Question

In terms of potential difference V; electric current I, permittivity ${\varepsilon _0}$ hand speed of light $c$, the dimensionally correct equation(s) is (are).
(This question has multiple correct options)
(A) ${\mu _0}{I^2} = {\varepsilon _0}{V^2}$
(B) ${\varepsilon _0}I = {\mu _0}V$
(C) $I = {\varepsilon _0}cV$
(D )${\mu _0}cI = {\varepsilon _0}V$

Answer 1

To solve this type of question we have to use the dimensional formula of a physical quantity.
(i) ${\text{I = }}\left[ {\text{A}} \right]$
(ii) $V = \left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$
(iii) $c = \left[ {L{T^{ - 1}}} \right]$
(iv) ${\mu _0} = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]$
(v) ${\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$

Complete answer:
Let us now check the dimensional formula for every option. Those options for which the right-hand side will be equal to the right-hand side will be the correct options.
Let us then first check option (A).
We have given the relation which is a,
${\mu _0}{I^2} = {\varepsilon _0}{V^2}$
Now, let us substitute their dimensional formula to check if LHS is equal to RHS, we get
$\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]{\left[ A \right]^2} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]{\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]^2}$
Let us simplify it by adding the powers of the same base/quantities ($\because {{\text{x}}^{\text{n}}}{\text{.}}{{\text{x}}^{ \pm {\text{m}}}}{\text{ = }}{{\text{x}}^{{\text{n}} \pm {\text{m}}}}$)
$\left[ {ML{T^{ - 2}}} \right] = \left[ {ML{T^{ - 2}}} \right]$
Here, RHS is equal to LHS. Therefore option (A) is dimensionally correct.

Now let's use check option (B).
We have given the relation which is a
${\varepsilon _0}I = {\mu _0}V$
Let us substitute their dimensional formula, we get
$\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]\left[ A \right] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$
Let us simplify by adding the powers of same quantities (using the law of powers)
$\left[ {ML{T^{ - 2}}{A^{ - 1}}} \right] = \left[ {{L^{ - 1}}TA} \right]$
Here we get,
RHS ≠LHS, therefore option (B) is dimensionally incorrect.
Now, Let us check option (C).

We have given the relation which is as follows,
$I = {\varepsilon _0}cV$
Let us substitute for their dimensional formula, we get,
$\left[ A \right] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\left[ {L{T^{ - 1}}} \right]\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$
Let us simply by adding the powers of the same quantity and we get,
$\left[ A \right] = \left[ A \right]$
i.e.LHS = RHS, therefore option (C) is dimensionally correct.

Now let us finally check option (D).
We have given the relation which is as follows,
${\mu _0}cI = {\varepsilon _0}V$
Let us substitute their dimensional formula. On substituting values we get,
$\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]\left[ {L{T^{ - 1}}} \right]\left[ A \right] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$
$\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right] = \left[ {{L^{ - 1}}TA} \right]$
Here we get,
$RHS$ ≠ $LHS$, therefore option (D) is dimensionally incorrect.

Hence, option (A) “${\mu _0}{I^2} = {\varepsilon _0}{V^2}$” and (C) “$I = {\varepsilon _0}cV$” are the correct options.

Additional information:
Based on a dimensional formula, the physical quantities are divided into four types:

1. Dimensional Variables: These are the physical quantities that do not have a constant value, for example, velocity, work, etc.
2. Dimensionless variables: These are physical quantities that do not have dimensions. For example, strain, plane angle, etc
3. Dimensional constant: quantities that have values but still no dimension. For example, Planck’s constant (h), gravitational constant (G), etc.
4. Dimensionless constant: Numbers like 1, 2, 3, π, etc. are non-dimensional constants.

Note:
The dimensional formula is the representation of quantities in terms of mass, length, time, and current, amount of substance, temperature, and luminous intensity.
Dimensional formulas are used to convert units of one system into units of other systems, to check the correctness of the equation, and to establish a relationship between various physical quantities.