Find the dimensions of inductance :-

Options

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

Explanation

To find the dimensions of inductance, we can start from the formula for the potential energy stored in an inductor, which is given by:

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

where:
- \( U \) is the potential energy,
- \( L \) is the inductance,
- \( I \) is the current.

### Step 1: Rearranging the formula
From the formula, we can express inductance \( L \) as:

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

### Step 2: Finding the dimensions of energy (U)
The dimensions of energy can be derived from the kinetic energy formula:

\[ U = \frac{1}{2} m v^2 \]

The dimensions of mass \( m \) are \( [M] \) and the dimensions of velocity \( v \) are \( [L T^{-1}] \). Therefore, the dimensions of energy \( U \) are:

\[ [U] = [M][L T^{-1}]^2 = [M][L^2 T^{-2}] \]

Thus, the dimensions of energy \( U \) are:

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

### Step 3: Finding the dimensions of current (I)
The dimensions of electric current \( I \) are represented as:

\[ [I] = [A] \]

where \( A \) is the unit of current (Ampere).

### Step 4: Substituting dimensions into the inductance formula
Now, substituting the dimensions of \( U \) and \( I \) into the equation for \( L \):

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

### Step 5: Simplifying the dimensions
Now, we simplify the dimensions:

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

### Final Result
Thus, the dimensions of inductance \( L \) are:

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