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(A) The magnetization $M$ is given by the formula $M = \chi H$,where $\chi$ is the magnetic susceptibility and $H$ is the magnetic field intensity.For

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$1.125 	imes 10^6$

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(A) The magnetization $M$ is given by the formula $M = \chi H$,where $\chi$ is the magnetic susceptibility and $H$ is the magnetic field intensity.For a solenoid,the magnetic field intensity is $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current.The magnetic susceptibility is related to the relative permeability $\mu_r$ by the relation $\chi = \mu_r - 1$.Given: $n = 900 \ m^{-1}$,$I = 2.5 \ A$,and $\mu_r = 501$.Substituting these values into the formula $M = nI(\mu_r - 1)$:$M = 900 	imes 2.5 	imes (501 - 1)$$M = 2250 	imes 500$$M = 1,125,000 \ A \ m^{-1} = 1.125 	imes 10^6 \ A \ m^{-1}$.

$A$ solenoid has a core of material with relative permeability $501$. The windings of the solenoid are insulated from the core and carry a current of $2.5 \ A$. If the number of turns is $900$ per metre,the magnetization in $A \ m^{-1}$ is:

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