Answer
$M = 7.51 A/m$
Work Step by Step
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:
$M = χ_{m}H = (3.13 \times 10^{-4})(24,000 A-turns/m) = 7.51 A/m$
