Magnetic Field of a Toroid Using Ampere's Law: Difference between revisions

From Physics Book
Jump to navigation Jump to search
Line 21: Line 21:
==Examples==
==Examples==
===Simple===
===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?
A toroid frame is made out of plastic of small square cross section and tightly wrapped uniformly with 100 turns of wire, so that the magnetic field has essentially the same magnitude throughout the plastic (radius R of the curved part is much larger than cross section width w). With a current of 2 A and radius of 5 m, what is the magnetic field inside the plastic.


===Middling - Difficult===
===Middling - Difficult===

Revision as of 21:52, 5 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 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

A toroid frame is made out of plastic of small square cross section and tightly wrapped uniformly with 100 turns of wire, so that the magnetic field has essentially the same magnitude throughout the plastic (radius R of the curved part is much larger than cross section width w). With a current of 2 A and radius of 5 m, what is the magnetic field inside the plastic.

Middling - Difficult

The toroid shown in the diagram has an inner radius of [math]\displaystyle{ R_{i} }[/math] and an outer radius of [math]\displaystyle{ R_{o} }[/math] and is centered at the origin in the diagram. The z-axis passes through the center of the doughnut hole. This toroid is wrapped with [math]\displaystyle{ N }[/math] loops of current [math]\displaystyle{ I }[/math] flowing up the outside surface of the toroid, radially inward, down the inner surface, and then radial outward. Assume that the magnetic field produced by this toroid has the form [math]\displaystyle{ \vec{B} = B(r,z)\hat{φ} }[/math] at every point in space where [math]\displaystyle{ r }[/math] is the perpendicular distance from the z-axis and [math]\displaystyle{ \hat{φ} }[/math] is a unit vector which "curls" around the z-axis, i.e., it is always tangent to any circle with rotational symmetry around the z-axis.


Middling

(a.) Consider a z-axis centered Amperian loop in the plane of the toroid, at [math]\displaystyle{ z = 0 }[/math], with a radius [math]\displaystyle{ r \lt R_{i} }[/math] and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.


(b.) Consider a z-axis centered Amperian loop in the plane of the toroid, at [math]\displaystyle{ z = 0 }[/math], with a radius [math]\displaystyle{ r \gt R_{o} }[/math] and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.


Difficult

(c.) Consider a z-axis centered Amperian loop in the plane of the toroid, at [math]\displaystyle{ z = 0 }[/math], with a radius [math]\displaystyle{ R_{i} \lt r \lt R_{o} }[/math] and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.


(d.) Consider a z-axis centered Amperian loop far above the toroid [math]\displaystyle{ z \gt \gt R_{o} }[/math], with a radius [math]\displaystyle{ R_{i} \lt r \lt R_{o} }[/math] and use it to find the magnitude of the magnetic field far above the toroid.

Connectedness

  1. How is this topic connected to something that you are interested in?
  2. How is it connected to your major?
  3. Is there an interesting industrial application?

See also

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

Further reading

Books, Articles or other print media on this topic

External links

Internet resources on this topic

References

This section contains the the references you used while writing this page