The magnetic field energy in an inductor changes from maximum value to minimum value in `5.0 ms` when connected to an AC source. The frequency of the source is

Options

A. `20 Hz
B. ``50 Hz`
C. `200 Hz`
D. `500 Hz`.

Answer

To find the frequency of the AC source when the magnetic field energy in an inductor changes from maximum to minimum value in 5.0 ms, we can follow these steps:

### Step 1: Understand the relationship between energy and current in an inductor
The energy stored in an inductor (E) is given by the formula:
\[ E = \frac{1}{2} L I^2 \]
where \(L\) is the inductance and \(I\) is the current.

### Step 2: Express current in terms of time
The current in an inductor connected to an AC source can be expressed as:
\[ I(t) = I_p \sin(\omega t) \]
where \(I_p\) is the peak current and \(\omega\) is the angular frequency.

### Step 3: Determine the maximum and minimum energy states
The maximum energy occurs when \(\sin^2(\omega t) = 1\) and the minimum energy occurs when \(\sin^2(\omega t) = 0\). The transition from maximum to minimum energy occurs over a quarter of the AC cycle.

### Step 4: Relate the time taken to the frequency
The time taken to go from maximum energy to minimum energy is:
\[ \Delta t = \frac{T}{4} \]
where \(T\) is the period of the AC signal. The period \(T\) is related to frequency \(f\) by:
\[ T = \frac{1}{f} \]
Thus,
\[ \Delta t = \frac{1}{4f} \]

### Step 5: Set up the equation with the given time
Given that \(\Delta t = 5.0 \, \text{ms} = 5.0 \times 10^{-3} \, \text{s}\), we can set up the equation:
\[ 5.0 \times 10^{-3} = \frac{1}{4f} \]

### Step 6: Solve for frequency \(f\)
Rearranging the equation gives:
\[ f = \frac{1}{4 \times 5.0 \times 10^{-3}} = \frac{1}{0.02} = 50 \, \text{Hz} \]

### Conclusion
The frequency of the AC source is \(50 \, \text{Hz}\).

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