Example: Ampere's Law - Revision history

Tyler.anderson: /* Solution */

2010-01-19T03:43:14Z

Solution

← Older revision		Revision as of 20:43, 18 January 2010
Line 11:		Line 11:
	The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>		The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>

	The mean path length encloses the current i N-times.[[Image:~~toroidish~~.~~jpg~~\|thumb\|widthpx\| ]]		The mean path length encloses the current i N-times.[[Image:Toroidish.JPG\|thumb\|widthpx\| ]]

	From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>		From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>

Tyler.anderson: /* Solution */

2010-01-19T03:42:57Z

Solution

← Older revision		Revision as of 20:42, 18 January 2010
Line 11:		Line 11:
	The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>		The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>

	The mean path length encloses the current i N-times.[[Image:toroidish\|thumb\|widthpx\| ]]		The mean path length encloses the current i N-times.[[Image:toroidish.jpg\|thumb\|widthpx\| ]]

	From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>		From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>

Tyler.anderson: /* Solution */

2010-01-19T03:42:48Z

Solution

← Older revision		Revision as of 20:42, 18 January 2010
Line 10:		Line 10:

	The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>		The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>

			The mean path length encloses the current i N-times.[[Image:toroidish\|thumb\|widthpx\| ]]

	From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>		From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>

Tyler.anderson: /* Solution */

2010-01-19T03:35:57Z

Solution

← Older revision		Revision as of 20:35, 18 January 2010
Line 12:		Line 12:

	From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>		From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>

			One can assume a uniform H<sub>m</sub> throughout the cross-section of the toroid because the width is smaller than the mean radius.

	===Author===		===Author===

Tyler.anderson at 03:34, 19 January 2010

2010-01-19T03:34:26Z

@@ Line 5: / Line 5: @@
 ===Solution===
-The magnetic field intensity H<sub>m</sub> is constant along the circular contour because of symmetry. The mean path length is <math>r_m=\frac{1}{2}*(\frac{OD+ID}{2})</math>
+The magnetic field intensity H<sub>m</sub> is constant along the circular contour because of symmetry.
 The mean path length is <math>L_m = 2*\pi*r_m = 2\pi*6.25</math>

Tyler.anderson: New page: ===Problem Statement=== Consider a toroid which has N = 50 turns. The toroid has an ID = 10cm and OD = 15cm. For a current i = 5A calculate the field intensity H along the mean path lengt...

2010-01-19T03:28:56Z

New page: ===Problem Statement=== Consider a toroid which has N = 50 turns. The toroid has an ID = 10cm and OD = 15cm. For a current i = 5A calculate the field intensity H along the mean path lengt...

New page

===Problem Statement===
Consider a toroid which has N = 50 turns. The toroid has an ID = 10cm and OD = 15cm. For a current i = 5A calculate the field intensity H along the mean path length within the toroid.
(this problem is similar to Example 5-1 in the text book)

===Solution===

The magnetic field intensity H<sub>m</sub> is constant along the circular contour because of symmetry. The mean path length is <math>r_m=\frac{1}{2}*(\frac{OD+ID}{2})</math>
The mean path length is <math>L_m = 2*\pi*r_m = 2\pi*6.25</math>

← Older revision		Revision as of 20:43, 18 January 2010
Line 11:		Line 11:
	The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>		The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>

	The mean path length encloses the current i N-times.[[Image:~~toroidish~~.~~jpg~~\|thumb\|widthpx\| ]]		The mean path length encloses the current i N-times.[[Image:Toroidish.JPG\|thumb\|widthpx\| ]]

	From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>		From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>

← Older revision		Revision as of 20:42, 18 January 2010
Line 11:		Line 11:
	The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>		The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>

	The mean path length encloses the current i N-times.[[Image:toroidish\|thumb\|widthpx\| ]]		The mean path length encloses the current i N-times.[[Image:toroidish.jpg\|thumb\|widthpx\| ]]

	From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>		From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>

← Older revision		Revision as of 20:42, 18 January 2010
Line 10:		Line 10:

	The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>		The mean path length is <math>L_m = 2\pir_m = 2\pi*6.25\approx .393m</math>

			The mean path length encloses the current i N-times.[[Image:toroidish\|thumb\|widthpx\| ]]

	From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>		From Eq. 5-1 <math>H_m = \frac{Ni}{L_m} = \frac{505}{.393}\approx 636</math>