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Question #43537

Difficulty: 8/10 | 🟢 Free

The Question

A thin, flexible insulating ring of radius RR and cross-sectional area AA has a uniform linear mass density λ\lambda and a uniform positive linear charge density μ\mu. The ring is rotated about its central axis, which is perpendicular to its plane, with a constant angular velocity ω\omega. A uniform magnetic field of magnitude BB is applied parallel to the axis of rotation such that the resulting Lorentz force acts radially outward. If the material of the ring has a breaking stress σb\sigma_b, which of the following expressions gives the minimum magnetic field BB required to break the ring? (Assume the ring remains perfectly circular until it breaks and neglect gravity).

Options

  • A.

    σbAμωR2−λωμ\frac{\sigma_b A}{\mu \omega R^2} - \frac{\lambda \omega}{\mu}

  • B.

    σbAμωR2\frac{\sigma_b A}{\mu \omega R^2}

  • C.

    σbAμωR2+λωμ\frac{\sigma_b A}{\mu \omega R^2} + \frac{\lambda \omega}{\mu}

  • D.

    2σbAμωR2−λωμ\frac{2\sigma_b A}{\mu \omega R^2} - \frac{\lambda \omega}{\mu}

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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

Breaking Stress Concept|Force on Current-Carrying Loop|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the superposition of mechanical and electromagnetic forces in determining structural equilibrium (Hoop Stress).

The primary trap lies in the calculation of the equivalent current of a rotating charged ring; students often incorrectly use I=μωI = \mu \omega instead of I=μωRI = \mu \omega R, leading to dimensional inconsistencies.

Additionally, students frequently fail to account for the non-electromagnetic centrifugal force acting simultaneously on the mass of the ring, or they misapply the differential force balance Tdθ=dFradialT d\theta = dF_{radial} by integrating over the entire loop and incorrectly concluding the net force (and thus tension) is zero.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMechanical Properties of Solids - Class 11 Part 2
ConceptBreaking Stress Concept
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree