Magnetic Field of a Toroid Using Ampere's Law: Difference between revisions
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<br />The magnetic field, <math>{\vec{B}}</math>, is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to <math>{d\vec{l}}</math>). The path of a toroid is circular, so <math>{\oint\,d\vec{l}}</math> is equal to <math>{2πr}</math>. Therefore, the path integral of the magnetic field is equal to <math>{B2πr}</math>. The amount of current piercing the soap film (i.e. <math>{∑I_{inside path}}</math>) is <math>NI</math>, where <math>N</math> is the number of piercings (i.e. turns in the coil) and <math>I</math> is the current. Ampere's law is now this: | <br />The magnetic field, <math>{\vec{B}}</math>, is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to <math>{d\vec{l}}</math>). The path of a toroid is circular, so <math>{\oint\,d\vec{l}}</math> is equal to <math>{2πr}</math>. Therefore, the path integral of the magnetic field is equal to <math>{B2πr}</math>. The amount of current piercing the soap film (i.e. <math>{∑I_{inside path}}</math>) is <math>NI</math>, where <math>N</math> is the number of piercings (i.e. turns in the coil) and <math>I</math> is the current. Ampere's law is now this: | ||
<br /> <div style="text-align: center;"><math>{B2πr = μ_{0}NI}</math></div> | <br /> <div style="text-align: center;"><math>{B2πr = μ_{0}NI}</math></div> | ||
<br /> From this, we can solve for the magnetic field: | <br /> From this, we can solve for the magnetic field for a toroid: | ||
<br /> <div style="text-align: center;"><math>{B = \frac{μ_{0}NI}{2πr}}</math></div> | <br /> <div style="text-align: center;"><math>{B = \frac{μ_{0}NI}{2πr}}</math></div> | ||
Revision as of 21:45, 1 December 2015
Claimed by Kevin McGorrey
This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.
Magnetic Field of a Toroid using Ampere's Law
Using Ampere's Law simplifies finding the magnetic field of a toroid.
Geometry of a Toroid
A toroid is essentially a solenoid whose ends meet. It is shaped like a doughnut (the base shape does not need to be a circle; it can be a triangle, square, quadrilateral, etc.), is symmetrical around an axis of rotation, and contains N loops around a closed, circular path with a radius of r inside of its loop.
A Mathematical Model
First we start with solving the path integral from Ampere's law:
The magnetic field, [math]\displaystyle{ {\vec{B}} }[/math], is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to [math]\displaystyle{ {d\vec{l}} }[/math]). The path of a toroid is circular, so [math]\displaystyle{ {\oint\,d\vec{l}} }[/math] is equal to [math]\displaystyle{ {2πr} }[/math]. Therefore, the path integral of the magnetic field is equal to [math]\displaystyle{ {B2πr} }[/math]. The amount of current piercing the soap film (i.e. [math]\displaystyle{ {∑I_{inside path}} }[/math]) is [math]\displaystyle{ NI }[/math], where [math]\displaystyle{ N }[/math] is the number of piercings (i.e. turns in the coil) and [math]\displaystyle{ I }[/math] is the current. Ampere's law is now this:
From this, we can solve for the magnetic field for a toroid:
A Computational Model
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