Let `[epsilon_(0)]` denote the dimensional formula of the permittivity of the vacuum, and `[mu_(0)]` that of the permeability of the vacuum. If `M = mass ,L = length, T = time and I = electric current`,

Options

A. `varepsilon_(0) = [M^(-1)L^(-3)T^(2)I]`
B. `varepsilon_(0) = [M^(-1)L^(-3)T^(4)I^(2)]`
C. `mu_(0) = [MLT^(-2)I^(-2)]`
D. `mu_(0) = [ML^(2)T^(-1)I]`

Answer

To find the dimensional formulas for the permittivity of vacuum \((\epsilon_0)\) and the permeability of vacuum \((\mu_0)\), we will follow these steps:

### Step 1: Find the dimensional formula for \(\epsilon_0\)

1. **Start with the formula for the Coulomb's law**:
\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \]
Rearranging gives:
\[ \epsilon_0 = \frac{q_1 q_2}{4\pi F r^2} \]

2. **Identify the dimensions**:
- The dimension of force \(F\) is:
\[ [F] = M L T^{-2} \]
- The dimension of charge \(q\) can be expressed as:
\[ [q] = I T \]
- The dimension of distance \(r\) is:
\[ [r] = L \]

3. **Substituting the dimensions into the formula**:
\[ \epsilon_0 = \frac{(I T)(I T)}{(M L T^{-2})(L^2)} = \frac{I^2 T^2}{M L^3 T^{-2}} = \frac{I^2 T^4}{M L^3} \]

4. **Final dimensional formula for \(\epsilon_0\)**:
\[ [\epsilon_0] = M^{-1} L^{-3} T^{4} I^{2} \]

### Step 2: Find the dimensional formula for \(\mu_0\)

1. **Start with the formula for the force between two parallel wires**:
\[ F = \frac{\mu_0 I_1 I_2}{2\pi d} \]
Rearranging gives:
\[ \mu_0 = \frac{F \cdot 2\pi d}{I_1 I_2} \]

2. **Identify the dimensions**:
- The dimension of force \(F\) is:
\[ [F] = M L T^{-2} \]
- The dimension of current \(I\) is:
\[ [I] = I \]
- The dimension of distance \(d\) is:
\[ [d] = L \]

3. **Substituting the dimensions into the formula**:
\[ \mu_0 = \frac{(M L T^{-2}) \cdot (L)}{I^2} = \frac{M L^2 T^{-2}}{I^2} \]

4. **Final dimensional formula for \(\mu_0\)**:
\[ [\mu_0] = M L^{-1} T^{-2} I^{-2} \]

### Summary of Dimensional Formulas
- \(\epsilon_0 = M^{-1} L^{-3} T^{4} I^{2}\)
- \(\mu_0 = M L^{-1} T^{-2} I^{-2}\)