What are the dimensions of χ, the magnetic susceptibility? Consider an H-atom. Gives an expression for χ, upto a constant by constructing a number 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.

Hint: Use the concept of dimensional analysis.
Step 1: As I and H have the same units and dimensions, hence, χ has no dimensions. Here, in this question, χ is to related e, m, v, R, and μ0. We know that the dimensions of μ0 = MLQ-2
From Biot-Savart's law;
       dB=μ04πIdlsin θr2   μ0=4πr2dBIdl sin θ=4πr2Idl sin θ×fqv sin θdB=Fqv sin θ
   Dimensions of μ0=L2×(MLT2)(QT1)(L)×1×(Q)(LT1)×(1)=[MLQ2]
where Q is the dimension of charge.
Step 2: As χ is dimensionless, it should have no involvement of charge Q in its dimensional formula. It will be so if μ0 and e together should have the value μ0e2, as e has the dimensions of charge.
Let,  χ=μ0e2mavbRc............(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:
[M0L0T0Q0]=[MLQ2]×[Q2][Ma]×(LT1)b×[L]c                    =[M1+a+L1+b+cTbQ0]
Equating the powers of M, L and T, we get;
                         0=1+aa=1,0=1+b+c0=bb=0, 0=1+0+c or c=1                     ...i
Putting values in Eq. (i), we get;
             χ = μ0e2m1v2R1=μ0e2mR                              ...ii
Step 3: Here    μ0=4π×107 TmA1
              e = 1.6×1019 C
               m=9.1×1031kg.R1010 m
          χXgiven solid=104105=10