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## Q. 7.4

Figure 7.8(a) shows a rectangular magnetic core with an air-gap. Find the exciting current needed to establish a flux density of $B_{g}$ = 1.2 T in the air-gap. Given N = 400 turns and $μ_{r}$(iron) = 4000.

## Verified Solution

It is a simple series magnetic circuit with its analog shown in Fig. 7.8(b).

Core length $l_{c}$ = 2 [(20 – 4) + (16 – 4)] – 0.2 = 55.8 cm

Cross-sectional area of core $A_{c}$ = 16 cm²

Core reluctance      $\mathcal{R}_{c} = \frac{55.8 × 10^{-2} }{4000 × 4π × 10^{-7} × 16 × 10^{-4}} = 0.694 × 10^{5}$ AT/Wb

= 0.694 × $10^{5}$ AT/Wb

Air-gap length $l_{g}$ = 0.2 cm
Area of air-gap $A_{g}$ = 16 cm²

Air gap reluctance    $\mathcal{R}_{g} = \frac{0.2 × 10^{-2} }{ 4π × 10^{-7} × 16 × 10^{-4}}$

= 9.95 × $10^{5}$ AT/Wb

$\mathcal{R}(total) = \mathcal{R}_{c} + \mathcal{R}_{g}$

= 0.694 × $10^{5}$ + 9.95 × $10^{5}$ = 10.64 × $10^{5}$ AT/Wb

Flux in the magnetic circuit, $\phi$ = B A = 1.2 × 16 × $10^{-4}$ = 1.92 mWb

Now            $Ni = \mathcal{F} = \phi \mathcal{R}$

= 1.92 × $10^{-3}$ × 10.64 × $10^{5}$

= 2043 AT

∴            Exciting current     i      =  $\frac{2043}{400}$ = 5.11 A

It is seen above that

$\mathcal{R} _{g} / \mathcal{R} _{c}$ = 14.34

Therefore for simplicity of computation, $\mathcal{R} _{c}$ (magnetic core reluctance) may be altogether neglected. Then

i = 1.2 × 16 × $10^{-4}$ × 9.95 × $10^{5}$ /400
= 4.77 A

This simplification has caused an error of 6.6% which can be easily tolerated in magnetic circuit calculations. 