Example: Ampere's Law

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_m 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.

File:Toroidish.jpg

From Eq. 5-1 $H_{m}={\frac {N*i}{L_{m}}}={\frac {50*5}{.393}}\approx 636$

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

Author

Tyler Anderson