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Question #43079

Difficulty: 8/10 | 🟢 Free

The Question

A small magnetic dipole of constant magnitude $\mu$ and mass $m$ is constrained to move along the $z$ -axis in a viscous medium with a linear drag coefficient $b$ . An external mechanism forces the dipole to rotate uniformly in the $x$ - $z$ plane such that its magnetic moment vector is given by $\vec{\mu}(t) = \mu \sin(\omega t) \hat{i} + \mu \cos(\omega t) \hat{k}$ . The system is permeated by an AC magnetic field with a spatial gradient, given by $\vec{B}(x, z, t) = (-\beta x \hat{i} + \beta z \hat{k}) \cos(\omega t)$ , where $\beta$ is a positive constant. Calculate the average power delivered by the magnetic force (given by $\vec{F} = \nabla(\vec{\mu} \cdot \vec{B})$ ) to the dipole in the steady state.

Options

A.
$\frac{\mu^2 \beta^2 (3b^2 + 8m^2\omega^2)}{8b(b^2 + 4m^2\omega^2)}$
B.
$\frac{\mu^2 \beta^2}{4b}$
C.
$\frac{\mu^2 \beta^2 b}{2(b^2 + m^2\omega^2)}$
D.
$\frac{\mu^2 \beta^2 (3b^2 + 2m^2\omega^2)}{8b(b^2 + m^2\omega^2)}$

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Correct Answer

Step-by-Step Solution

Explanation

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

Average Power over Cycle|Magnetic Dipole Moment of Revolving Electron|Faraday’s Law Vector Form

Why This Question Is Tricky

Common Mistakes

Students incorrectly apply the average power formula by calculating the average force $\langle F \rangle$ and average velocity $\langle v \rangle$ independently, assuming $\langle P \rangle = \langle F \rangle \langle v \rangle$ (leading to Option B).

This trap ignores the non-conservative cross-coupling between the time-varying component of the gradient force and the induced AC drift velocity.

Because the force oscillates at $2\omega$ (due to the $\cos^2(\omega t)$ term from the dot product of the rotating dipole and the AC field), the product of the AC force and AC velocity yields a non-zero DC power term that must be added to the baseline DC power dissipation.

Option D traps students who miss the frequency doubling and use $\omega$ instead of $2\omega$ in the impedance term.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Alternating Current - Class 12
Concept	Average Power over Cycle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free