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Question #42806

Difficulty: 8/10 | 🟢 Free

The Question

A rigid, frictionless wire is bent into a quarter-circular arc of radius $R$ . In the reference frame of the wire, the arc lies in the $xy$ -plane, spanning from point $A(0, R)$ to point $B(R, 0)$ . The wire translates with a constant velocity $\vec{v}_0 = v_0 \hat{i}$ relative to the laboratory frame. A uniform, time-independent magnetic field $\vec{B} = B_0 \hat{k}$ exists throughout the space. A bead of mass $m$ and positive charge $q$ is threaded on the wire. An ideal, massless spring of spring constant $k$ and zero natural length connects the bead to point $B$ of the wire. At $t=0$ , the bead is released from rest relative to the wire at point $A$ . Assuming no gravitational forces act on the system, what is the speed of the bead relative to the wire at the instant it reaches point $B$ ?

Options

A.
$\sqrt{\frac{2}{m} \left( k R^2 + q v_0 B_0 R \right)}$
B.
$\sqrt{\frac{2 k R^2}{m}}$
C.
$\sqrt{\frac{2}{m} \left( k R^2 - q v_0 B_0 R \right)}$
D.
$\sqrt{\frac{2}{m} \left( k R^2 + \frac{\pi}{2} q v_0 B_0 R \right)}$

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

Work against Conservative Force|Non-Uniform Circular Motion|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Work-Energy Theorem in different reference frames and the transformation of electromagnetic fields.

Students often incorrectly apply the Work-Energy Theorem in the laboratory frame by assuming the normal force from the wire does zero work, leading them to equate the change in kinetic energy solely to the change in the spring's potential energy (Option B).

However, because the wire is translating, the normal force does non-zero work in the lab frame.

The most robust method is to shift to the inertial frame of the moving wire.

In this frame, the normal force does zero work, but students must correctly account for the induced motional electric field $\vec{E}' = \vec{v}_0 \times \vec{B}$ , which acts as an effective conservative force field and contributes to the total work done on the bead.

Quick Facts About This Question

Question Information

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