In terms of basic units of mass (M), length (L), time (T), and charge (Q), the dimensions of magnetic permeability of vacuum `(mu_0)` would be

Options

A. `["MLQ"^(-2)]`
B. `["LT"^(-1)"Q"^(-1)]`
C. `["ML"^(2)"T"^(-1)"Q"^(-2)]`
D. `["LTQ"^(-1)]`

Explanation

To find the dimensions of the magnetic permeability of vacuum (µ₀) in terms of the basic units of mass (M), length (L), time (T), and charge (Q), we can use the relationship derived from the force between two parallel current-carrying wires.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The force per unit length (F/L) between two parallel wires carrying currents I₁ and I₂ separated by a distance r is given by the formula:
\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi r} \]
Rearranging this gives:
\[ \mu_0 = \frac{F \cdot 2 \pi r}{I_1 I_2} \]

2. **Identify Dimensions**:
We need to find the dimensions of each term in the equation:
- **Force (F)** has the dimensions of mass times acceleration:
\[ [F] = MLT^{-2} \]
- **Distance (r)** has the dimensions of length:
\[ [r] = L \]
- **Current (I)** is defined as the rate of flow of charge, which gives it the dimensions:
\[ [I] = QT^{-1} \]

3. **Substitute Dimensions into the Equation**:
Now substituting the dimensions into the equation for µ₀:
\[ \mu_0 = \frac{(MLT^{-2}) \cdot (L)}{(QT^{-1}) \cdot (QT^{-1})} \]

4. **Simplify the Expression**:
This simplifies to:
\[ \mu_0 = \frac{MLT^{-2} \cdot L}{Q^2T^{-2}} = \frac{ML^2T^{-2}}{Q^2} \]

5. **Final Dimensions**:
Rearranging gives us:
\[ \mu_0 = ML^2Q^{-2}T^{-2} \]
Thus, the dimensions of magnetic permeability of vacuum (µ₀) are:
\[ [\mu_0] = MLQ^{-2} \]

### Final Answer:
The dimensions of magnetic permeability of vacuum (µ₀) are:
\[ [\mu_0] = MLQ^{-2} \]