Answer
$B = 0.0302 tesla $
Work Step by Step
Given:
0.25 m long coil with 400 turns
Current = 15 A
H = 24,000 A-turns-m
Required:
flux density inside a bar of chromium positioned within the coil
Solution:
Using a combination of Equation 20.5 and 20.6 and using the magnetic susceptibility of $χ_{m} = 3.13 \times 10^{-4} $ from Table 20.2, it follows:
$B = μ_{0}H + μ_{0}M = μ_{0}H + μ_{0}χ_{m}H = μ_{0}H (1+χ_{m})\\ = (1.257 \times 10^{-6} H/m)(24,000 A-turns/m)(1 + 3.13 \times 10^{-4})= 0.0302 tesla $
