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Question #42666

Difficulty: 8/10 | 🟢 Free

The Question

A smooth, rigid, insulating circular ring of radius aa is fixed in the xyxy-plane, with its equation given by (x−a)2+y2=a2(x-a)^2 + y^2 = a^2. A point charge +Q+Q is fixed at the point (−a,0)(-a, 0). A uniform but time-varying magnetic field B⃗=αtk^\vec{B} = \alpha t \hat{k} exists in a cylindrical region (x−a)2+y2≤R2(x-a)^2 + y^2 \le R^2 (where R>2aR > 2a), where α\alpha is a positive constant and tt is time. A small bead of mass mm and charge +q+q is released from rest at the origin (0,0)(0,0). Due to the induced electric field, the bead begins to slide along the upper semicircle (y≥0y \ge 0) of the ring. What is the kinetic energy of the bead when it reaches the point (2a,0)(2a, 0) for the first time? (Ignore gravity)

Options

  • A.

    qQ6πϵ0a+qαπa22\frac{qQ}{6\pi\epsilon_0 a} + \frac{q \alpha \pi a^2}{2}

  • B.

    qQ6πϵ0a\frac{qQ}{6\pi\epsilon_0 a}

  • C.

    qQ6πϵ0a+qαπa2\frac{qQ}{6\pi\epsilon_0 a} + q \alpha \pi a^2

  • D.

    qQ4πϵ0a+qαπa22\frac{qQ}{4\pi\epsilon_0 a} + \frac{q \alpha \pi a^2}{2}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Equipotential Surface Geometry|Trajectory of Charged Particle|Induced Electric Fields|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between conservative and non-conservative electric fields.

A common trap is to assume that the total work done on a charged particle is simply qΔVq\Delta V, relying solely on the static potential difference and ignoring the path-dependent work done by the induced electric field.

Because the induced electric field forms closed loops (∮E⃗⋅dl⃗≠0\oint \vec{E} \cdot d\vec{l} \neq 0), it cannot be represented by a scalar potential.

Success requires applying the generalized Work-Energy Theorem, calculating the electrostatic work via potential difference, and elegantly determining the induced work using Faraday's Law and Stokes' theorem over a cleverly chosen closed loop where the return path contributes zero work.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptEquipotential Surface Geometry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree