Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimensions of permeability are :

Options

A. `[M^(-1)LT^(-2)A]`
B. `[ML^(-2)T^(-2)A^(-1)]`
C. `[MLT^(-2)A^(-2)]`
D. `[MLT^(-1)A^(-1)]`

Explanation

To find the dimensions of permeability using mass (M), length (L), time (T), and current (A) as fundamental quantities, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Unit of Permeability**:
The unit of permeability is given as Newton per Ampere squared (N/A²).

2. **Express Newton in Fundamental Units**:
We know that force (Newton) can be expressed in terms of mass, length, and time. The formula for force is:
\[ \text{Force} = \text{Mass} \times \text{Acceleration} = M \cdot L \cdot T^{-2} \]
Therefore, the dimension of Newton (N) is:
\[ [N] = M \cdot L \cdot T^{-2} \]

3. **Substitute the Unit of Permeability**:
Now substituting the expression for Newton into the unit of permeability:
\[ \text{Permeability} = \frac{N}{A^2} = \frac{M \cdot L \cdot T^{-2}}{A^2} \]

4. **Combine the Dimensions**:
Thus, the dimensions of permeability (μ) can be written as:
\[ [\mu] = M \cdot L \cdot T^{-2} \cdot A^{-2} \]

5. **Final Expression**:
Therefore, the final dimensional formula for permeability is:
\[ [\mu] = M^1 \cdot L^1 \cdot T^{-2} \cdot A^{-2} \]

### Final Answer:
The dimensions of permeability are:
\[ [M^1 L^1 T^{-2} A^{-2}] \]