Example: Ampere's Law: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
(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...)
 
No edit summary
Line 5: Line 5:
===Solution===
===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>
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\approx .393m</math>

From Eq. 5-1 <math>H_m = \frac{N*i}{L_m} = \frac{50*5}{.393}\approx 636</math>

===Author===
Tyler Anderson

===Reviewers===

===Readers===

Revision as of 19:34, 18 January 2010

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 Hm is constant along the circular contour because of symmetry.

The mean path length is

The mean path length is

From Eq. 5-1

Author

Tyler Anderson

Reviewers

Readers