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What are the dimensions of chi, the magn...

What are the dimensions of `chi`, the magnetic susceptibility? Consider an H-atom. Guess an expression for `chi` upto a constant by constructing a quantity of dimensions of `chi`, out of parameters of the atom: e, m, v, R and `mu_0`. 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 `|chi|~10^5` for many solid materials.

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Magnetic susceptibility, `chi_m=I/H (("intensity of magnetisation"))/(("magnetising force"))`
As I and H have same units and dimensions, hence `chi` has no dimensions.
In this question, `chi` is to be related with e, m, v, R and `mu_0`. We know that dimensions of `mu_0=[MLQ^-2]`
[From Biot Savart's law, `dB=(mu_0)/(4pi)(Idlsintheta)/(r^2)`
or `mu_0=(4pir^2dB)/(Idlsintheta)=(4pir^2)/(Idlsintheta)xx(F)/(qvsintheta^')` `[:' dB=(F)/(qvsin theta^')]`
`:.` Dimensions of `mu_0=(L^2xx(MLT^-2))/((QT^-1)(L)xx1xx(Q)(LT^-1)xx(1))=[MLQ^-2]` where Q is for charge.
As `chi` is dimensionless, it should have no involvement of charge Q in its dimensional formula. It will be so if `mu_0` and e together should have the value `mu_0e^2`, as e has the dimensions of charge.
Let `chi=mu_0e^2m^av^bR^c` ...(i)
where a, b, c are the powers of m, v and R such that relation (i) is satisfied.
Dimensional equation of (i) is `M^0L^0T^0Q^0=(M^1L^1Q^-2)xx(Q^2)(M^a)xx(LT^-1)^bxx(L)^c`
`=M^(1+a)L^(1+b+c)T^(-b)Q^0`
Equating the powers of M, L and T, we get
`0=1+a` or `a=-1`, `0=1+b+c`...(ii)
`0=-b` or `b=0`, From (ii), `0=1+0+c` or `c=-1`
Putting values in (i), we get
`chi=mu_0e^2m^-1v^0R^-1=(mu_0e^2)/(mR)` ...(iii)
Here, `mu_0=4pixx10^-7TmA^-1, e=1*6xx10^(-19)C`,
`m=9*1xx10^(-31)kg, R=10^(-10)m`
`:. chi=((4pixx10^-7)xx(1*6xx10^(-19))^2)/((9*1xx10^(-31))xx10^(-10))~~10^-4`
`:. (chi)/(chi_((given solid)))=10^-4/10^-5=10`
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