Before You Begin

Question #43379

Difficulty: 8/10 | 🟢 Free

The Question

A solid insulating cylinder of mass $m$ , radius $R$ , and total positive charge $Q$ uniformly distributed over its lateral surface is released from rest on a rough inclined plane making an angle $\theta$ with the horizontal. The coefficient of static friction between the cylinder and the incline is $\mu_s$ (where $\mu_s > \frac{1}{3}\tan\theta$ ). A uniform magnetic field of magnitude $B$ is present parallel to the central axis of the cylinder. If the direction down the incline is taken as the $+x$ -axis and the outward normal to the incline as the $+y$ -axis, the magnetic field is given by $\vec{B} = -B\hat{k}$ (where $\hat{k} = \hat{i} \times \hat{j}$ ). As the cylinder rolls down without slipping, its speed $v$ increases until it eventually begins to slip. What is the speed $v$ of the cylinder at the instant slipping begins?

Options

A.
$\frac{mg}{QB} \left( \cos\theta - \frac{\sin\theta}{3\mu_s} \right)$
B.
$\frac{mg}{QB} \left( \cos\theta - \frac{\sin\theta}{\mu_s} \right)$
C.
$\frac{mg}{QB} \left( \cos\theta - \frac{\sin\theta}{2\mu_s} \right)$
D.
$\frac{mg}{QB} \left( \frac{\sin\theta}{3\mu_s} - \cos\theta \right)$

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

Lorentz Force Vector Direction|Rolling on Inclines|Static Friction Threshold

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to decouple the translational and rotational effects of the Lorentz force on a rigid body.

A common trap is assuming that because the magnetic force is velocity-dependent and the velocity of the cylinder's surface varies (from $0$ at the contact point to $2v$ at the top), the magnetic force must exert a net torque about the center of mass.

In reality, the magnetic force component arising from the rotational velocity is purely radial, meaning the net magnetic torque about the COM is strictly zero.

The magnetic field only provides a translational lift ( $QvB$ ) that reduces the normal force.

Students who incorrectly calculate a magnetic torque will get the wrong required friction, while those who ignore the rolling constraint will incorrectly use $f = mg\sin\theta$ instead of $f = \frac{1}{3}mg\sin\theta$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Lorentz Force Vector Direction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free