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Question #42792

Difficulty: 8/10 | 🟢 Free

The Question

A smooth rigid circular wire of radius $R$ is fixed in the horizontal $x-y$ plane with its center at the origin $O(0,0)$ . A small bead of mass $m$ and positive charge $q$ is threaded on the wire. The bead is attached to one end of an insulating light ideal spring of spring constant $k$ and natural length $R$ . The other end of the spring is pivoted freely at the point $P(-R, 0)$ on the wire. A uniform time-varying magnetic field $\vec{B} = \alpha t \hat{k}$ (where $\alpha > 0$ is a constant and $t$ is time) exists only in a cylindrical region of radius $a$ ( $a < R$ ) coaxial with the $z$ -axis. An external agent slowly moves the bead from point $A(R, 0)$ to point $C(0, R)$ along the wire in the counter-clockwise direction. What is the work done by the external agent?

Options

A.
$k R^2 (1 - \sqrt{2}) + \frac{\pi q \alpha a^2}{4}$
B.
$k R^2 (1 - \sqrt{2}) - \frac{\pi q \alpha a^2}{4}$
C.
$k R^2 (1 - \sqrt{2}) + \frac{\pi q \alpha R^2}{4}$
D.
$- k R^2 + \frac{\pi q \alpha a^2}{4}$

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

Symmetry and Virtual Work|Conservative Field Path Independence|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the limits of the conservative field assumption and the application of the work-energy theorem in the presence of non-conservative fields.

Students frequently fall into the trap of assuming that the work done by an external agent is simply the change in potential energy ( $\Delta U$ ) of the system, ignoring the induced electric field because it lacks a scalar potential.

The time-varying magnetic field creates a non-conservative induced electric field ( $\nabla \times \vec{E} \neq 0$ ) that does path-dependent work.

The options are designed to catch students who either completely ignore the induced field, apply Lenz's law incorrectly resulting in a sign error for the induced work, or mistakenly calculate the magnetic flux using the full loop radius $R$ instead of the confined field radius $a$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Symmetry and Virtual Work
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free