<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:content="http://purl.org/rss/1.0/modules/content/"><channel><title>Magnetic Field on Blog of Brian</title><link>https://changzh.me/en/tags/magnetic-field/</link><description>Recent content in Magnetic Field on Blog of Brian</description><generator>Hugo</generator><language>en</language><lastBuildDate>Thu, 16 Jul 2026 11:50:00 +0800</lastBuildDate><atom:link href="https://changzh.me/en/tags/magnetic-field/index.xml" rel="self" type="application/rss+xml"/><item><title>Bounded Magnetic Field with Gravity: Two Solutions for the "Just-Exits" Condition</title><link>https://changzh.me/en/posts/bounded-field-gravity-particle/</link><pubDate>Thu, 16 Jul 2026 11:50:00 +0800</pubDate><guid>https://changzh.me/en/posts/bounded-field-gravity-particle/</guid><description>&lt;p&gt;A classic high-school physics problem: a charged particle enters a bounded uniform magnetic field perpendicularly, and we ask how strong the field must be so the particle &amp;ldquo;just exits through the right boundary.&amp;rdquo; The standard solution ignores gravity, and the trajectory is a clean circular arc. But what happens if &lt;strong&gt;gravity is included and the problem is flipped — the field is given, and the width is what we seek&lt;/strong&gt;? The answer is: the circle becomes a &lt;strong&gt;cycloid&lt;/strong&gt;, and the problem can still be solved elegantly — in two quite different ways. This post documents that extended problem (B6 Extension) and its two solutions.&lt;/p&gt;</description></item></channel></rss>