The mass of an iron is $80$ $g m$ and its magnetic moment is $10 A m^{2}$ . If the density of iron is $8 g / c m^{3}$ then the value of intensity of magnetization will be :

10^{6} A / m

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10^{4} A / m

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10^{2} A / m

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10 A / m

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Solution

The correct option is A $10^{6} A / m$
$m a s s = 80$ , $d = 8 g m / c c = 8000 k g / m^{3}$ ., $m = 10 A - m^{2}$
$I = \frac{m}{v}$
$d = \frac{m a s s}{v}$
$v = \frac{80 \times 10^{- 3}}{8000} = 10^{- 5} m^{3}$
$I = \frac{10}{10^{- 5}}$
$I = 10^{6} A / m$

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The mass of an iron is 80gm and its magnetic moment is 10Am2. If the density of iron is 8 g/cm3 then the value of intensity of magnetization will be :

The correct option is A 106A/mmass=80, d=8gm/cc=8000kg/m3., m=10A−m2I=mvd=massvv=80×10−38000=10−5m3I=1010−5I=106A/m

The mass of an iron is $80$ $g m$ and its magnetic moment is $10 A m^{2}$ . If the density of iron is $8 g / c m^{3}$ then the value of intensity of magnetization will be :

The correct option is A $10^{6} A / m$
$m a s s = 80$ , $d = 8 g m / c c = 8000 k g / m^{3}$ ., $m = 10 A - m^{2}$
$I = \frac{m}{v}$
$d = \frac{m a s s}{v}$
$v = \frac{80 \times 10^{- 3}}{8000} = 10^{- 5} m^{3}$
$I = \frac{10}{10^{- 5}}$
$I = 10^{6} A / m$