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## Q. 10.3

Consider a Micrometals toroidal iron powder core with 52/μ75 material. The initial permeability versus magnetic field intensity curve is given in Figure 10.5.

## Verified Solution

An inductor was constructed using a toroid with core length l$_{c}$ = 4.23 cm, core cross-sectional area $A_{c}$ = 0.179 $cm^{2}$ and with N = 72 turns. Calculate the inductance L and the effective inductance $L_{\mathrm{eff}}$ as a function of current for the range 1 to 4 A.

For H = 30 Oe, μ$_{r}$ is read from Figure 10.5:

i = $\frac{Hl_{c}}{N}= \frac{(2387)(4.23\times10^{-2})}{72} = 1.403$ A

L(i) = $\mu_{r}(i)\times \mu_{0}\times N^{2}\times \frac{A_{c}}{l_{c}} = (56.1)(4\pi\times10^{-7})(72)^{2}\left(\frac{0.179\times 10^{-4}}{4.23\times 10^{-2}} \right) = 154.7 \mu$H

L$_{\mathrm{eff}}=L(i)+ i\frac{\Delta L}{\Delta i} = (154.7\times10^{-6})+ (1.403)\frac{(154.7-176.7)\times10^{-6}}{(1.403-0.935)} = 88.7 \mu$H

The full set of calculations are summarized in Table 10.1, and L and L$_{\mathrm{eff}}$ are plotted in Figure 10.6.

 Table 10.1 Calculations for inductance and effective inductance H (Oe) H (A/m) $\mu_{r}$ i (A) L ($\mu$H) $L_{\mathrm{eff}}(\mu H)$ 1 80 75 0.047 206.8 206.8 1.4 111 75 0.065 206.8 206.8 2 159 74.9 0.094 206.5 205.6 3 239 74 0.14 204.0 196.6 5 398 73.5 0.234 202.6 199.2 7 557 72.5 0.327 199.9 190.2 10 796 71.1 0.468 196.0 183.1 14 1114 68.4 0.655 188.6 162.5 20 1592 64.1 0.935 176.7 137.2 30 2387 56.1 1.403 154.7 88.7 50 3979 44 2.338 121.3 37.9 70 5571 35 3.273 96.5 9.6 100 7958 27 4.675 74.4 0.9