The dimensions of coefficient of self inductances are

Options

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

Answer

To find the dimensions of the coefficient of self-inductance (L), we start with the formula for the potential energy (U) stored in an inductor:

\[ U = \frac{1}{2} L i^2 \]

Where:
- \( U \) is the potential energy,
- \( L \) is the self-inductance,
- \( i \) is the current.

### Step 1: Determine the dimensions of potential energy (U)
The potential energy has the same dimensions as work, which is given by the formula:

\[ \text{Work} = \text{Force} \times \text{Displacement} \]

The dimension of force (F) is:

\[ F = \text{mass} \times \text{acceleration} = M \cdot L \cdot T^{-2} \]

Where:
- \( M \) is mass,
- \( L \) is length,
- \( T \) is time.

Thus, the dimension of work (or potential energy) is:

\[ \text{Dimension of } U = [F] \cdot [\text{Displacement}] = (M \cdot L \cdot T^{-2}) \cdot L = M \cdot L^2 \cdot T^{-2} \]

### Step 2: Relate the dimensions of self-inductance (L) to potential energy and current
From the formula for potential energy in the inductor, we can express the dimensions of self-inductance (L) as:

\[ L = \frac{U}{i^2} \]

### Step 3: Determine the dimensions of current (i)
The dimension of electric current (I) is represented as:

\[ [I] = A \]

Where \( A \) is the unit of electric current (Ampere).

### Step 4: Substitute the dimensions into the equation
Now substituting the dimensions of potential energy and current into the equation for self-inductance:

\[ [L] = \frac{[U]}{[i]^2} = \frac{M \cdot L^2 \cdot T^{-2}}{A^2} \]

### Step 5: Final expression for dimensions of self-inductance
Thus, we can write the dimensions of self-inductance (L) as:

\[ [L] = M \cdot L^2 \cdot T^{-2} \cdot A^{-2} \]

### Conclusion
The dimensions of the coefficient of self-inductance are:

\[ [L] = M \cdot L^2 \cdot T^{-2} \cdot A^{-2} \]