The dimensions of `mu_(0)` are

Options

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

Answer

To find the dimensions of \( \mu_0 \) (the permeability of free space), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of \( \mu_0 \)**:
- \( \mu_0 \) is known as the permeability of free space. It relates magnetic induction (B) to magnetic field intensity (H). The relationship can be expressed as:
\[ B = \mu_0 H \]

2. **Identify the SI Units**:
- The SI unit of \( \mu_0 \) is given as henry per meter (H/m).

3. **Determine the Dimensions of Henry**:
- The unit henry (H) can be expressed in terms of base SI units:
\[ 1 \, \text{H} = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \cdot \text{A}^{-2} = \text{ML}^2\text{T}^{-2}\text{A}^{-2} \]
- Therefore, the dimensions of henry are:
\[ [H] = \text{ML}^2\text{T}^{-2}\text{A}^{-2} \]

4. **Express the Dimensions of \( \mu_0 \)**:
- Since \( \mu_0 \) has the unit of henry per meter, we can write:
\[ [\mu_0] = \frac{[H]}{[\text{m}]} = \frac{\text{ML}^2\text{T}^{-2}\text{A}^{-2}}{\text{L}} = \text{ML}^1\text{T}^{-2}\text{A}^{-2} \]

5. **Final Result**:
- Thus, the dimensions of \( \mu_0 \) are:
\[ [\mu_0] = \text{ML}^1\text{T}^{-2}\text{A}^{-2} \]