Ncert Exercises & Solutions

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 $μ_{0}$ . 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 | $χ$ | $~ 10^{- 5}$ 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

μ_{0}

. We know that the dimensions of

μ_{0} = [{MLQ}^{- 2}]

From Biot-Savart's law;

\begin{array}{l} dB = \frac{μ_{0}}{4 π} \frac{Idlsin θ}{r^{2}} \\ \Rightarrow μ_{0} = \frac{4 {πr}^{2} dB}{Idl \sin θ} = \frac{4 {πr}^{2}}{Idl \sin θ} \times \frac{f}{qv \sin θ} [∵ dB = \frac{F}{qv \sin θ}] \end{array}

∴ Dimensions of μ_{0} = \frac{L^{2} \times ({MLT}^{- 2})}{({QT}^{- 1}) (L) \times 1 \times (Q) ({LT}^{- 1}) \times (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

μ_{0}

and e together should have the value

μ_{0} e^{2}

, as e has the dimensions of charge.

Let,

χ = μ_{0} e^{2} m^{a} v^{b} R^{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:

\begin{array}{l} [M^{0} L^{0} T^{0} Q^{0}] = [{MLQ}^{- 2}] \times [Q^{2}] [M^{a}] \times ({LT}^{- 1})^{b} \times {[L]}^{c} \\ = [M^{1 + a} + L^{1 + b + c} T^{- b} Q^{0}] \end{array}

Equating the powers of M, L and T, we get;

\begin{array}{l} 0 = 1 + a \Rightarrow a = - 1, 0 = 1 + b + c \\ 0 = - b \Rightarrow b = 0, 0 = 1 + 0 + c or c = - 1 . . . (i) \end{array}

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

χ = μ_{0} e^{2} m^{- 1} v^{2} R^{- 1} = \frac{μ_{0} e^{2}}{mR} . . . (ii)

Step 3 : Here μ_{0} = 4 π \times 10^{- 7} {TmA}^{- 1}

e = 1 .6 \times 10^{- 19} C

m = 9 .1 \times 10^{- 31} kg . R - 10^{- 10} m

χ = \frac{(4 π \times 10^{- 7}) \times (1 .6 \times 10^{- 19})^{2}}{(9 .1 \times 10^{- 31}) \times 10^{- 10}} \approx 10^{- 4}

∴ \frac{χ}{X_{(given solid)}} = \frac{10^{- 4}}{10^{- 5}} = 10

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