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
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. All information is derived from Matters and Interactions 4th Edition Textbook, chapter 19, section 8.
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{ {\vec{B}_{loop}} = \frac{{µ}_{0}}{4π} * \frac{2πIR^2}{[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{{µ}_{0}}{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{{µ}_{0}}{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 2π. The integration is as follows: [math]\displaystyle{ \int_0^{2π}\! {Δ\vec{B}_{z}} = \frac{{µ}_{0}}{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{{µ}_{0}}{4π} * \frac{I{Δ\vec{l}}}{R^2}=\frac{{µ}_{0}}{4π} * \frac{I{2πR}}{R^2} }[/math]=[math]\displaystyle{ \frac{{µ}_{0}}{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{{µ}_{0}}{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
Simple
A thin circular loop of wire of radius 5 cm consists of 100 turns of wire. If the conventional current of the wire is 4 amperes, what is the magnitude and direction of the magnetic field at the center of the loop?
The first step, and possibly the easiest step, is to determine the direction of [math]\displaystyle{ {\vec{B}} }[/math]. This is determined by the Right Hand Rule. Curving your fingers in the direction of the conventional current around the wire has the right thumb pointing out of the page, [math]\displaystyle{ {+z} }[/math] direction.
The next step is to calculate the value for [math]\displaystyle{ {\vec{|B|}} }[/math]. Using the formula for finding the magnetic field of a circular loop of a wire at the center, the value is calculated below:
[math]\displaystyle{ {\vec{|B|}} }[/math] = [math]\displaystyle{ \frac{{µ}_{0}}{4π} * \frac{2πINR^2}{z^3} }[/math] where z will also be the radius of the loop, since it is the distance to the center. Since there are multiple turns in the wire, the equation must be multiplied by the number of turns, N, in order to account for the magnetic field of the whole.
Plugging in the known values returns:
[math]\displaystyle{ {\vec{|B|}} }[/math] = [math]\displaystyle{ {1x10^{-7}} * \frac{2π(4A)(.05m)^2}{(.05m)^3} }[/math]
Solve.
[math]\displaystyle{ {\vec{|B|}} = 5.03x10^{-3} T }[/math]
The vector notation of the magnetic field is: [math]\displaystyle{ {\vec{B}} = (0,0, 5.03x10^{-3}) T }[/math]
Middling
a.)A loop of wire carries a conventional current of 2 amperes. The radius of the loop is .05 m. Calculate the magnitude of the magnetic field at a distance .25 m from the center of the loop along the axis of the loop. b.)Then, what would the magnitude of the magnetic field at the same location if there were 200 turns of wire in a coil instead of a loop?
a.) Since the observation location is at a distance .25 m from the center of the loop, the equation [math]\displaystyle{ {\vec{B}_{loop}} = \frac{{µ}_{0}}{4π} * \frac{2πIR^2}{[R^2 + z^2]^{3/2}} }[/math], will be used. R is the radius of the loop and z if the distance from the center of the loop.
Plugging in the known values, obtains:
[math]\displaystyle{ {\vec{B}_{loop}} = {1e-7} * \frac{2π(2A)(.05)^2}{[(.05)^2 + (.25)^2]^{3/2}} }[/math]
Solve:
[math]\displaystyle{ {\vec{B}_{loop}} = 1.89 e-7 T }[/math]
b.) Now that there are multiple turns of wire, the number of turns,N, must be multiplied to the [math]\displaystyle{ {\vec{B}_{loop}} }[/math] in order to account for the [math]\displaystyle{ {\vec{B}} }[/math] of the coil. [math]\displaystyle{ {\vec{B}_{coil}} = {\vec{B}_{loop}}*N }[/math]
Plugging in known values:
[math]\displaystyle{ {\vec{B}_{coil}} = 1.89 e-7 T * 200 }[/math]
Solve:
[math]\displaystyle{ {\vec{B}_{coil}} =3.78 e-5 T }[/math]
Difficult
A very long wire carrying a conventional current of 3.5 amperes is straight except for a circular loop of radius 5.8cm. Calculate the approximate magnitude and the direction of the magnetic field at the center of the loop.
This problem is more difficult in that you must calculate the magnetic field of a straight wire and the magnetic field of the circular loop. The summation of these two field obtains the net magnetic field, which the the field "felt" at the center of the loop.
Using the magnetic field equation of a wire [math]\displaystyle{ {\vec{|B|_{wire}}} }[/math] = [math]\displaystyle{ \frac{{µ}_{0}}{4π} * \frac{2I}{r} }[/math], one can calculate the magnitude of the magnetic field. Using the Right Hand Rule, the direction of the magnetic field is into the page, [math]\displaystyle{ {-z} }[/math] direction.
Plugging in known values:
[math]\displaystyle{ {\vec{|B|}_{wire}} }[/math] =[math]\displaystyle{ {1e-7}* \frac{2(3.5A)=}{(.058m)} }[/math]
Solve:
[math]\displaystyle{ {\vec{|B|}_{wire}} }[/math] = 1.21 e -5 T
so [math]\displaystyle{ {\vec{B}_{wire}} }[/math] = (0,0,-1.21e-5) T
Next to solve the magnetic field of the loop. Using the equation:
[math]\displaystyle{ {\vec{|B|}_{loop}} }[/math] = [math]\displaystyle{ \frac{{µ}_{0}}{4π} * \frac{2πIR^2}{z^3} }[/math] , the magnitude of the magnetic field of the loop is obtained. Using the Right Hang Rule, the direction is into the page, [math]\displaystyle{ {-z} }[/math] direction.
Plugging in known values:
[math]\displaystyle{ {\vec{|B|}_{loop}} }[/math] = [math]\displaystyle{ {1e-7} * \frac{2π(3.5A)(.058m)^2}{(.058m)^3} }[/math]
Solve:
[math]\displaystyle{ {\vec{|B|}_{loop}} }[/math] = 3.79 e -5 T
So [math]\displaystyle{ {\vec{B}_{loop}} }[/math] = (0,0, -3.79 e-5) T
[math]\displaystyle{ {\vec{B}_{net}} }[/math] = [math]\displaystyle{ {\vec{B}_{loop}} + {\vec{B}_{wire}} }[/math]
[math]\displaystyle{ {\vec{B}_{net}} }[/math] = (0,0,-3.79e-5) T + (0,0,-1.21e-5)T [math]\displaystyle{ {\vec{B}_{net}} }[/math] = (0,0,-5.00e-5) T
Connectedness
By calculating the magnetic field of a loop, one can calculate the magnetic field and current of coils and solenoids. This has resonance in the mechanical engineering world, my major, in that engineers are constantly determining the fields of current through transformers, which are a series of coils, in order to properly account for the distribution electricity to an area. Without this knowledge of the magnetic field of a loop, engineers will not be able to redistribute electricity and current to various locations.
Resources
Text
Matter and Interactions 4E, Chapter 19, Section 8.