Dimension of `mu_0 in_0//c^2` will be

Options

A. `M^0LT^(-2)`
B. `M_0L^4T^(-4)`
C. `M_0L^(-4)T^4`
D. none of the above

Answer

To find the dimension of \(\frac{\mu_0 \epsilon_0}{c^2}\), we will follow these steps:

### Step 1: Understand the Constants
- \(\mu_0\) is the permeability of free space.
- \(\epsilon_0\) is the permittivity of free space.
- \(c\) is the speed of light.

### Step 2: Use the Relationship Between the Constants
The speed of light in free space is given by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
Squaring both sides gives:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]
From this, we can express \(\mu_0 \epsilon_0\) in terms of \(c^2\):
\[ \mu_0 \epsilon_0 = \frac{1}{c^2} \]

### Step 3: Substitute into the Expression
Now we need to find the dimensions of \(\frac{\mu_0 \epsilon_0}{c^2}\):
\[ \frac{\mu_0 \epsilon_0}{c^2} = \frac{1/c^2}{c^2} = \frac{1}{c^4} \]

### Step 4: Find the Dimensions of \(c^4\)
The speed of light \(c\) has the dimensions of velocity:
\[ [c] = [L][T^{-1}] \]
Thus, the dimensions of \(c^4\) are:
\[ [c^4] = [L^4][T^{-4}] \]

### Step 5: Find the Dimensions of \(\frac{1}{c^4}\)
Taking the reciprocal gives:
\[ \left[\frac{1}{c^4}\right] = [L^{-4}][T^{4}] \]

### Final Answer
Thus, the dimensions of \(\frac{\mu_0 \epsilon_0}{c^2}\) are:
\[ [L^{-4}][T^{4}] \]