Solveeit Logo

Question

Physics Question on Magnetism and matter

A magnetising field of 1600Am11600\, A\, m^{-1} produces a magnetic flux of 2.4?105Wb2.4 ? 10^{-5}\, Wb in an iron bar of cross-sectional area 0.2cm20.2\, cm^2. The susceptibility of an iron bar is

A

298

B

596

C

1192

D

1788

Answer

596

Explanation

Solution

Here, H=1600Am1,H = 1600\, A\, m^{-1}, Magnetic flux (ϕ)=2.4×105Wb\left(\phi\right)=2.4\times10^{-5}\,Wb A=0.2cm2=0.2×104m2A=0.2\,cm^{2}=0.2\times10^{-4}\,m^{2} B=ϕA=2.4×1050.2×104=1.2Wbm2\therefore B=\frac{\phi}{A}=\frac{2.4\times10^{-5}}{0.2\times10^{-4}}=1.2\,Wb\,m^{2} μ=BH=1.21600=7.5×104NA2\therefore \mu=\frac{B}{H}=\frac{1.2}{1600}=7.5\times10^{-4}\,N\,A^{-2} Hence, susceptibility χ=μμ01=7.5×1044π×1071=596\chi=\frac{\mu}{\mu_{0}}-1=\frac{7.5\times10^{-4}}{4\pi\times10^{-7}}-1=596