An inductor in connected to an AC source . Magnetic field energy stored in inductor is found to change from its minimum value to maximum value in 10 ms. Frequency of AC source is

Options

A. 25 Hz.
B. 50 Hz.
C. 75 Hz.
D. 100 Hz.

Explanation

To solve the problem, we need to determine the frequency of the AC source connected to the inductor based on the information provided about the magnetic field energy stored in the inductor.

### Step-by-Step Solution:

1. **Understanding the Energy in the Inductor**:
The energy (E) stored in an inductor is given by the formula:
\[ E = \frac{1}{2} L I^2 \]
where \(L\) is the inductance and \(I\) is the current flowing through the inductor.

2. **Identifying Minimum and Maximum Energy**:
- The minimum energy (\(E_{\text{min}}\)) occurs when the current \(I = 0\). Thus, \(E_{\text{min}} = 0\).
- The maximum energy (\(E_{\text{max}}\)) occurs when the current reaches its maximum value (\(I = I_{\text{max}}\)). Therefore, \(E_{\text{max}} = \frac{1}{2} L I_{\text{max}}^2\).

3. **Time to Change from Minimum to Maximum Energy**:
According to the problem, the energy changes from its minimum value to maximum value in 10 ms. This transition occurs as the current goes from 0 to \(I_{\text{max}}\).

4. **Relating Time to the AC Cycle**:
In an AC circuit, the current varies sinusoidally. The time taken for the current to go from 0 to its maximum value is \(T/4\), where \(T\) is the time period of the AC source. Given that this time is 10 ms, we can write:
\[ \frac{T}{4} = 10 \text{ ms} \]

5. **Calculating the Time Period (T)**:
To find the total time period \(T\), we multiply both sides of the equation by 4:
\[ T = 4 \times 10 \text{ ms} = 40 \text{ ms} \]

6. **Calculating the Frequency (f)**:
The frequency \(f\) is the reciprocal of the time period \(T\):
\[ f = \frac{1}{T} = \frac{1}{40 \times 10^{-3} \text{ s}} = \frac{1}{0.04 \text{ s}} = 25 \text{ Hz} \]

### Final Answer:
The frequency of the AC source is \(25 \text{ Hz}\).