JEE Challenger
More from Units and Measurements
PhyMedAdv 2015 • Paper 2

Dimensional Analysis of Electromagnetism Relations Involving Current and Potential Difference

With potential difference represented by VV, electric current by II, permittivity of free space by ε0\varepsilon_0, permeability of free space by μ0\mu_0, and speed of light in vacuum by cc, identify the dimensionally valid formula(s):

Options

A

μ0I2=ε0V2\mu_0 I^2 = \varepsilon_0 V^2

Correct
B

ε0I=μ0V\varepsilon_0 I = \mu_0 V

C

I=ε0cVI = \varepsilon_0 c V

Correct
D

μ0cI=ε0V\mu_0 c I = \varepsilon_0 V

Topics & Concepts

Units and MeasurementsElectric PotentialElectromagnetic Waves

Step-by-Step Solution

To determine the dimensionally valid formula(s), we can relate the given quantities using the dimensions of impedance (resistance).

  1. Relation between μ0\mu_0, ε0\varepsilon_0, and the speed of light cc: c=1μ0ε0c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

  2. Characteristic Impedance of Free Space (η0\eta_0): The intrinsic impedance of free space has the dimension of resistance (i.e., [VI]\left[\frac{V}{I}\right]) and is given by: η0=μ0ε0\eta_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} Therefore, the dimensional relation is: [VI]=[μ0ε0]\left[\frac{V}{I}\right] = \left[\sqrt{\frac{\mu_0}{\varepsilon_0}}\right]

Now, let us evaluate each option:

  • Option (A): μ0I2=ε0V2\mu_0 I^2 = \varepsilon_0 V^2 Rearranging gives: V2I2=μ0ε0  ⟹  (VI)2=μ0ε0\frac{V^2}{I^2} = \frac{\mu_0}{\varepsilon_0} \implies \left(\frac{V}{I}\right)^2 = \frac{\mu_0}{\varepsilon_0} Taking the square root: [VI]=[μ0ε0]\left[\frac{V}{I}\right] = \left[\sqrt{\frac{\mu_0}{\varepsilon_0}}\right] This matches the dimensional relation for impedance. Hence, Option (A) is dimensionally valid.

  • Option (B): ε0I=μ0V\varepsilon_0 I = \mu_0 V Rearranging gives: VI=ε0μ0=1η02\frac{V}{I} = \frac{\varepsilon_0}{\mu_0} = \frac{1}{\eta_0^2} The dimensions of VI\frac{V}{I} are Resistance\text{Resistance} (Ω\Omega), whereas the dimensions of ε0μ0\frac{\varepsilon_0}{\mu_0} are Ω−2\Omega^{-2}. Hence, Option (B) is incorrect.

  • Option (C): I=ε0cVI = \varepsilon_0 c V Rearranging gives: VI=1ε0c\frac{V}{I} = \frac{1}{\varepsilon_0 c} Substituting c=1μ0ε0c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}: 1ε0c=1ε0(1μ0ε0)=μ0ε0ε0=μ0ε0=η0\frac{1}{\varepsilon_0 c} = \frac{1}{\varepsilon_0 \left(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right)} = \frac{\sqrt{\mu_0 \varepsilon_0}}{\varepsilon_0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} = \eta_0 Since [1ε0c]=[η0]=[VI]\left[\frac{1}{\varepsilon_0 c}\right] = [\eta_0] = \left[\frac{V}{I}\right], the relation is dimensionally consistent. Hence, Option (C) is dimensionally valid.

  • Option (D): μ0cI=ε0V\mu_0 c I = \varepsilon_0 V Rearranging gives: VI=μ0cε0=μ0ε01μ0ε0=μ0ε03/2\frac{V}{I} = \frac{\mu_0 c}{\varepsilon_0} = \frac{\mu_0}{\varepsilon_0} \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = \frac{\sqrt{\mu_0}}{\varepsilon_0^{3/2}} This does not have the dimension of impedance [μ0ε0]\left[\sqrt{\frac{\mu_0}{\varepsilon_0}}\right]. Hence, Option (D) is incorrect.

Thus, the dimensionally valid equations are (A) and (C).

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