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Question #42839

Difficulty: 8/10 | 🟢 Free

The Question

An infinitely long solid insulating cylinder of radius $R$ has a volume charge density $\rho(r) = \rho_0 \frac{r}{R}$ , where $r$ is the radial distance from the cylinder's axis and $\rho_0$ is a positive constant. A thin, infinitely long straight wire lies along the axis of the cylinder and carries a steady current $I$ . A particle of mass $m$ and charge $+q$ is projected from the surface of the cylinder ( $r = R$ ) with an initial velocity $\vec{v}_0 = v_0 \hat{k}$ parallel to the axis. Assuming the initial velocity $v_0$ and current $I$ are sufficient for the particle to reach a radial distance of $r = R/2$ , what is the total speed of the particle at this instant? (Ignore gravity and edge effects).

Options

A.
$\sqrt{v_0^2 - \frac{7 q \rho_0 R^2}{36 m \epsilon_0}}$
B.
$\sqrt{v_0^2 - \frac{7 q \rho_0 R^2}{36 m \epsilon_0} + \frac{q \mu_0 I v_0}{\pi m} \ln(2)}$
C.
$\sqrt{v_0^2 - \frac{3 q \rho_0 R^2}{8 m \epsilon_0}}$
D.
$\sqrt{v_0^2 - \frac{7 q \rho_0 R^2}{72 m \epsilon_0}}$

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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

Gauss Law Applicability|Biot-Savart Law Vector Symmetry|Lorentz Force Vector Direction

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to navigate a complex superposition of non-uniform electric and magnetic fields without falling into the trap of solving coupled differential equations.

Because the magnetic field from the Biot-Savart law creates a Lorentz force that continuously changes the velocity vector's direction and magnitude (coupling the radial and axial velocity components), students often incorrectly attempt to integrate the equations of motion or mistakenly assume the magnetic field does work on the radial component.

The core physical insight required is recognizing that the magnetic Lorentz force ( $q(\vec{v} \times \vec{B})$ ) is strictly perpendicular to the velocity vector and therefore does zero work.

By applying the Work-Energy Theorem, the problem elegantly collapses into calculating only the work done by the conservative electric field (derived via Gauss's Law for a non-uniform charge distribution), completely bypassing the complex trajectory mechanics.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electric Charges and Fields - Class 12
Concept	Gauss Law Applicability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free