Example Problem - Toroid: Difference between revisions

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By:  Kirk Betz
'''Problem:'''
'''Problem:'''
 
Concerning Ampere's law
Let look at a coil around a toroid shown in the figure below. The coil has N = 20 turns around the toroid. The toroid has an inside diameter of ID = 4 cm and an outside diameter OD = 5 cm. Determine the field intensity H along the mean path length within the toroid with a current i = 2.5 A.
Let look at a coil around a toroid shown in the figure below. The coil has N = 20 turns around the toroid. The toroid has an inside diameter of ID = 4 cm and an outside diameter OD = 5 cm. Determine the field intensity H along the mean path length within the toroid with a current i = 2.5 A.


[[Image:toroid.jpg]]
[[Image:toroid.jpg]]
''Figure created by Kirk Betz''


'''Solution:'''
'''Solution:'''


Do symmetry the magnetic field intensity Hm along a circular contour within the toroid is constant. We can find the radius by
Do symmetry the magnetic field intensity Hm along a circular contour within the toroid is constant. We can find the mean radius by
 


<math>r_m=(\frac{1}{2})\frac{OD + ID}{2} = 2.25\ cm</math>


<math>r_m=(\frac{1}{2})\frac{OD + ID}{2}</math>
Using the mean radius the mean path of length <math>l_m</math> can be calculated.




<math> l_m = { 2 \pi r_m} = 0.141\ </math>
<math> l_m = { 2 \pi r_m} = 0.141\ </math>
With Ampere's Law (below) the field intensity along the mean path can be Found.


<math>H_m=(\frac{Ni}{l_m})</math>
<math>H_m=(\frac{Ni}{l_m})</math>
Finally teh H_m can be calculated.


<math>H_m=\frac{20 x 2.5}{.141}= 354.6\ A /m)</math>
<math>H_m=\frac{20 x 2.5}{.141}= 354.6\ A /m)</math>
Since the width of the toroid is much smaller than the mean radius <math>r_m</math> we can assume a uniform <math>H_m</math> throughout teh cross-section of the toroid.
== Reviewed by ==
[[Will Griffith]]
[[Matthew Fetke]]
== Read by  ==
== Points for page ==
Worth 117
Kirk total for paper 117 + 50 conference = 167

Latest revision as of 16:41, 20 January 2010

By: Kirk Betz

Problem: Concerning Ampere's law Let look at a coil around a toroid shown in the figure below. The coil has N = 20 turns around the toroid. The toroid has an inside diameter of ID = 4 cm and an outside diameter OD = 5 cm. Determine the field intensity H along the mean path length within the toroid with a current i = 2.5 A.


Figure created by Kirk Betz


Solution:

Do symmetry the magnetic field intensity Hm along a circular contour within the toroid is constant. We can find the mean radius by


rm=(12)OD+ID2=2.25cm

Using the mean radius the mean path of length lm can be calculated.


lm=2πrm=0.141

With Ampere's Law (below) the field intensity along the mean path can be Found.

Hm=(Nilm)

Finally teh H_m can be calculated.

Hm=20x2.5.141=354.6A/m)

Since the width of the toroid is much smaller than the mean radius rm we can assume a uniform Hm throughout teh cross-section of the toroid.

Reviewed by

Will Griffith

Matthew Fetke

Read by

Points for page

Worth 117

Kirk total for paper 117 + 50 conference = 167