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Question #43008

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m$ and positive charge $q$ is projected from an initial position $\vec{r}_0$ (with magnitude $r_0$ ) with an initial velocity $\vec{v}_0$ perpendicular to $\vec{r}_0$ . The space contains a magnetic field $\vec{B} = \frac{k}{r^2} \hat{r}$ , where $\hat{r}$ is the radial unit vector from the origin and $k$ is a positive constant. It is observed that the particle's trajectory is confined to the surface of a cone with its vertex at the origin. Identify the correct expression for the magnitude of the particle's areal velocity about the origin, and the semi-vertical angle $\alpha$ of this cone.

Options

A.
Areal Velocity = $\frac{1}{2} r_0 v_0$ , $\alpha = \tan^{-1}\left(\frac{m v_0 r_0}{q k}\right)$
B.
Areal Velocity = $\frac{m v_0^2 r_0^2}{2 q k}$ , $\alpha = \tan^{-1}\left(\frac{q k}{m v_0 r_0}\right)$
C.
Areal Velocity = $\frac{1}{2} r_0 v_0$ , $\alpha = \sin^{-1}\left(\frac{m v_0 r_0}{q k}\right)$
D.
Areal Velocity = $\frac{m v_0^2 r_0^2}{2 q k}$ , $\alpha = \cos^{-1}\left(\frac{q k}{m v_0 r_0}\right)$

Check Your Answer

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

Trajectory of Charged Particle|Kepler’s Second Law (Areal Velocity)|Centripetal Acceleration Vector

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze motion in a non-uniform, monopole-like magnetic field using fundamental conservation laws rather than memorized formulas.

The primary trap is the assumption that a magnetic force, being perpendicular to velocity, will cause standard circular or helical motion with a radius $R = mv/qB$ , leading students to calculate an incorrect 'effective' areal velocity based on a fictitious circular path.

Furthermore, students might incorrectly assume that because the force is not central, the magnitude of angular momentum must change.

By deriving the conservation of the Poincaré vector ( $\vec{J} = \vec{L} - qk\hat{r}$ ), the student must recognize that while the direction of $\vec{L}$ changes, its magnitude remains strictly constant, resulting in a constant areal velocity and motion restricted to a conical surface.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Trajectory of Charged Particle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free