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Karnataka Board PUCPUC Science 2nd PUC Class 12

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

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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.

Long Answer
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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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Chapter 5: Magnetism And Matter - MCQ I [Page 32]

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NCERT Exemplar Physics Exemplar [English] Class 12
Chapter 5 Magnetism And Matter
MCQ I | Q 5.22 | Page 32

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