Circular Loop of Wire
Claimed by Rachel B.
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.
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.
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].
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
Examples
Be sure to show all steps in your solution and include diagrams whenever possible