Before You Begin

Question #43511

Difficulty: 8/10 | 🟢 Free

The Question

A uniform solid conducting cylinder of mass $m$ and radius $r$ is placed on two long, parallel, horizontal conducting rails separated by a distance $l$ . The rails are connected at one end by a series combination of a resistor with resistance $R$ and an inductor with inductance $L$ . A uniform vertical magnetic field of magnitude $B$ is present everywhere. At $t=0$ , the cylinder is given an initial horizontal velocity $v_0$ parallel to the rails. Assume the cylinder rolls without slipping at all times, and the electrical resistance of the rails and the cylinder is negligible. What is the net displacement $S$ of the cylinder from its initial position when it finally comes to rest ( $t \to \infty$ )?

Options

A.
$\frac{3 m R v_0}{2 B^2 l^2}$
B.
$\frac{m R v_0}{B^2 l^2}$
C.
$\frac{3 m R v_0}{2 B^2 l^2} + \frac{L v_0}{R}$
D.
$\frac{m R v_0}{B^2 l^2} + \frac{3 L v_0}{2 R}$

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

Pure Rolling Condition|Transient Current in RL Circuit|Force on Current-Carrying Loop

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple electromagnetic damping with rigid body dynamics without falling into the trap of solving complex differential equations.

The 'lethal collision' occurs on two fronts.

First, students often misapply the pure rolling condition by forgetting that static friction provides the necessary torque to decelerate the cylinder's rotation, leading them to use the bare mass $m$ instead of the effective mass $\frac{3}{2}m$ (Option B).

Second, the presence of the inductor $L$ baits students into attempting to solve a second-order differential equation for $v(t)$ .

Even if they attempt to integrate, they may incorrectly assume the inductor stores residual energy that permanently affects the total stopping distance, adding an erroneous dimensionally-consistent $\frac{L v_0}{R}$ term (Options C and D).

Recognizing that $\int L \frac{di}{dt} dt = 0$ over the full motion is the key to bypassing the mathematical trap.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	System of Particles and Rotational Motion - Class 11 Part 1
Concept	Pure Rolling Condition
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free