The average dipole moment of Fe atoms is 1.8 | vedclass

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(D) $1$. Calculate the volume of the iron rod: $V = \pi r^2 L = \pi (0.5 \ cm)^2 	imes 10 \ cm = 2.5 \pi \ cm^3 \approx 7.85 \ cm^3$.$2$. Calculate t

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$12$

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(D) $1$. Calculate the volume of the iron rod: $V = \pi r^2 L = \pi (0.5 \ cm)^2 	imes 10 \ cm = 2.5 \pi \ cm^3 \approx 7.85 \ cm^3$.$2$. Calculate the mass of the rod: $m = 	ext{density} 	imes V = 7.87 \ g/cm^3 	imes 7.85 \ cm^3 \approx 61.8 \ g$.$3$. Calculate the number of atoms $(N)$: $N = \frac{m}{M} 	imes N_A = \frac{61.8}{55.87} 	imes 6.023 	imes 10^{23} \approx 6.66 	imes 10^{23}$ atoms.$4$. Calculate the total magnetic moment: $M_{total} = N 	imes \mu_{atom} = (6.66 	imes 10^{23}) 	imes (1.8 	imes 10^{-23} \ A-m^2) \approx 12 \ A-m^2$.

The average dipole moment of $Fe$ atoms is $1.8 \times 10^{-23} \ A-m^2$. The magnetic moment of an iron rod of length $10 \ cm$ and diameter $1 \ cm$ is........$A-m^2$: (density and atomic weight of $Fe$ are $7.87 \ g/cm^3$ and $55.87$ respectively).

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