Example: Ampere's Law

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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 r_m=\frac{1}{2}*(\frac{OD+ID}{2})

The mean path length is L_m = 2*\pi*r_m = 2\pi*6.25\approx .393m

The mean path length encloses the current i N-times.
Toroidish.JPG

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

One can assume a uniform Hm throughout the cross-section of the toroid because the width is smaller than the mean radius.

Author

Tyler Anderson

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