Circular Loop of Wire

Claimed by Rachel B.

Circular Loop of Wire

This page is about calculating the magnetic field of a circular loop of wire. It uncovers the importance of this calculation as well as the formulas and examples associated with it.

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The Main Idea

State, in your own words, the main idea for this topic

A Mathematical Model

What are the mathematical equations that allow us to model this topic. For example [math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net} }[/math] where p is the momentum of the system and F is the net force from the surroundings.

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Steps to Solving

Step 1

"Cut Up Into Pieces" The first step in solving the magnetic field of a circular loop is to cut up the loop into pieces in order to understand the area better.

In this picture, [math]\displaystyle{ {Δ\vec{l}} }[/math] is the length of the short sections to be cut. The angle between [math]\displaystyle{ {Δ\vec{l}} }[/math] and [math]\displaystyle{ {Δ\vec{r}} }[/math] are always 90°, perpendicular to each other. By cutting the loop into sections, [math]\displaystyle{ {Δ\vec{l}} }[/math], the direction [math]\displaystyle{ {Δ\vec{B}} }[/math] can be found.

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Circular Loop of a Wire.

This picture shows the segment of wire cut from the whole. Here, you can see that [math]\displaystyle{ {\vec{l}} }[/math] is perpendicular to [math]\displaystyle{ {\hat{r}} }[/math] at every point in the loop. [math]\displaystyle{ {Δ\vec{B}} }[/math] is also perpendicular to [math]\displaystyle{ {Δ\vec{l}} }[/math] and [math]\displaystyle{ {\vec{r}} }[/math].

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Cut Up Circular Loop.

Step 2

The second step to solving the magnetic field of a circular loop of wire is to 'Write an Expression for One Piece'. Symmetry greatly simplifies the expression as [math]\displaystyle{ {Δ\vec{B}}_{x} }[/math] and [math]\displaystyle{ {Δ\vec{B}}_{y} }[/math] will cancel each other out due to the equal values on both sides of the loop. However, [math]\displaystyle{ {Δ\vec{B}_{z}} }[/math] will be the only contributing factor, and we can assume that this value is the same around the entire wire. This allows for the calculation portion to be relatively simple due to the fact that calculating the magnetic field of one section of the loop is the same for the entire loop.

[math]\displaystyle{ {\vec{r}}: {\vec{r}} = (obs. location) - (source) = (0,0,z) - (0,R,0) = (0,-R,z) }[/math]

Magnitude of [math]\displaystyle{ {\vec{r}}: r = [R^2 + z^2]^{1/2} }[/math]

Unit vector [math]\displaystyle{ {\hat{r}}: \frac{\vec{r}}{\|r\|} }[/math]

The location of the piece is dependent upon θ, which will be the integrable factor. Therefore [math]\displaystyle{ {Δ\vec{l}}: {\|\vec{l}\|}= (-RΔθ,0,0) }[/math]

The magnetic field due to one piece is [math]\displaystyle{ {Δ\vec{B}} = \frac{µ0}{4π} * I* \frac{(-RΔθ,0,0)×(0,-R,z)}{[R^2 + z^2]^{1/2}} }[/math]

By only taking the z-component of the solved cross product, the equation derived is [math]\displaystyle{ {Δ\vec{B}_{z}} = \frac{µ0}{4π} * \frac{IR^2Δθ}{[R^2 + z^2]^{3/2}} }[/math]

Step 3

The third step requires the summation of all the pieces of the loop and their contributions. This involves integrating over the circumference of the loop, which is 2. The integration is as follows: [math]\displaystyle{ \int_0^{2π}\! {Δ\vec{B}_{z}} = \frac{µ0}{4π} * \frac{IR^2Δθ}{[R^2 + z^2]^{3/2}} }[/math] From here the formula for the magnetic field of a loop is derived.

Examples

Be sure to show all steps in your solution and include diagrams whenever possible