---
title: "A positive ion of mass \\(m\\) and charge \\(+q\\) enters a region of uniform magnetic field \\(\\vec{B}\\) perpendicular to its velocity vector with speed \\(v_0\\). The ion travels in a circular arc of radius \\(R\\). A second positive ion with charge \\(+q\\) and mass \\(2m\\) enters the same magnetic field region with the exact same speed \\(v_0\\). What is the radius of the circular path of the second ion in terms of \\(R\\)?"
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url: "https://nerd-notes.com/ubq/118548/"
date_modified: "2026-08-04T08:11:18+00:00"
---

# A positive ion of mass \(m\) and charge \(+q\) enters a region of uniform magnetic field \(\vec{B}\) perpendicular to its velocity vector with speed \(v_0\). The ion travels in a circular arc of radius \(R\). A second positive ion with charge \(+q\) and mass \(2m\) enters the same magnetic field region with the exact same speed \(v_0\). What is the radius of the circular path of the second ion in terms of \(R\)?

A positive ion of mass \(m\) and charge \(+q\) enters a region of uniform magnetic field \(\vec{B}\) perpendicular to its velocity vector with speed \(v_0\). The ion travels in a circular arc of radius \(R\). A second positive ion with charge \(+q\) and mass \(2m\) enters the same magnetic field region with the exact same speed \(v_0\). What is the radius of the circular path of the second ion in terms of \(R\)?

![A schematic showing a region of uniform magnetic field directed perpendicularly into the page, represented by a four-by-four grid of small x symbols. A positive ion of mass m and charge +q enters horizontally from the left edge into the field region with a rightward velocity vector labeled v_0. Inside the field, the ion's path curves upward in a circular arc of radius R indicated by a curved dashed line and a radius arrow labeled R. No other labels or text appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785831078-Z1mMTe.jpg)

- **A.** \(2R\)
- **B.** \(\sqrt{2}R\)
- **C.** \(\dfrac{1}{2}R\)
- **D.** \(\dfrac{1}{\sqrt{2}}R\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118548/*