---
title: "A long wire carrying a steady current \\(I\\) lies in a plane and is bent into two straight sections connected by a semi-circular arc of radius \\(R\\). The straight sections lie along a horizontal line passing through point \\(P\\), which is the center of curvature of the semi-circular arc. Which of the following gives the magnitude of the net magnetic field at point \\(P\\) and provides the correct physical justification for the contribution from the straight sections?"
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url: "https://nerd-notes.com/ubq/118560/"
date_modified: "2026-08-04T08:11:22+00:00"
---

# A long wire carrying a steady current \(I\) lies in a plane and is bent into two straight sections connected by a semi-circular arc of radius \(R\). The straight sections lie along a horizontal line passing through point \(P\), which is the center of curvature of the semi-circular arc. Which of the following gives the magnitude of the net magnetic field at point \(P\) and provides the correct physical justification for the contribution from the straight sections?

A long wire carrying a steady current \(I\) lies in a plane and is bent into two straight sections connected by a semi-circular arc of radius \(R\). The straight sections lie along a horizontal line passing through point \(P\), which is the center of curvature of the semi-circular arc. Which of the following gives the magnitude of the net magnetic field at point \(P\) and provides the correct physical justification for the contribution from the straight sections?

![A single thin wire carrying current \(I\) lies in the plane of the page. The wire consists of a long straight segment extending horizontally from the left, a semi-circular arc of radius \(R\) curving upward above the horizontal axis, and another long straight segment extending horizontally to the right along the same axis. Point \(P\) is located at the center of curvature of the semi-circular arc, directly on the horizontal axis. Black arrows along the wire indicate current \(I\) flowing from left to right. A dashed radial line labeled \(R\) connects point \(P\) to the top of the arc. No other labels, lines, or vectors appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785831081-QXOzsx.jpg)

- **A.** \(B = \dfrac{\mu_0 I}{2R} + \dfrac{\mu_0 I}{\pi R}\), because both the arc and the two straight sections contribute to the net magnetic field at point \(P\).
- **B.** \(B = \dfrac{\mu_0 I}{2R}\), because the straight sections create equal and opposite magnetic fields that cancel each other at point \(P\).
- **C.** \(B = \dfrac{\mu_0 I}{4R} + \dfrac{\mu_0 I}{2\pi R}\), because each straight section acts as a semi-infinite wire contributing to the field at point \(P\).
- **D.** \(B = \dfrac{\mu_0 I}{4R}\), because current elements in the straight sections are collinear with point \(P\), yielding \(d\vec{\ell} \times \hat{r} = 0\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118560/*