The energy stored in an inductor of self inductance L carrying current l, is given by

Options

A. `Ll^(2)`
B. `-Ll^(2)`
C. `(1)/(2)Ll^(2)`
D. `-(1)/(2)Ll^(2)`

Answer

To find the energy stored in an inductor of self-inductance \( L \) carrying a current \( I \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Inductance**:
The self-inductance \( L \) of an inductor is a measure of how much magnetic flux is generated per unit current. The magnetic flux \( \Phi \) linked with the inductor is given by:
\[ \Phi = L \cdot I \]

2. **Induced EMF**:
The induced electromotive force (emf) \( \mathcal{E} \) in the inductor is related to the rate of change of magnetic flux. According to Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Substituting the expression for \( \Phi \):
\[ \mathcal{E} = -\frac{d(L \cdot I)}{dt} = -L \frac{dI}{dt} \]

3. **Power in the Inductor**:
The power \( P \) delivered to the inductor is given by the product of the current \( I \) and the induced emf \( \mathcal{E} \):
\[ P = I \cdot \mathcal{E} = I \cdot (-L \frac{dI}{dt}) = -L I \frac{dI}{dt} \]
Since we are interested in the energy stored, we can consider the positive value of power.

4. **Work Done**:
The power can also be expressed as the rate of work done \( \frac{dW}{dt} \):
\[ P = \frac{dW}{dt} \]
Thus, we can equate:
\[ \frac{dW}{dt} = L I \frac{dI}{dt} \]

5. **Integrating to Find Total Work Done**:
To find the total work done \( W \), we integrate both sides with respect to time. We can express the current changing from \( 0 \) to \( I \):
\[ W = \int_0^W dW = \int_0^I L I \, dI \]
This gives:
\[ W = L \int_0^I I \, dI \]

6. **Calculating the Integral**:
The integral \( \int_0^I I \, dI \) evaluates to:
\[ \int_0^I I \, dI = \frac{I^2}{2} \]
Therefore, substituting back, we have:
\[ W = L \cdot \frac{I^2}{2} = \frac{1}{2} L I^2 \]

7. **Conclusion**:
The energy stored in the inductor is:
\[ U = \frac{1}{2} L I^2 \]

### Final Answer:
The energy stored in an inductor of self-inductance \( L \) carrying current \( I \) is given by:
\[ U = \frac{1}{2} L I^2 \]