# Example: Ampere's Law

## Contents

### 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.

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.

Tyler Anderson