Each atom of an iron bar (5 , cm times 1 , cm | vedclass

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(C) $1$. Calculate the number of atoms per unit volume $(n)$: $n = \frac{ho N_A}{M} = \frac{7.78 	imes 10^3 	imes 6.02 	imes 10^{23}}{56 	imes 1

Vedclass · Accepted Answer

$7.54$

Vedclass · Answer

(C) $1$. Calculate the number of atoms per unit volume $(n)$: $n = \frac{ho N_A}{M} = \frac{7.78 	imes 10^3 	imes 6.02 	imes 10^{23}}{56 	imes 10^{-3}} \approx 8.36 	imes 10^{28} \, atoms/m^3$. $2$. Calculate the volume of the bar $(V)$: $V = 5 \, cm 	imes 1 \, cm 	imes 1 \, cm = 5 	imes 10^{-6} \, m^3$. $3$. Calculate the total number of atoms $(N)$: $N = n 	imes V = 8.36 	imes 10^{28} 	imes 5 	imes 10^{-6} = 4.18 	imes 10^{23} \, atoms$. $4$. Calculate the total magnetic moment $(M_{total})$: $M_{total} = N 	imes \mu_{atom} = 4.18 	imes 10^{23} 	imes 1.8 	imes 10^{-23} \approx 7.52 \, A \cdot m^2$. Rounding to the nearest provided option,the result is $7.54 \, A \cdot m^2$.

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