## Materials Science and Engineering: An Introduction

$M = 7.51 A/m$
Given: 0.25 m long coil with 400 turns Current = 15 A H = 24,000 A-turns-m $χ_{m} = 3.13 \times 10^{-4}$ Required: magnetization Solution: Using Equation 20.6: $M = χ_{m}H = (3.13 \times 10^{-4})(24,000 A-turns/m) = 7.51 A/m$