## Materials Science and Engineering: An Introduction

Given: Bar of iron-silicon alloy within a coil of wire that is 0.40 m long, having 50 turns with a current of 0.1 A. Required: B field within the bar Solution: Using Equation 20.1: $H = \frac{NI}{l} = \frac{(50 turns)(0.1 A)}{0.40 m} = 12.5 A/m$ From the given Figure 20.29, the B value is about 1.07 tesla.