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The dipole moment of each molecule of para magnetic gas is 1.5 x10^–23 Am². The temperature of gas is 27°C and the number of molecules per unit volume in it is 2×10²⁶m^–3. The maximum possible intensity of magnetisation in the gas will be (in A/m) is

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3×10³

4×10^–3

5×10⁵

6×10^–4

answer is A.

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Detailed Solution

$I = \frac{M_{Total}}{V} = \frac{{NM}_{each}}{V} = \frac{2 \times 10^{26} \times 1.5 \times 10^{- 23}}{1} = 3 \times 10^{3} A / m$

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