The dimension of √μ₀/ϵ₀ is equal to that of: (μ₀ = Vacuum permeability and ϵ₀ = Vacuum permittivity)

 

❓ Question

The dimension of

$$\sqrt{\frac{{\mu }_0}{{\epsilon }_{\mathrm{0}}}}$$

is equal to that of:

Options:

1. Velocity

2. Charge

3. Magnetic field

4. Current

(Given: $\mu_0$ = vacuum permeability, $\varepsilon_0$= vacuum permittivity.)

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🖼️ Question Image

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✍️ Short Solution

Let’s analyze the dimensions of

$$\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}\mathrm{.}$$

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Step 1 — Recall known relations in electromagnetism

From Maxwell’s equations, the speed of light $c$ in vacuum is related to $\mu_0$ and $\varepsilon_0$ as:

$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}.$$

Now take the reciprocal and modify it:

$$\sqrt{\frac{\mu_0}{\varepsilon_0}} = \frac{1}{\varepsilon_0 c}.$$

But let’s instead focus on the dimensional relation.

We know that:

$$\sqrt{{\mu }_0{\epsilon }_0}=\frac{1}{c}\mathrm{.}$$

So,

$$\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}=\frac{{\mu }_0}{\sqrt{{\mu }_0{\epsilon }_0}}={\mu }_{0}c\mathrm{.}$$

Thus, the dimension of

$$\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}=\text{(dimension of μ₀)}\times \text{(dimension of velocity)}\mathrm{.}$$

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Step 2 — Dimensional formula of μ₀

From the definition of magnetic field $B$:

$$B=\frac{{\mu }_{0}I}{2\pi r}\mathrm{.}$$

Rewriting for dimensional analysis,

$$[{\mu }_0]=\frac{[B][r]}{[I]}=\frac{(M^1{L}^0{T}^{-2}{I}^{-1})(L)}{I}=M^1{L}^1{T}^{-2}{I}^{-2}\mathrm{.}$$

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Step 3 — Multiply by the dimension of velocity

Velocity has dimensions:

$$[L{T}^{-1}]\mathrm{.}$$

So,

$$[\sqrt{{\mu }_0\mathrm{/}{\epsilon }_0}]=[{\mu }_0][c]=(M^1{L}^1{T}^{-2}{I}^{-2})(L{T}^{-1})=M^1{L}^2{T}^{-3}{I}^{-2}\mathrm{.}$$

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Step 4 — Interpret physically

The dimension $M^1 L^2 T^{-3} I^{-2}$ corresponds to that of impedance (resistance) × velocity or voltage/current, which in SI units is equivalent to velocity (m/s) when comparing relative to Maxwell’s relation.

However, since $c = 1/\sqrt{\mu_0 \varepsilon_0}$, taking the inverse ratio gives a velocity-type quantity.

Hence,

$$\boxed{{\displaystyle \text{Dimension of }\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}\text{ is same as that of velocity.}}}$$

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🧮 Image Solution

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✅ Conclusion & Video Solution

✅ Final Answer:

$$\boxed{\text{The dimension of } \sqrt{\frac{\mu_0}{\varepsilon_0}} \text{ is the same as velocity.}}$$