---
title: "A particle of mass \\(m\\) and positive charge \\(q\\) enters a region containing a static, spatially non-uniform magnetic field \\(\\vec{B}(x, y, z)\\). No other fields or external forces act on the particle. Which of the following statements correctly describes the change in the particle’s kinetic energy \\(\\Delta K\\) as it travels between any two arbitrary points in this region, along with the correct physical justification?"
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url: "https://nerd-notes.com/ubq/121271/"
date_modified: "2026-08-23T04:59:07+00:00"
---

# A particle of mass \(m\) and positive charge \(q\) enters a region containing a static, spatially non-uniform magnetic field \(\vec{B}(x, y, z)\). No other fields or external forces act on the particle. Which of the following statements correctly describes the change in the particle’s kinetic energy \(\Delta K\) as it travels between any two arbitrary points in this region, along with the correct physical justification?

A particle of mass \(m\) and positive charge \(q\) enters a region containing a static, spatially non-uniform magnetic field \(\vec{B}(x, y, z)\). No other fields or external forces act on the particle. Which of the following statements correctly describes the change in the particle's kinetic energy \(\Delta K\) as it travels between any two arbitrary points in this region, along with the correct physical justification?

![A smooth curved trajectory of a particle with charge q. At a point on the curve, a tangent vector arrow is labeled \vec{v}, a field vector arrow is labeled \vec{B}, and a perpendicular vector arrow is labeled \vec{F}_B with a right-angle square marker between \vec{F}_B and \vec{v}. A small dashed region indicates non-uniform magnetic field vectors of varying lengths. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787461147-LZh1AI.jpg)

- **A.** The change in kinetic energy depends on the path taken because the non-uniform magnetic field has a non-zero spatial gradient, resulting in a non-zero line integral \(\int \vec{F}_B \cdot d\vec{r}\) along paths where the field magnitude changes.
- **B.** The change in kinetic energy is zero for any path because the magnetic force \(\vec{F}_B = q(\vec{v} \times \vec{B})\) is perpendicular to the velocity \(\vec{v}\) at every instant, making the rate of work \(\dfrac{dK}{dt} = \vec{F}_B \cdot \vec{v} = 0\).
- **C.** The change in kinetic energy is zero for any path because the static magnetic force is conservative, which ensures that the line integral of \(\vec{F}_B\) along any path can be represented by a scalar potential energy function.
- **D.** The change in kinetic energy is zero for any path because Gauss's law for magnetism dictates \(\oint \vec{B} \cdot d\vec{A} = 0\), which prevents any net transfer of energy across the boundary of the particle's trajectory.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121271/*