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Question #42831

Difficulty: 8/10 | 🟢 Free

The Question

A thin non-conducting ring of mass $M$ (in $\text{kg}$ ), radius $R$ (in $\text{m}$ ), and uniformly distributed charge $Q$ (in $\text{C}$ ) is pivoted at its center on a horizontal surface. The surface exerts a viscous frictional torque $\tau_v = -b \omega$ on the ring, where $\omega$ is its angular velocity (in $\text{rad/s}$ ) and $b$ is a positive constant (in $\text{kg}\cdot\text{m}^2/\text{s}$ ). A uniform magnetic field $\vec{B}$ is applied perpendicular to the plane of the ring. The magnetic field magnitude varies with time as $B(t) = B_0 (1 - e^{-t/t_0})$ for $t \ge 0$ , where $B_0$ is in Tesla and $t_0$ is a time constant in seconds. The ring is initially at rest. What is the total angular displacement of the ring from $t=0$ until it finally comes to rest?

Options

A.
$\frac{Q R^2 B_0}{2 b}$
B.
$\frac{Q R^2 B_0}{b}$
C.
$\frac{Q R^2 B_0}{2 b} \left( 1 - e^{-\frac{b t_0}{M R^2}} \right)$
D.
$\frac{Q R^2 B_0}{2 b} \left( \frac{M R^2}{b t_0 + M R^2} \right)$

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

Dipole in Uniform Field|Friction on Inclined Planes|SHM Definition via Differential Equation

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Angular Impulse Theorem over an extended time interval, specifically trapping students who attempt to solve the full differential equation for constrained motion with damping.

Students often misapply Newton's Second Law by trying to find the explicit time-domain solution $\omega(t)$ for the forced, damped system, which is mathematically tedious and leads them to incorrectly assume that the mass $M$ (moment of inertia) or the time constant $t_0$ must dictate the final displacement (as seen in the distractors).

The core physical insight is recognizing that because the system begins and ends at rest, the net angular impulse is zero.

By integrating the equation of motion over all time, the total viscous angular impulse ( $b \Delta \theta$ ) can be directly equated to the total driving angular impulse from the magnetic field, completely bypassing the transient dynamics.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electric Charges and Fields - Class 12
Concept	Dipole in Uniform Field
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free