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Question #43261

Difficulty: 8/10 | 🟢 Free

The Question

A rigid, non-conducting rod of length LL is pivoted at one end (x=0x=0) and is free to rotate in a horizontal plane. The rod has a non-uniform mass density λm(x)=λ0x\lambda_m(x) = \lambda_0 x and a non-uniform charge density λq(x)=αx\lambda_q(x) = \alpha x, where xx is the distance from the pivot. The total mass of the rod is MM and its total charge is QQ. Initially, the rod is at rest. A uniform magnetic field, perpendicular to the plane of rotation, is switched on and slowly increased from 00 to a steady value BB. Assuming no friction, no radiation losses, and that all quantities are in SI units, find the final kinetic energy KK acquired by the rod and the motional electromotive force E\mathcal{E} developed across its ends in the steady state.

Options

  • A.

    K=Q2B2L216M,E=QB2L24MK = \frac{Q^2 B^2 L^2}{16 M}, \quad \mathcal{E} = \frac{Q B^2 L^2}{4 M}

  • B.

    K=Q2B2L224M,E=QB2L24MK = \frac{Q^2 B^2 L^2}{24 M}, \quad \mathcal{E} = \frac{Q B^2 L^2}{4 M}

  • C.

    K=Q2B2L216M,E=QB2L23MK = \frac{Q^2 B^2 L^2}{16 M}, \quad \mathcal{E} = \frac{Q B^2 L^2}{3 M}

  • D.

    K=Q2B2L224M,E=QB2L23MK = \frac{Q^2 B^2 L^2}{24 M}, \quad \mathcal{E} = \frac{Q B^2 L^2}{3 M}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Gyromagnetic Ratio Invariance|Motional EMF in Rotating Rods|Work-Energy Theorem for Rotation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple electromagnetic induction with rigid body dynamics in a non-uniform system.

The primary trap lies in the miscalculation of the moment of inertia and the motional EMF.

Students often reflexively use I=13ML2I = \frac{1}{3}ML^2 for a rod, ignoring the linear density gradient, which leads to an incorrect kinetic energy (Option B).

Furthermore, a common misconception in calculating motional EMF is using the velocity of the center of mass (vcm=ωxcmv_{cm} = \omega x_{cm}) instead of integrating ∫vBdx\int v B dx.

For this non-uniform rod, xcm=23Lx_{cm} = \frac{2}{3}L, which yields an incorrect EMF of QB2L23M\frac{Q B^2 L^2}{3 M} if the integration principle is bypassed (Option C).

Option D catches students who make both conceptual errors.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMagnetism and Matter - Class 12
ConceptGyromagnetic Ratio Invariance
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree