What are the dimensions of $\mathrm{\chi }$, the magnetic susceptibility? Consider an H-atom. Gives an expression for $\mathrm{\chi }$, upto a constant by constructing a number of dimensions of $\mathrm{\chi }$, out of parameters of the atom e, m, v, R, and ${\mathrm{\mu }}_{0}$. 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 |$\mathrm{\chi }$| for many solid materials.

Hint: Use the concept of dimensional analysis.
Step 1: As I and H have the same units and dimensions, hence, $\mathrm{\chi }$ has no dimensions. Here, in this question, $\mathrm{\chi }$ is to related e, m, v, R, and ${\mathrm{\mu }}_{0}$. We know that the dimensions of
From Biot-Savart's law;
where Q is the dimension of charge.
Step 2: As $\mathrm{\chi }$ is dimensionless, it should have no involvement of charge Q in its dimensional formula. It will be so if ${\mathrm{\mu }}_{0}$ and e together should have the value ${\mathrm{\mu }}_{0}{\mathrm{e}}^{2}$, as e has the dimensions of charge.
Let,  $\mathrm{\chi }={\mathrm{\mu }}_{0}{\mathrm{e}}^{2}{\mathrm{m}}^{\mathrm{a}}{\mathrm{v}}^{\mathrm{b}}{\mathrm{R}}^{\mathrm{c}}$............(i)
where a, b, c are the power of m, v, and R respectively, such that relation (i) is satisfied.
The dimensional equation of (i) is:
Equating the powers of M, L and T, we get;

Putting values in Eq. (i), we get;