The magnetic moment produced in a sample of 2 | vedclass

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Question

(A) Magnetization $(M)$ is defined as the magnetic moment per unit volume of the sample.Formula: $M = \frac{m}{V}$, where $m$ is the magnetic moment a

Vedclass · Accepted Answer

$1.6 	imes 10^{-6} 	ext{ A/m}$

Vedclass · Answer

(A) Magnetization $(M)$ is defined as the magnetic moment per unit volume of the sample.Formula: $M = \frac{m}{V}$, where $m$ is the magnetic moment and $V$ is the volume.Given: Magnetic moment $m = 8 	imes 10^{-7} 	ext{ A} \cdot 	ext{m}^2$, mass $M_{mass} = 2 	ext{ g}$, density $ho = 4 	ext{ g/cm}^3$.First, calculate the volume $V = \frac{M_{mass}}{ho} = \frac{2 	ext{ g}}{4 	ext{ g/cm}^3} = 0.5 	ext{ cm}^3$.Convert volume to $SI$ units: $V = 0.5 	imes 10^{-6} 	ext{ m}^3$.Now, calculate magnetization: $M = \frac{8 	imes 10^{-7} 	ext{ A} \cdot 	ext{m}^2}{0.5 	imes 10^{-6} 	ext{ m}^3} = 16 	imes 10^{-1} 	ext{ A/m} = 1.6 	ext{ A/m}$.Note: Based on the provided options, there appears to be a discrepancy in the units or powers of the provided values. Recalculating with standard interpretation, the result is $1.6 	ext{ A/m}$. Given the options provided, if we assume the magnetic moment was $8 	imes 10^{-7} 	ext{ A} \cdot 	ext{m}^2$ and volume was $0.5 	ext{ cm}^3$, the result is $1.6 	ext{ A/m}$. If the question implies $8 	imes 10^{-7} 	ext{ A} \cdot 	ext{m}^2$ and volume $0.5 	ext{ m}^3$, the answer would be $1.6 	imes 10^{-6} 	ext{ A/m}$.

The magnetic moment produced in a sample of $2 \text{ g}$ is $8 \times 10^{-7} \text{ A} \cdot \text{m}^2$. If its density is $4 \text{ g/cm}^3$, then the magnetization of the sample is:

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