The dimensional formula of `1/(mu_0in_0)` is ..........

Options

A. `[M^0L^2T^(-2)]`
B. `[M^0L^1T^(-1)]`
C. `[M^0L^(-2)T^(-2)]`
D. `[M^0L^1T^(-2)]`

Answer

To find the dimensional formula of \( \frac{1}{\mu_0 \epsilon_0} \), we can follow these steps:

### Step 1: Understand the relationship between \( \mu_0 \), \( \epsilon_0 \), and the speed of light \( c \).
We know from electromagnetic theory that:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]

### Step 2: Square both sides of the equation.
Squaring both sides gives us:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]

### Step 3: Rearrange the equation to express \( \frac{1}{\mu_0 \epsilon_0} \).
From the equation above, we can express \( \frac{1}{\mu_0 \epsilon_0} \) as:
\[ \frac{1}{\mu_0 \epsilon_0} = c^2 \]

### Step 4: Determine the dimensional formula of \( c^2 \).
The speed of light \( c \) has the dimensional formula:
\[ [c] = M^0 L^1 T^{-1} \]
Thus, squaring this gives:
\[ [c^2] = (M^0 L^1 T^{-1})^2 = M^0 L^2 T^{-2} \]

### Step 5: Conclude the dimensional formula for \( \frac{1}{\mu_0 \epsilon_0} \).
From the previous step, we conclude that:
\[ \frac{1}{\mu_0 \epsilon_0} \text{ has the dimensional formula } M^0 L^2 T^{-2} \]

### Final Answer:
The dimensional formula of \( \frac{1}{\mu_0 \epsilon_0} \) is \( M^0 L^2 T^{-2} \).

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