Dimension of Mu0 Epsilon0 and Its Physical Meaning

 

❓ 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 Explanation

This problem is based on:

👉 Units & dimensions
👉 Electromagnetic constants
👉 Wave relation in vacuum.

Main idea:

We use relation:

$$\boxed{ c=\frac1{\sqrt{\mu_0\varepsilon_0}} }$$

From this:

$$\boxed{ \sqrt{\frac{\mu_0}{\varepsilon_0}} =\mu_0 c }$$

This quantity is called:

$$\boxed{ \text{Characteristic impedance of free space} }$$

whose unit is ohm $(\Omega)$.

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🔷 Step 1 — Use EM Wave Relation 💯

Known relation:

$$c=\frac1{\sqrt{\mu_0\varepsilon_0}}$$

Rearrange:

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

Now check dimensions.

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🔷 Step 2 — Unit of $\mu_0$

Vacuum permeability:

$$\mu_0$$

has unit:

$$\boxed{ \text{N A}^{-2} }$$

Speed of light:

$$c \rightarrow \text{m/s}$$

Thus:

$$\mu_0 c$$

has unit equivalent to:

$$\boxed{ \Omega }$$

which is resistance.

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🔷 Step 3 — Final Identification

$$\boxed{ \sqrt{\frac{\mu_0}{\varepsilon_0}} }$$

has dimensions of:

$$\boxed{ \text{Resistance} }$$

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🔷 Step 4 — JEE Trap Alert 🚨

❌ $\mu_0$ and $\varepsilon_0$ dimensions directly manipulate karke confuse ho jaana

❌ EM wave relation bhool jaana

Remember:

$$\boxed{ c=\frac1{\sqrt{\mu_0\varepsilon_0}} }$$

is the shortcut.

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✅ Final Answer

$$\boxed{ \text{Resistance} }$$

(Option 2)

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