The relation between $\mu$ and $H$ for a specimen of iron is $\mu=\left[\frac{0.4}{H}+12 imes 10^{-4} ight] \mathrm{Hm}^{-1}$. The value of H which produces flux density of 1 T will be ( $\mu=$ magnetic permeability, $\mathrm{H}=$ magnetic intensity)

Question

The relation between $\mu$ and $H$ for a specimen of iron is $\mu=\left[\frac{0.4}{H}+12 	imes 10^{-4}ight] \mathrm{Hm}^{-1}$. The value of H which produces flux density of 1 T will be ( $\mu=$ magnetic permeability, $\mathrm{H}=$ magnetic intensity)

Aicharya · Accepted Answer

For iron specimen, $\begin{aligned} & \mu=\left[\frac{0.4}{\mathrm{H}}+12 	imes 10^{-4}ight] \mathrm{H}_{\mathrm{m}^{-1}}, \mathrm{~B}=1 \mathrm{~T} \ & 	herefore \quad \mathrm{~B}=\mu \mathrm{H}=0.4+\left(12 	imes 10^{-4}ight) \mathrm{H} \ & \Rightarrow 1=0.4+12 	imes 10^{-4} \mathrm{H} \ & 	herefore \mathrm{H}=500 \mathrm{Am}^{-1}\end{aligned}$

The relation between $\mu$ and $H$ for a specimen of iron is $\mu=\left[\frac{0.4}{H}+12 \times…

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