Question

Find the quantity the dimension of which is equal to the dimension of the ratio of magnetic flux and the resistance.
(A) Induced emf
(B) Charge
(C) Inductance
(D) Current

Answer 1

Hint
Here, we will use the principle of homogeneity of dimension i.e. we will equate the dimension of L.H.S to the dimension of R.H.S. By comparing we can get the dimension of the quantity. If we have dimension of a physical quantity then we can get the unit of quantity and finally we can get the quantity.

Complete step by step answer
Since the magnetic flux is defined as the magnetic field passes through the given area.
Therefore, the dimensional formula of magnetic flux is given by, $\left[ \phi \right] = \left[ {\rm{B}} \right] \times \left[ {{\rm{Area}}} \right]$
Now, the magnetic field in terms of magnetic force is given by, ${\rm{F}} = {\rm{q}}\left( {{\rm{\vec v}} \times {\rm{\vec B}}} \right)$
Therefore, $\left[ \phi \right] = \dfrac{{\left[ {\rm{F}} \right]\left[ {\rm{A}} \right]}}{{\left[ {\rm{q}} \right]\left[ {\rm{v}} \right]}}$
On putting dimensional formula for each term, we get
$ = \dfrac{{\left[ {{{\rm{M}}^1}{{\rm{L}}^1}{{\rm{T}}^{ - 2}}} \right]\left[ {{{\rm{L}}^2}} \right]}}{{\left[ {{{\rm{A}}^1}{{\rm{T}}^1}} \right]\left[ {{\rm{L}}{{\rm{T}}^{ - 1}}} \right]}}$
$ = \left[ {{{\rm{M}}^1}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}{{\rm{A}}^{ - 1}}} \right]$ … (1)
Now, according to ohm’s law, resistance is given by, ${\rm{R}} = \dfrac{{\rm{V}}}{{\rm{I}}}$
And, the dimensional formula of resistance is given by, $\left[ {\rm{R}} \right] = \dfrac{{\left[ {\rm{V}} \right]}}{{\left[ {\rm{I}} \right]}}$
$ = \dfrac{{\left[ {{{\rm{M}}^1}{{\rm{L}}^2}{{\rm{A}}^{ - 1}}{{\rm{T}}^{ - 3}}} \right]}}{{\left[ {\rm{A}} \right]}}$
$ = \left[ {{\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 3}}{{\rm{A}}^{ - 2}}} \right]$ … (2)
On dividing equations (1) by (2), we get
i.e. $\dfrac{{\left[ \phi \right]}}{{\left[ {\rm{R}} \right]}} = \dfrac{{\left[ {{{\rm{M}}^1}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}{{\rm{A}}^{ - 1}}} \right]}}{{\left[ {{{\rm{M}}^1}{{\rm{L}}^2}{{\rm{T}}^{ - 3}}{{\rm{A}}^{ - 2}}} \right]}} = \left[ {{\rm{AT}}} \right]$
Since the charge is given by, ${\rm{q}} = {\rm{It}}$ (Here ‘I’ represents the current)
This implies that the dimensional formula of charge is $\left[ {{\rm{AT}}} \right]$.
Therefore, (B), the charge is the required solution.

Additional Information
According to the principle of homogeneity, the dimensions of the physical quantities on both sides i.e. L.H.S. and R.H.S of an equation should be the same. It also states that the physical quantities with the same dimensions should be added or subtracted. Therefore, this principle is used to check the correctness of the equations used in physics.

Note
While calculating the dimensional formula of the quantity, the formula of the physical quantity should be known and then express it in terms of basic dimensional formulas.