Magnetic Field of a Toroid Using Ampere's Law - Revision history

Nkannan6 at 03:12, 19 April 2018

2018-04-19T03:12:00Z

← Older revision		Revision as of 23:12, 18 April 2018
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	~~Claimed by Kevin McGorrey~~

	This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.		This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

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	===Geometry of a Toroid===		===Geometry of a Toroid===
	[[File:Geometry Fig.JPG \|frame\|right\|alt=Alt text\| A toroid's geometry.]]		[[File:Geometry Fig.JPG \|frame\|right\|alt=Alt text\| A toroid's geometry.]]
	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 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. Mathematically, it can be described as an object or surface generated by revolving a closed plane around an external axis that is parallel to it so it does not intersect. Toroids are commonly used as electronic compartments since they generate a magnetic field or frequency depending on the material of the ring and the type of wire and the number of loops the wire is wound around the ring.

	===A Mathematical Model===		===A Mathematical Model===
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	====Another Example====		====Another Example====
	(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.		(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.
			<br/>
	[[File:ToroidalExample.JPG]]		[[File:ToroidalExample.JPG]]

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	<br /> <div style="text-align: center;"><math>{B2πr ={μ_{0}NI}{2πr}}</math></div>		<br /> <div style="text-align: center;"><math>{B2πr ={μ_{0}NI}{2πr}}</math></div>
	<br /> <div style="text-align: center;"><math>{B ={μ_{0}NI}}</math></div>		<br /> <div style="text-align: center;"><math>{B ={μ_{0}NI}}</math></div>
			The final answer is then <math>{B ={μ_{0}NI}}</math>.

	== See also ==		== See also ==

Nkannan6 at 03:02, 19 April 2018

2018-04-19T03:02:56Z

Nkannan6 at 02:36, 19 April 2018

2018-04-19T02:36:57Z

← Older revision		Revision as of 22:36, 18 April 2018
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	====Another Example====		====Another Example====
	(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.		(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.
	[[File:ToroidalExample.JPG]]		[[File:ToroidalExample.JPG\| right\|frame\|none\|alt=Alt text\| The toroidal coil]]
	<br>		<br />
	Solution:		Solution:
	</br>		We determine the magnitude of the magnetic B-field by using Ampere’s law:
			<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>
			We chose a circle running through the center of the hole of the toroid (the blue line in the figure) for Ampere’s loop. The length of this circle is:
			<br /> <div style="text-align: center;"><math>l = 2\pi r<math/>
			The total current flowing through the circle bordered by the Ampere’s loop is:
			<br /> <div style="text-align: center;"><math>I_c = NlI = N(2\pi r)I<math/>
			The magnetic field vector <math>B<math/> is parallel to the vector of the length element <math>dl<math/> all the way around the integration path. Thus we can express the right-hand side of Ampere’s law as:
			<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = B2\pi r<math/>
			The left-hand and the right-hand sides of the Ampere’s law must be equal and we can express the magnitude of the <math>B</math>-field as:
			<br /> <div style="text-align: center;"><math> B2\pi r = N(2\pi r)I(\mu _0)<math/>
			<br /> <div style="text-align: center;"><math> B = \mu _0NI<math/>


	== See also ==		== See also ==
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	*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/		*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/
	*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf		*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf
			*http://physicstasks.eu/1784/magnetic-field-inside-a-toroid

	==References==		==References==

Nkannan6: /* Another Example */

2018-04-19T02:19:30Z

Another Example

← Older revision		Revision as of 22:19, 18 April 2018
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	====Another Example====		====Another Example====
	(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current I flows through it.		(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.
			[[File:ToroidalExample.JPG]]
			<br>
			Solution:
			</br>

	== See also ==		== See also ==

Nkannan6 at 02:16, 19 April 2018

2018-04-19T02:16:53Z

← Older revision		Revision as of 22:16, 18 April 2018
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	====Another Example====		====Another Example====
	(a.) ~~Consider~~ a ~~toroid which consists~~ of <math>N</math> turns~~, as shown in the figure below. Find the magnetic field everywhere~~.		(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current I flows through it.
	~~[[File:Example.jpg]]~~

