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Question #43007

Difficulty: 8/10 | 🟢 Free

The Question

A smooth, non-conducting circular ring of radius $R$ is fixed in a vertical plane. A uniform magnetic field is confined to a cylindrical region of radius $R/2$ whose axis passes through the center of the ring and is perpendicular to its plane. The magnetic field is zero outside this region. The magnetic field is directed into the page and its magnitude varies with time $t$ as $B(t) = \alpha t$ , where $\alpha$ is a positive constant. A small bead of mass $m$ and positive charge $q$ is released from rest at the highest point of the ring. Assuming the induced electric field is the only non-conservative field present, what is the speed of the bead when it reaches the lowest point of the ring?

Options

A.
$\sqrt{ 4gR + \frac{\pi q \alpha R^2}{4m} }$
B.
$\sqrt{ 4gR + \frac{\pi q \alpha R^2}{m} }$
C.
$\sqrt{ 4gR + \frac{q \alpha R^2}{2m} }$
D.
$\sqrt{ 4gR }$

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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

Trajectory of Charged Particle|Induced Electric Fields|Average vs. Instantaneous Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Work-Energy Theorem in the presence of non-conservative induced electric fields, a common blind spot for students accustomed to static conservative fields.

The primary trap is twofold: first, students often ignore the work done by the induced electric field entirely, treating it as a standard vertical circle problem.

Second, if they do account for the induced field, they frequently miscalculate the flux by using the full area of the ring ( $\pi R^2$ ) instead of the confined region ( $\pi (R/2)^2$ ), leading to an incorrect electric field magnitude.

Additionally, students might incorrectly use the diameter ( $2R$ ) instead of the arc length ( $\pi R$ ) for the path integral of the tangential electric force.

The problem forces a rigorous application of Faraday's Law in integral form and the realization that the induced electric force does continuous positive work along the curved path.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Trajectory of Charged Particle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free