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  • What are the dimensions of X, the magnetic susceptibility?

What are the dimensions of χ, the magnetic susceptibility? Consider an H-atom. Guess an expression for χ, up to a constant by constructing a quantity 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 \left | x \right |\sim 10^{5} for many solid materials.

Answers (1)

X_m=\frac{I}{}H=\frac{\text{Intensity of magnetisation}}{\text{Magnetising force }}

 As I and H both have same units and dimensions, hence χ has no dimensions. 

 Here, χ is related with e, m, v, R and \mu _{0} 

  From Biot-Savart'slaw,

 dB=\frac{\mu _0}{4\pi} Idl\frac{\sin \theta }{r^2 }

\mu _0=\frac{4\pi r^2dB}{Idl\sin \theta }=\frac{4\pi r^2}{Idl \sin\theta }\times \frac{F}{qv \sin \theta }

 Dimensions of \mu _{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   

 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 \mu _0e^2, as e has the dimensions of charge.

 Let X= \mu _0e^2 m^av^bR^c .......i

where a, b ,c are the power of m,v and R respectively  

 [M^0L^0T^0A^0T^0]=[MLA^{-2}T^{-2}]\times [A^2T^2][M]^a\times [LT^{-1}]^b\times [L]^c =[M^{1+a}L^{1+b+c}T^{-b}A^0 ]

Equating the powers we get 0=1+a 

  a=-1 

  0=1+b+c   

  0=-b  

 b=0  

 1+0+c=0 

 c=-1 

 Putting values in equation i we get  X=\mu _{0}e^2m^{-1}v^2R^{-1}=\frac{\mu _0e^2}{mR }

Here, \mu _0=4\pi \times 10^{-7} Tm A^{-1 }

 e=1.6\times 10^{-19} C

m=9.1\times 10^{-31} kg

R=10^{-10}m X=\left (4\pi \times 10^{-7} \right )\times \frac{(1.6\times 10^{-19})^{2}}{9.1\times 10^{-31}\times 10^{-10}}\approx 10^{-4}


 

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infoexpert24

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