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Question #42089

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m$ and charge $q > 0$ is projected from the origin $(0,0,0)$ with an initial velocity $\vec{v} = v_0 \hat{i}$ . The region $x > 0$ contains a non-uniform magnetic field given by $\vec{B} = \alpha x \hat{k}$ , where $\alpha$ is a positive constant. The region $x < 0$ has no magnetic field. Neglecting gravity, what is the maximum positive x-coordinate ( $x_{max}$ ) reached by the particle?

Options

A.
$\sqrt{\frac{2m v_0}{q \alpha}}$
B.
$\sqrt{\frac{m v_0}{q \alpha}}$
C.
$\sqrt{\frac{m v_0}{2q \alpha}}$
D.
$2\sqrt{\frac{m v_0}{q \alpha}}$

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

Change in Velocity (Δv)|Moving Charge in Non-Uniform B|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze motion in a spatially varying magnetic field, specifically targeting the misconception that standard cyclotron formulas ( $R = mv/qB$ ) can be applied locally to find macroscopic trajectories.

The trap (Option B) lures students into equating the instantaneous centripetal force to $mv_0^2/x$ , falsely assuming a circular path with radius $x$ .

The correct approach requires recognizing that while the magnetic force does no work (conserving kinetic energy), the trajectory is non-circular.

Students must use Newton's second law to find the differential change in the velocity vector, specifically integrating the y-component of acceleration over $x$ to find the velocity components, effectively utilizing the conservation of canonical momentum in the y-direction due to the 1D gradient of the magnetic field.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Change in Velocity (Δv)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free