A current of 1 A through a coil of inductance of 200 mH is increasing at a rate of `0.5As^(-1)`. The energy stored in the inductor per second is

Options

A. `0.5Js^(-1)`
B. `5.0Js^(-1)`
C. `0.1Js^(-1)`
D. `2.0Js^(-1)`

Explanation

To solve the problem of finding the energy stored in the inductor per second, we will follow these steps:

### Step 1: Understand the formula for energy stored in an inductor
The energy (U) stored in an inductor is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
where:
- \( U \) is the energy in joules,
- \( L \) is the inductance in henries,
- \( I \) is the current in amperes.

### Step 2: Differentiate the energy with respect to time
To find the energy stored in the inductor per second, we need to differentiate the energy with respect to time:
\[ \frac{dU}{dt} = \frac{d}{dt} \left( \frac{1}{2} L I^2 \right) \]
Using the chain rule, we get:
\[ \frac{dU}{dt} = L I \frac{dI}{dt} \]
This equation tells us how the energy changes with time based on the inductance, current, and the rate of change of current.

### Step 3: Substitute the known values
We have the following values:
- \( L = 200 \, \text{mH} = 200 \times 10^{-3} \, \text{H} \)
- \( I = 1 \, \text{A} \)
- \( \frac{dI}{dt} = 0.5 \, \text{A/s} \)

Now, substituting these values into the differentiated equation:
\[ \frac{dU}{dt} = (200 \times 10^{-3}) \times (1) \times (0.5) \]

### Step 4: Calculate the energy stored per second
Now, we perform the calculation:
\[ \frac{dU}{dt} = 200 \times 10^{-3} \times 0.5 \]
\[ \frac{dU}{dt} = 100 \times 10^{-3} \]
\[ \frac{dU}{dt} = 0.1 \, \text{J/s} \]

### Final Answer
The energy stored in the inductor per second is:
\[ \frac{dU}{dt} = 0.1 \, \text{J/s} \]

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