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title: "A planar conducting loop lies in the \\(xy\\)-plane and carries a steady counterclockwise current \\(I\\). The bottom portion of the loop is formed by the parabolic curve \\(y = d + kx^2\\), and the top is closed by the horizontal line segment \\(y = H\\), where \\(d\\), \\(k\\), and \\(H\\) are positive constants such that \\(H > d\\). Which of the following expressions correctly gives the \\(z\\)-component of the magnetic field, \\(B_z\\), at the origin \\((0,0)\\) due only to the parabolic segment of the loop?"
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url: "https://nerd-notes.com/ubq/124956/"
date_modified: "2026-09-28T14:12:12+00:00"
---

# A planar conducting loop lies in the \(xy\)-plane and carries a steady counterclockwise current \(I\). The bottom portion of the loop is formed by the parabolic curve \(y = d + kx^2\), and the top is closed by the horizontal line segment \(y = H\), where \(d\), \(k\), and \(H\) are positive constants such that \(H > d\). Which of the following expressions correctly gives the \(z\)-component of the magnetic field, \(B_z\), at the origin \((0,0)\) due only to the parabolic segment of the loop?

A planar conducting loop lies in the \(xy\)-plane and carries a steady counterclockwise current \(I\). The bottom portion of the loop is formed by the parabolic curve \(y = d + kx^2\), and the top is closed by the horizontal line segment \(y = H\), where \(d\), \(k\), and \(H\) are positive constants such that \(H > d\). Which of the following expressions correctly gives the \(z\)-component of the magnetic field, \(B_z\), at the origin \((0,0)\) due only to the parabolic segment of the loop?

![A Cartesian coordinate plane with horizontal axis labeled x and vertical axis labeled y, intersecting at the origin labeled (0,0). A closed planar loop lies entirely in the upper half-plane. The bottom boundary of the loop is an upward-opening parabolic curve labeled y = d + kx^2, with its vertex on the y-axis at (0,d). The top boundary of the loop is a horizontal straight line segment at height H, labeled y = H. The parabola and horizontal line intersect at two points, one at negative x and one at positive x. Two counterclockwise current arrows are shown on the perimeter of the loop: one arrow on the horizontal segment pointing to the left, and one arrow on the parabolic segment near its vertex pointing to the right, labeled I. A small dot marks the origin (0,0). No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790604732-iT3H1y.jpg)

- **A.** \(B_z = \dfrac{\mu_0 I}{4\pi} \int_{-\sqrt{\frac{H}{k}}}^{\sqrt{\frac{H}{k}}} \dfrac{-(d + kx^2)}{\left[x^2 + (d + kx^2)^2\right]^{3/2}} \, dx\)
- **B.** \(B_z = \dfrac{\mu_0 I}{4\pi} \int_{-\sqrt{\frac{H-d}{k}}}^{\sqrt{\frac{H-d}{k}}} \dfrac{kx^2 - d}{\left[x^2 + (d + kx^2)^2\right]^{3/2}} \, dx\)
- **C.** \(B_z = \dfrac{\mu_0 I}{4\pi} \int_{-\sqrt{\frac{H-d}{k}}}^{\sqrt{\frac{H-d}{k}}} \dfrac{\sqrt{1 + 4k^2 x^2}}{x^2 + (d + kx^2)^2} \, dx\)
- **D.** \(B_z = \dfrac{\mu_0 I}{4\pi} \int_{-\sqrt{\frac{H-d}{k}}}^{\sqrt{\frac{H-d}{k}}} \dfrac{-(d + 3kx^2)}{\left[x^2 + (d + kx^2)^2\right]^{3/2}} \, dx\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124956/*