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Question #43266

Difficulty: 8/10 | 🟢 Free

The Question

A spherical ferromagnetic rotor of volume V=1.0×10−3 m3V = 1.0 \times 10^{-3} \text{ m}^3 and moment of inertia I=5.0×10−3 kg m2I = 5.0 \times 10^{-3} \text{ kg m}^2 is free to rotate about a fixed vertical axis. A uniform horizontal magnetic field of magnitude B=2.0 TB = 2.0 \text{ T} is applied and forced to rotate about the same vertical axis at a constant angular velocity ωB=150 rad/s\omega_B = 150 \text{ rad/s}. Due to magnetic hysteresis, the rotor's magnetization vector M⃗\vec{M} maintains a constant magnitude M=1.0×104 A/mM = 1.0 \times 10^4 \text{ A/m} but lags behind the rotating magnetic field B⃗\vec{B} by a constant angle δ=30∘\delta = 30^\circ. The rotor experiences a viscous drag torque τd=−bω\tau_d = -b \omega, where b=0.2 N m sb = 0.2 \text{ N m s} and ω\omega is the rotor's angular velocity. Once the rotor reaches its steady-state terminal angular velocity ω0\omega_0, what is ω0\omega_0 and the rate of internal heat generation PhP_h within the rotor strictly due to magnetic hysteresis?

Options

  • A.

    ω0=50 rad/s,  Ph=1000 W\omega_0 = 50 \text{ rad/s}, \; P_h = 1000 \text{ W}

  • B.

    ω0=50 rad/s,  Ph=1500 W\omega_0 = 50 \text{ rad/s}, \; P_h = 1500 \text{ W}

  • C.

    ω0=50 rad/s,  Ph=500 W\omega_0 = 50 \text{ rad/s}, \; P_h = 500 \text{ W}

  • D.

    ω0=150 rad/s,  Ph=0 W\omega_0 = 150 \text{ rad/s}, \; P_h = 0 \text{ W}

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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

Torque and Potential Energy in B-Field|Energy Partitioning|Elastic Hysteresis Loop

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to partition energy in a non-conservative magnetic system, specifically modeling the physics of a hysteresis motor.

Students commonly fall into the trap of assuming the magnetic potential energy is a simple state function (U=−m⃗⋅B⃗U = -\vec{m}\cdot\vec{B}) and incorrectly apply conservation of mechanical energy.

They may calculate the total power input Pin=τmωBP_{in} = \tau_m \omega_B and assume it all goes into viscous dissipation (Option B), or they might confuse the mechanical power Pmech=τmω0P_{mech} = \tau_m \omega_0 with the hysteresis loss (Option C).

The core physical insight required is that the magnetic field does work on the material at a rate proportional to the field's rotation ωB\omega_B, but only a fraction (ω0/ωB\omega_0/\omega_B) becomes mechanical work; the 'slip' relative rotation (ωB−ω0\omega_B - \omega_0) drives the hysteretic cycle, dissipating energy as internal thermodynamic heat.

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Question Information

SubjectPhysics
ChapterMagnetism and Matter - Class 12
ConceptTorque and Potential Energy in B-Field
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree