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 θ[∵dB=Fqv sin θ]
∴ Dimensions of μ0=L2×(MLT−2)(QT−1)(L)×1×(Q)(LT−1)×(1)=[MLQ−2]
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]=[MLQ−2]×[Q2][Ma]×(LT−1)b×[L]c =[M1+a+L1+b+cT−bQ0]
Equating the powers of M, L and T, we get;
0=1+a⇒a=−1,0=1+b+c0=−b⇒b=0, 0=1+0+c or c=−1 ...(i)
Putting values in Eq. (i), we get;
χ = μ0e2m−1v2R−1=μ0e2mR ...(ii)
Step 3: Here μ0=4π×10−7 TmA−1
e = 1.6×10−19 C
m=9.1×10−31kg.R−10−10 m
χ=(4π×10−7)×(1.6×10−19)2(9.1×10−31)×10−10≈10−4
∴ χX(given solid)=10−410−5=10
