An iron rod of cross-sectional area 0.2 , cm^ | vedclass

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(B) The magnetic flux $\phi$ is given by $\phi = B \cdot A$, where $B$ is the magnetic field induction and $A$ is the cross-sectional area.$B = \frac{

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$596$

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(B) The magnetic flux $\phi$ is given by $\phi = B \cdot A$, where $B$ is the magnetic field induction and $A$ is the cross-sectional area.$B = \frac{\phi}{A} = \frac{2.4 	imes 10^{-5} \, Wb}{0.2 	imes 10^{-4} \, m^2} = 1.2 \, T$.The relation between magnetic field $B$, magnetic intensity $H$, and relative permeability $\mu_r$ is $B = \mu_0 \mu_r H$.Thus, $\mu_r = \frac{B}{\mu_0 H} = \frac{1.2}{(4\pi 	imes 10^{-7}) 	imes 1600}$.$\mu_r = \frac{1.2}{2.01 	imes 10^{-3}} \approx 596.8$.Since $\mu_r = 1 + \chi_m$, the magnetic susceptibility $\chi_m = \mu_r - 1$.$\chi_m = 596.8 - 1 = 595.8 \approx 596$.

An iron rod of cross-sectional area $0.2 \, cm^2$ is subjected to a magnetizing field intensity of $1600 \, A \cdot m^{-1}$. If the magnetic flux produced in the rod is $2.4 \times 10^{-5} \, Wb$, what is the magnetic susceptibility of the material of the rod?

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