Example: Ampere's Law: Difference between revisions
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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... |
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===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 20:34, 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