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Question
What are the dimensions of χ, the magnetic susceptibility? Consider an H-atom. Guess an expression for χ, upto a constant by constructing a quantity of dimensions of χ, out of parameters of the atom: e, m, v, R and µ0. Here, m is the electronic mass, v is electronic velocity, R is Bohr radius. Estimate the number so obtained and compare with the value of |χ| ~ 10–5 for many solid materials.
Solution
As the intensity of magnetization (M) and magnetizing field (H) both has the same unit, i.e. ampere per metre and `chi = M/H` so `ch` (susceptibility) has no unit.
We have to relate `chi` with e, v, vecm`, R and `mu_2`. We will relate these physical quantities by using dimension and Biot-Savart's law.
By Biot = Savart's law `dB = mu_0/(4pi) = (I.dl sin theta)/r^2` can be used to find out the dimension of `mu_0`.
`mu_0 = (dB.4pir^2)/(Idl sin theta)` for dB
For the dimension of dB
F = Bqv sin θ
`B = F/(qv sin theta) = (MLT^-2)/(QLT^-1) = [ML^0T^-1Q^-1]`
∴ `mu_0 = ([MT^1Q^-1]L^2)/(QT^-1L) - [MLQ^-2]`
Where Q is dimension of charge.
X depends on the magnetic moment induced when H is turned on. H couple to atomic electrons through its charge e. The effect on m is via current I which involves another factor of e. The combination `mu_0e^2` does not depend on the "charge" Q. Dimension `chi = mu_0^ae^2m^bv^cR^d`
`[M^0L^0T^0Q^0] = [MLQ^-2]^a Q^2M^b[LT^-1]^c L^d`
`[M^0L^0T^0Q^0] = M^(a-b) L^(a+c+d) T^c Q^(-2a+2)`
Comparing the powers
c = 0, a + b = 0, – 2a + 2 = 0, a + c + d = 0
1 + b = 0, – 2a = – 2, 1 + 0 + d = 0
b = – 1, a = + 1, d = – 1
`chi = (mu_0)e^2m^-1v^0R^-1`
`chi = (mu_0e^2)/(mR)`
`mu_0 = 4pi xx 10^-7 TmA^-1, e = 1.6 xx 10^-19`
`m = 9.1 xx 10^-31 kg`, `R = 10^-10 m`
`chi = ((4pi xx 10^-7)(1.6 xx 10^-19)(1.6 xx 10^-19))/((9.1 xx 10^-31) xx 10^-10)`
= `(4 xx 3.1 xx 1.6 xx 1.6)/9.1 xx 10^(-7-19-19 + 31 + 10)`
= `(124 xx 256 xx 10^(-45 + 41))/9.1 = 317.4/91 xx 10^-4 = 3.5 xx 10^-4`
`|chi^'| = 10^-5` ....(Given)
`chi/|chi^'| = (3.5 xx 10^-4)/10^-5 = (3.5 xx 10^-4)/(10^-1 xx 10^-4)` = 35
`chi = 35|chi^'|`
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