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

Answer

To solve the problem, we need to derive the dimensional formulas for the permittivity of vacuum (ε₀) and the permeability of vacuum (μ₀).

### Step 1: Determine the Dimensional Formula for Permittivity (ε₀)

1. **Start with Coulomb's Law**: The force (F) between two charges (q₁ and q₂) separated by a distance (r) is given by:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
Rearranging this gives:
\[ \epsilon_0 = \frac{q_1 q_2}{F \cdot r^2} \]

2. **Identify the dimensions**:
- Charge (q) has the dimension of current (I) multiplied by time (T):
\[ [q] = [I][T] \]
- Thus,
\[ [q^2] = [I^2][T^2] \]
- The dimension of force (F) is:
\[ [F] = [M][L][T^{-2}] \]
- The dimension of distance squared (r²) is:
\[ [r^2] = [L^2] \]

3. **Substituting the dimensions into the formula**:
\[ [\epsilon_0] = \frac{[I^2][T^2]}{[M][L][T^{-2}] \cdot [L^2]} = \frac{[I^2][T^2]}{[M][L^3][T^{-2}]} \]
Simplifying this gives:
\[ [\epsilon_0] = [M^{-1}][L^{-3}][T^4][I^2] \]

### Step 2: Determine the Dimensional Formula for Permeability (μ₀)

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:
\[ [I] \]

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

### Final Results

- The dimensional formula for permittivity of vacuum (ε₀) is:
\[ [\epsilon_0] = [M^{-1}][L^{-3}][T^4][I^2] \]
- The dimensional formula for permeability of vacuum (μ₀) is:
\[ [\mu_0] = [M][L^2][T^{-2}][I^{-2}] \]