Magnetic Field of a Toroid Using Ampere's Law

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 since using the Biot-Savart law would be extremely difficult due to having to integrate over all the current elements in the 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:

[math]\displaystyle{ {\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside path}} }[/math]

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:

[math]\displaystyle{ {B2πr = μ_{0}NI} }[/math]

From this, we can solve for the magnetic field for a toroid:

[math]\displaystyle{ {B = \frac{μ_{0}NI}{2πr}} }[/math]

Examples

Simple

Use Ampere's law to calculate the magnetic field inside a toroid with 100 turns that has an inner radius of 5 m, an outer radius of 10 m, and carries a current of 2 A?

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