Nkannan6 at 00:33, 19 April 2018

2018-04-19T00:33:09Z

← Older revision		Revision as of 20:33, 18 April 2018
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	(d.) Consider a z-axis centered Amperian loop far above the toroid <math>z >> R_{o}</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field far above the toroid.		(d.) Consider a z-axis centered Amperian loop far above the toroid <math>z >> R_{o}</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field far above the toroid.
	[[File:Mid - Diff Example Sol cd.JPG \| center\|frame\|none\|alt=Alt text\| The solution for the difficult example.]]		[[File:Mid - Diff Example Sol cd.JPG \| center\|frame\|none\|alt=Alt text\| The solution for the difficult example.]]
			<br />

			====Another Example====
			(a.) Consider a toroid which consists of <math>N</math> turns, as shown in the figure below. Find the magnetic field everywhere.
			[[File:Example.jpg]]



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	*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf		*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf
	*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/		*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/
			*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf

	==References==		==References==

Ycho78 at 00:16, 28 November 2016

2016-11-28T00:16:06Z

← Older revision		Revision as of 20:16, 27 November 2016
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	Claimed by Kevin McGorrey ~~Claimed by Yeon Jae Cho (2016)~~		Claimed by Kevin McGorrey

	This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.		This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

Ycho78 at 00:13, 28 November 2016

2016-11-28T00:13:34Z

← Older revision		Revision as of 20:13, 27 November 2016
Line 1:		Line 1:
	Claimed by Kevin McGorrey		Claimed by Kevin McGorrey Claimed by Yeon Jae Cho (2016)

	This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.		This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

Kmcgorrey3: /* External links */

2015-12-06T02:41:44Z

External links

← Older revision		Revision as of 22:41, 5 December 2015
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	===External links===		===External links===
			*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
	~~Internet resources on this topic~~		*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf
			*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/

	==References==		==References==

Kmcgorrey3: /* References */

2015-12-06T02:37:38Z

References

← Older revision		Revision as of 22:37, 5 December 2015
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	==References==		==References==
			Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. Pg 888-889.
	~~This section contains the the references you used while writing this page~~		<br />
			Fall 2014 Test 4 from Phys 2212.
	[[Category:Maxwell's Equations]]		[[Category:Maxwell's Equations]]

← Older revision		Revision as of 23:02, 18 April 2018
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	====Another Example====		====Another Example====
	(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.		(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.
	[[File:ToroidalExample.JPG~~\| right\|frame\|none\|alt=Alt text\| The toroidal coil~~]]		[[File:ToroidalExample.JPG]]
	~~<br />~~
	Solution:		Solution:
	We determine the magnitude of the magnetic B-field by using Ampere’s law:		<br />We determine the magnitude of the magnetic B-field by using Ampere’s law
	<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>		<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>
	We chose a circle running through the center of the hole of the toroid ~~(the blue line in the figure)~~ for Ampere’s loop. The length of this circle is:		We chose a circle running through the center of the hole of the toroid for Ampere’s loop. The length of this circle is
	<br /> <div style="text-align: center;"><math>l = 2\pi r<math/>		<br /> <div style="text-align: center;"><math>{l ={2πr}}</math></div>
	The total current flowing through the circle bordered by the Ampere’s loop is:		The total current flowing through the circle bordered by the Ampere’s loop is
	<br /> <div style="text-align: center;"><math>~~I_c~~ = NlI = N(2\pi r)I<math/>		<br /> <div style="text-align: center;"><math>{I_{c} = NlI = {2πr}NI}</math></div>
	The magnetic field vector <math>B<math/> is parallel to the vector of the length element <math>dl<math/> all the way around the integration path. Thus we can express the right-hand side of Ampere’s law as:		The magnetic field vector <math>B</math> is parallel to the vector of the length element <math>dl</math> all the way around the integration path. Thus we can express the right-hand side of Ampere’s law as
	<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = ~~B2\~~pi r<math/>		<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = B{2πr}}</math></div>
	The left-hand and the right-hand sides of the Ampere’s law must be equal and we can express the magnitude of the <math>B</math>-field as:		The left-hand and the right-hand sides of the Ampere’s law must be equal and we can express the magnitude of the <math>B</math>-field as
	<br /> <div style="text-align: center;"><math> B2\pi r = N(2\pi r~~)I(\mu _0)~~<math/>		<br /> <div style="text-align: center;"><math>{B2πr ={μ_{0}NI}{2πr}}</math></div>
	<br /> <div style="text-align: center;"><math> B = \mu ~~_0NI~~<math/>		<br /> <div style="text-align: center;"><math>{B ={μ_{0}NI}}</math></div>