A magnetising field of $1600 \frac{A}{m}$ produces a magnetic flux $2.4 X 10^{- 5} W b$ in a bar of iron of cross section $0.2 c m^{2}$ .Calculate permeability and susceptibility of the bar:

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Solution

Magnetic Induction $B = \frac{F l u x}{A r e a} = \frac{2.4 \times 10^{- 5}}{0.2 \times 10^{- 4}}$
$= 1.2 T$
Permeability $μ = \frac{B}{H} = \frac{1.2}{7.5 \times 10^{- 4}} = 0.75 \times 10^{- 3}$
Susceptibility $X = 624$
$\frac{u}{u_{0}} = 1 + X$
$⟹ \frac{0.75 \times 10^{- 3}}{4 π \times 10^{- 7}} = 1 + X ⟹ X = 624$

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A magnetising field of 1600Amproduces a magnetic flux 2.4X10−5Wb in a bar of iron of cross section 0.2cm2 .Calculate permeability and susceptibility of the bar:

Magnetic Induction B=FluxArea=2.4×10−50.2×10−4=1.2TPermeability μ=BH=1.27.5×10−4=0.75×10−3Susceptibility X=624uu0=1+X⟹0.75×10−34π×10−7=1+X⟹X=624

A magnetising field of $1600 \frac{A}{m}$ produces a magnetic flux $2.4 X 10^{- 5} W b$ in a bar of iron of cross section $0.2 c m^{2}$ .Calculate permeability and susceptibility of the bar: