2)

Options

  • A.

    0.8 J0.8 \text{ J}

  • B.

    1.0 J1.0 \text{ J}

  • C.

    1.8 J1.8 \text{ J}

  • D.

    2.8 J2.8 \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

Soft vs Hard Magnetic Materials|Impulsive Torque and Angular Momentum

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to reconcile macroscopic rigid-body dynamics (impulse and momentum) with microscopic continuum energy dissipation (hysteresis and eddy currents).

The primary trap is assuming that the given angular impulse JJ is solely due to eddy currents, leading students to incorrectly attribute the entire 1.8 J1.8 \text{ J} kinetic energy loss to eddy currents.

In reality, the non-conservative nature of the hysteresis loop (where magnetization lags the external field) produces a macroscopic retarding force.

Thus, the total angular impulse and the corresponding total kinetic energy loss account for both hysteresis and eddy current dissipation.

Students must apply the Work-Energy theorem to find the total loss and subtract the hysteresis loss (V⋅AhV \cdot A_h) to isolate the eddy current contribution.

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Question Information

SubjectPhysics
ChapterMagnetism and Matter - Class 12
ConceptSoft vs Hard Magnetic Materials
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree