Dimension of `(1/(mu_0varepsilon_0))` is

Options

A. `L/T`
B. `T/L`
C. `L^2/T^2`
D. `T^2/L^2`

Answer

To find the dimension of \( \frac{1}{\mu_0 \varepsilon_0} \), we can follow these steps:

### Step 1: Understand the constants involved
- \( \mu_0 \) is the magnetic permeability of free space.
- \( \varepsilon_0 \) is the electric permittivity of free space.

### Step 2: Use the relationship between \( \mu_0 \), \( \varepsilon_0 \), and the speed of light \( c \)
We know from electromagnetic theory that:
\[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \]
This implies:
\[ c^2 = \frac{1}{\mu_0 \varepsilon_0} \]

### Step 3: Find the dimensions of \( c^2 \)
The speed of light \( c \) has the same dimensions as any speed, which is:
\[ [c] = \frac{[L]}{[T]} \quad \text{(length over time)} \]
Thus, the dimensions of \( c^2 \) are:
\[ [c^2] = \left(\frac{[L]}{[T]}\right)^2 = \frac{[L]^2}{[T]^2} \]

### Step 4: Equate the dimensions
From the relationship \( c^2 = \frac{1}{\mu_0 \varepsilon_0} \), we can equate the dimensions:
\[ \left[\frac{1}{\mu_0 \varepsilon_0}\right] = [c^2] = \frac{[L]^2}{[T]^2} \]

### Step 5: Conclusion
Thus, the dimension of \( \frac{1}{\mu_0 \varepsilon_0} \) is:
\[ \frac{[L]^2}{[T]^2} \]

### Final Answer
The dimension of \( \frac{1}{\mu_0 \varepsilon_0} \) is \( \frac{L^2}{T^2} \).

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