# Discussion: The Hoola-Field

A ring is of the shape of a hoola-hoop of negligible thickness. A ring of radius $$R$$ carries a current $$I$$. Prove that the magnetic field at a given point in the plane of the ring at a distance $$a$$ from the center, due to the magnetic field of the ring, is $B = \dfrac {\mu_0}{2\pi} \cdot IR \cdot \displaystyle\int_{0}^{\pi} \dfrac {R - a \cos \theta}{\sqrt{\left( a^2 + R^2 - 2aR \cos \theta \right)^3}} \, \mathrm{d}\theta.$

Note: This problem originally appeared on the IPhOO.

Note by Ahaan Rungta
4 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. list

1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Check the solution to : Rotating Rod In Magnetic Field

- 4 years, 3 months ago