Question

The dimensional formula of \[\dfrac{1}{2}{\mu _o}{H^2}\] (\[{\mu _o}\]-Permeability of free space and \[H\]-magnetic field intensity) is:
A. \[{\text{ML}}{{\text{T}}^{{\text{ - 1}}}}\]
B. \[{\text{M}}{{\text{L}}^2}{{\text{T}}^{{\text{ - 2}}}}\]
C. \[{\text{M}}{{\text{L}}^{ - 1}}{{\text{T}}^{{\text{ - 2}}}}\]
D. \[{\text{M}}{{\text{L}}^2}{{\text{T}}^{{\text{ - 1}}}}\]

Answer 1

Hint:We are asked to find the dimensional formula of a given quantity. To find the dimensional formula of the quantity, first find the dimensions of each term of the quantity. Then combine them to find the dimensional formula of the whole quantity.

Complete step by step answer:
Given the quantity, \[\dfrac{1}{2}{\mu _o}{H^2}\], where \[{\mu _o}\] is permeability of free space and \[H\] is magnetic field intensity.
Let \[A = \dfrac{1}{2}{\mu _o}{H^2}\] ………………....(i)
To find the dimension of the above quantity, we check the dimension of \[{\mu _o}\] and \[H\].The unit of \[{\mu _o}\] is,
\[\left[ {{\mu _o}} \right] = \dfrac{{\left[ {{\text{Newton}}} \right]}}{{\left[ {{\text{Ampere}}} \right]}} = \dfrac{{\left[ {\text{N}} \right]}}{{\left[ {\text{A}} \right]}}\] …………....(ii)
The dimension of Newton is \[\left[ {\text{N}} \right] = \left[ {{\text{ML}}{{\text{T}}^{{\text{ - 2}}}}} \right]\]and ampere is \[\left[ {\text{A}} \right] = \left[ {\text{I}} \right]\]
where \[{\text{M}}\] is the dimension of mass, \[{\text{L}}\] is the dimension of length,\[{\text{T}}\] is the dimension of time and \[{\text{I}}\] is dimension of current.

Putting the values of \[\left[ {\text{N}} \right]\] and \[\left[ {\text{A}} \right]\] in equation (i), we get
\[\left[ {{\mu _o}} \right] = \dfrac{{\left[ {{\text{ML}}{{\text{T}}^{{\text{ - 2}}}}} \right]}}{{\left[ {\text{I}} \right]}}\]
\[ \Rightarrow \left[ {{\mu _o}} \right] = \left[ {{\text{ML}}{{\text{T}}^{{\text{ - 2}}}}{{\text{I}}^{{\text{ - 1}}}}} \right]\] ………………(iii)
The unit of \[H\] is,
\[\left[ {\text{H}} \right] = \dfrac{{\left[ {{\text{Ampere}}} \right]}}{{\left[ {{\text{Metre}}} \right]}}\]
Now putting the dimension of ampere and metre we get,
\[\left[ {\text{H}} \right] = \dfrac{{\left[ {\text{I}} \right]}}{{\left[ {\text{L}} \right]}} = \left[ {{\text{I}}{{\text{L}}^{{\text{ - 1}}}}} \right]\] ……………....(iv)
Putting the values of dimension from equations (iii) and (iv) in equation (i) we get
\[\left[ A \right] = \left[ {\dfrac{1}{2}{\mu _o}{H^2}} \right] = \left[ {{\text{ML}}{{\text{T}}^{{\text{ - 2}}}}{{\text{I}}^{ - 1}}} \right]{\left[ {{\text{I}}{{\text{L}}^{{\text{ - 1}}}}} \right]^2}\]
\[ \Rightarrow \left[ A \right] = \left[ {{\text{ML}}{{\text{T}}^{{\text{ - 2}}}}{{\text{I}}^{ - 1}}{{\text{I}}^2}{{\text{L}}^{{\text{ - 2}}}}} \right]\]
\[ \therefore \left[ A \right] = \left[ {{\text{M}}{{\text{L}}^{ - 1}}{{\text{T}}^{{\text{ - 2}}}}} \right]\]
Therefore the dimension of \[\dfrac{1}{2}{\mu _o}{H^2}\] is \[{\text{M}}{{\text{L}}^{ - 1}}{{\text{T}}^{{\text{ - 2}}}}\].

Hence, the correct answer is option C.

Note: Whenever it is given to find out the dimension of a quantity, then first find the dimension of each term in the given quantity and combine them to find the dimension of the quantity. To find dimension of a quantity, we use dimensional analysis and see their dependency on seven fundamental units which are mass [M], length [L], time [T], electric current [I], thermodynamic temperature [K], luminous intensity [cd] and amount of substance [mol].