The dimension of sqrt(mu_0/epsilon_0) is equal to that of : (mu_0 = Vacuum permeability and epsilon_0 = Vacuum permittivity)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q26 – JEE-MAIN 2025

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Home Physics EM WavesQuestion #175

The dimension of

\sqrt{\frac{μ_{0}}{ϵ_{0}}}

is equal to that of : (

μ_{0}

= Vacuum permeability and

ϵ_{0}

= Vacuum permittivity)

Clue 1 of 3: Conceptual Principle

The quantity $\sqrt{\frac{μ_{0}}{ϵ_{0}}}$ represents the characteristic impedance of free space, often denoted as $Z_{0}$ .

🥷

Ninja Strategy•Recognize Characteristic Impedance

The expression $\sqrt{\frac{μ_{0}}{ϵ_{0}}}$ is the characteristic impedance of free space, which by definition has the same dimensions as resistance.

🧠 Full Solution Path

Step 1: Determine Dimensions of

μ_{0}

and

ϵ_{0}

✦ Active

The dimension of vacuum permeability $μ_{0}$ can be derived from the force between two parallel current-carrying wires or from the magnetic field formula. Using $F = q v B$ and $B = \frac{μ_{0} I}{2 π r}$ :

[B] = \frac{[F]}{[q] [v]} = \frac{[M L T^{- 2}]}{[A T] [L T^{- 1}]} = [M T^{- 2} A^{- 1}]

[μ_{0}] = \frac{[B] [r]}{[I]} = \frac{[M T^{- 2} A^{- 1}] [L]}{[A]} = [M L T^{- 2} A^{- 2}]

The dimension of vacuum permittivity $ϵ_{0}$ can be derived from Coulomb's law $F = \frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{r^{2}}$ :

[ϵ_{0}] = \frac{[q^{2}]}{[F] [r^{2}]} = \frac{[A^{2} T^{2}]}{[M L T^{- 2}] [L^{2}]} = [M^{- 1} L^{- 3} T^{4} A^{2}]

Step 2: Calculate Dimension of

\sqrt{\frac{μ_{0}}{ϵ_{0}}}

○ Expand

Now, we find the dimension of the ratio $\frac{μ_{0}}{ϵ_{0}}$ and then take its square root:

[\frac{μ_{0}}{ϵ_{0}}] = \frac{[M L T^{- 2} A^{- 2}]}{[M^{- 1} L^{- 3} T^{4} A^{2}]} = [M^{1 - (- 1)} L^{1 - (- 3)} T^{- 2 - 4} A^{- 2 - 2}] = [M^{2} L^{4} T^{- 6} A^{- 4}]

[\sqrt{\frac{μ_{0}}{ϵ_{0}}}] = \sqrt{[M^{2} L^{4} T^{- 6} A^{- 4}]} = [M L^{2} T^{- 3} A^{- 2}]

Step 3: Compare with Options○ Expand

Let's find the dimensions of the given options:

Voltage (V): [V] = \frac{Work}{Charge} = \frac{[M L^{2} T^{- 2}]}{[A T]} = [M L^{2} T^{- 3} A^{- 1}]

Resistance (R): [R] = \frac{Voltage}{Current} = \frac{[M L^{2} T^{- 3} A^{- 1}]}{[A]} = [M L^{2} T^{- 3} A^{- 2}]

Capacitance (C): [C] = \frac{Charge}{Voltage} = \frac{[A T]}{[M L^{2} T^{- 3} A^{- 1}]} = [M^{- 1} L^{- 2} T^{4} A^{2}]

Inductance (L): [L] = \frac{Magnetic Flux}{Current} = \frac{[B A]}{[I]} = \frac{[M T^{- 2} A^{- 1} L^{2}]}{[A]} = [M L^{2} T^{- 2} A^{- 2}]

Comparing the dimension of $\sqrt{\frac{μ_{0}}{ϵ_{0}}}$ ( $[M L^{2} T^{- 3} A^{- 2}]$ ) with the options, it matches the dimension of Resistance.

💡 Teacher's Secret Hint

Remember that the quantity $\sqrt{\frac{μ_{0}}{ϵ_{0}}}$ is also known as the characteristic impedance of free space, which has the same dimensions as resistance.

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Physics Question 26 – JEE-MAIN 2025

Clue 1 of 3: Conceptual Principle

🧠 Full Solution Path