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Question #43010

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m$ and charge $+q$ is projected from the origin $(0,0,0)$ with an initial velocity $\vec{v} = \sqrt{3}v_0 \hat{i} + v_0 \hat{j}$ . A non-uniform magnetic field $\vec{B} = B_0 x \hat{k}$ exists in the region $x \ge 0$ , where $B_0$ is a positive constant. Neglecting gravity, the maximum x-coordinate reached by the particle is:

Options

A.
$\sqrt{\frac{2mv_0}{q B_0}}$
B.
$\sqrt{\frac{4mv_0}{q B_0}}$
C.
$\sqrt{\frac{6mv_0}{q B_0}}$
D.
$\sqrt{\frac{8mv_0}{q B_0}}$

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

Cyclotron Frequency Invariance|Lorentz Force Vector Direction|Moving Charge in Non-Uniform B

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze motion in a non-uniform magnetic field, where standard cyclotron formulas ( $R = mv/qB$ ) completely fail.

The primary trap lies in assuming the particle's velocity simply reverses its initial y-component ( $v_y = -v_0$ ) at maximum displacement, or incorrectly assuming the total speed is just $v_0$ .

By introducing a spatial gradient ( $B \propto x$ ), students are forced to apply the fundamental Lorentz force integration ( $m dv_y = -q B(x) dx$ ).

Furthermore, they must correctly apply the Work-Energy theorem to deduce that the velocity vector rotates continuously until $v_y = -2v_0$ at the turning point, accounting for the full initial kinetic energy.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Cyclotron Frequency Invariance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free