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## Q. 2.4

The magnetic circuit of Fig.2.10 has cast steel core. The cross-sectional area of the central limb is 800 $mm^{2}$ and that of each outer limb is 600$mm^{2}$. Calculate the exciting current needed to set up a flux of 0.8 mWb in the air gap.Neglect magnetic leakage and fringing.The magnetization characteristic of cast steel is given in Fig. 2.16.  ## Verified Solution

Air gap

$B_{g}=\frac{0\cdot 8}{800}\times \frac{10^{-3} }{10^{-6} } =1 T$ and $H_{g} =\frac{1}{4\pi\times 10^{-7} }$ AT/ m
$F_{g}=\frac{1}{4\pi \times 10^{-7} }\times 1\times 10^{-3} =796 AT$
Central limb

$B_{c}=B_{g}=1$T

From Fig. 2.16
$H_{c} =1000$ AT/ m
$F_{c} =1000\times 160\times 10^{-3} =160$ AT

Because of symmetry, flux divides equally between the two outer limbs. So

$\phi$(outer limb)$=0\cdot 8/ 2=0\cdot 4$mWb
$B$(outer limb)$=\frac{0\cdot 4\times 10^{-3} }{600\times 10^{-6} }=0\cdot 667$AT
$F$(outer limb)$=375\times 400\times 10^{-3} =150$ AT
$F (total)=796+ 160+ 150=1106$ AT
Exciting current $=1106/ 500=2\cdot 21$ A 