The relation between mu and H for a specimen | vedclass

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(B) The relationship between magnetic flux density $B$,magnetic permeability $\mu$,and magnetic intensity $H$ is given by $B = \mu H$.Given $\mu = [\f

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$500 \ A m^{-1}$

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(B) The relationship between magnetic flux density $B$,magnetic permeability $\mu$,and magnetic intensity $H$ is given by $B = \mu H$.Given $\mu = [\frac{0.4}{H} + 12 	imes 10^{-4}] \ H m^{-1}$ and $B = 1 \ T$.Substituting the expression for $\mu$ into the formula $B = \mu H$:$B = [\frac{0.4}{H} + 12 	imes 10^{-4}] 	imes H$$B = 0.4 + (12 	imes 10^{-4}) H$Given $B = 1 \ T$,we have:$1 = 0.4 + (12 	imes 10^{-4}) H$$0.6 = 12 	imes 10^{-4} H$$H = \frac{0.6}{12 	imes 10^{-4}} = \frac{0.6 	imes 10^4}{12} = \frac{6000}{12} = 500 \ A m^{-1}$.

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

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