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Numericals · Q14

Q.A rod of magnetic material of cross section 0.25 cm2^2 is located in a 4000 A m−1^{-1} magnetising field. The magnetic flux passing through the rod is 25×10−625\times10^{-6} Wb. Find

(a) the relative permeability
(b) the magnetic susceptibility and
(c) the magnetisation of the rod.
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Converting the cross-section, A=0.25 cm2=0.25×10−4 m2=2.5×10−5 m2A=0.25\ \text{cm}^2=0.25\times10^{-4}\ \text{m}^2=2.5\times10^{-5}\ \text{m}^2. From the given flux, B=ϕA=25×10−62.5×10−5=1B=\dfrac{\phi}{A}=\dfrac{25\times10^{-6}}{2.5\times10^{-5}}=1 T. (a) The permeability is μ=BH=14000=2.5×10−4\mu=\dfrac{B}{H}=\dfrac{1}{4000}=2.5\times10^{-4} T m A−1^{-1}, so the relative permeability is μr=μμ0=2.5×10−44π×10−7≈199\mu_r=\dfrac{\mu}{\mu_0}=\dfrac{2.5\times10^{-4}}{4\pi\times10^{-7}}\approx199.

(b) The susceptibility is χ=μr−1≈198\chi=\mu_r-1\approx198.

(c) The magnetisation is M=χH=198×4000=7.92×105M=\chi H=198\times4000=7.92\times10^5 A m−1^{-1}. [!ANSWER] μr≈199\mu_r\approx199, χ≈198\chi\approx198, M≈7.92×105M\approx7.92\times10^5 A m−1^{-1}

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