Circular Loop of Wire: Difference between revisions
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===Magnetic Field at the Center of the Loop=== | ===Magnetic Field at the Center of the Loop=== | ||
In order to find the magnetic field at the center of the loop, take the cross product of <math>{Δ\vec{l}}</math> and <math>{\hat{r}}</math>. This will leave the equation to be <math>{Δ\vec{B}_{loop}} = | In order to find the magnetic field at the center of the loop, take the cross product of <math>{Δ\vec{l}}</math> and <math>{\hat{r}}</math>. This will leave the equation to be <math>{Δ\vec{B}_{loop}} = {Δ\vec{B}_{z}} = ∑\frac{µ0}{4π} * \frac{I{Δ\vec{l}}}{R^2}=\frac{µ0}{4π} * \frac{I{2πR}}{R^2}</math>=<math> \frac{µ0}{4π} * \frac{2πI}{R}</math> | ||
===Magnetic Field Far from the Loop=== | ===Magnetic Field Far from the Loop=== | ||
Revision as of 17:22, 29 November 2015
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
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.
A Mathematical Model
Equations associated with calculating the magnetic field of a loop.
Magnetic Field of a Loop
The formula to find the magnetic field of a loop is [math]\displaystyle{ {Δ\vec{B}_{loop}} = \frac{µ0}{4π} * \frac{2πIR^2Δθ}{[R^2 + z^2]^{3/2}} }[/math], where R is the radius of the loop, I is the conventional current throughout the loop , and z is the distance from the center of the loop along the axis.
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] where [math]\displaystyle{ {µ}_{0} }[/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 (see Mathematical Model)
Special Cases
Magnetic Field at the Center of the Loop
In order to find the magnetic field at the center of the loop, take the cross product of [math]\displaystyle{ {Δ\vec{l}} }[/math] and [math]\displaystyle{ {\hat{r}} }[/math]. This will leave the equation to be [math]\displaystyle{ {Δ\vec{B}_{loop}} = {Δ\vec{B}_{z}} = ∑\frac{µ0}{4π} * \frac{I{Δ\vec{l}}}{R^2}=\frac{µ0}{4π} * \frac{I{2πR}}{R^2} }[/math]=[math]\displaystyle{ \frac{µ0}{4π} * \frac{2πI}{R} }[/math]
Magnetic Field Far from the Loop
When "z",or the distance from the center of the loop, is much larger than the "R", the radius of the loop, then the magnetic field equation is [math]\displaystyle{ \frac{µ0}{4π} * \frac{2πIR^2}{z^3} }[/math]
Right Hand Rule
For loops, there is a special Right Hand Rule. Curl the right finger in the direction of the conventional current and the direction of the magnetic field should point in the direction of your thumb. This gives the same result as the result of the cross product of [math]\displaystyle{ {Δ\vec{l}} }[/math] and [math]\displaystyle{ {\vec{r}} }[/math].
Examples
Be sure to show all steps in your solution and include diagrams whenever possible