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Question
A paramagnetic gas has 2.0 × 1026 atoms/m with atomic magnetic dipole moment of 1.5 × 10−23 A m2 each. The gas is at 27°C.
- Find the maximum magnetization intensity of this sample.
- If the gas in this problem is kept in a uniform magnetic field of 3 T, is it possible to achieve saturation magnetization? Why?
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Solution
Data: `"N"/"V" = 2.0 xx 10^26` atoms/m3,
µ = 1.5 × 10-23 Am2, T = 27 + 273 = 300 K, B = 3 T, kB = 1.38 × 10-23 J/K, 1 eV = 1.6 × 10-19 J
(a) The maximum magnetization of the material,
`"M"_"z" = "N"/"V" mu = (2.0 xx 10^26)(1.5 xx 10^-23)`
= 3 × 103 A/m
(b) The maximum orientation energy per atom is
`upsilon_"max" = - mu"B" cos 180^circ = mu"B"`
`= (1.5 xx 10^-23)(3) = (4.5 xx 10^-23)/(1.6 xx 10^-19)`
= 2.8 × 10-4 eV
The average thermal energy of each atom,
E = `3/2 "k"_"B" "T"`
where kB is the Botzmann constant.
∴ E = 1.5(1.38 × 10-23)(300)
= 6.21 × 10-21 J = `(6.21 xx 10^-21)/(1.6 xx 10^-19)`
= 3.9 × 10-2 eV
Since the thermal energy of randomization is about two orders of magnitude greater than the magnetic potential energy of orientation, saturation magnetization will not be achieved at 300 K.
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