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 given that the magnetic field energy in an inductor changes from maximum to minimum in 5.0 ms, we can follow these steps:

### Step 1: Understand the relationship between magnetic energy and current
The magnetic field 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 through the inductor.

### Step 2: Analyze the behavior of current in an AC circuit
In an AC circuit, the current varies sinusoidally. The current reaches its maximum value and then goes to zero, which corresponds to the maximum and minimum energy stored in the inductor.

### Step 3: Determine the time taken for energy change
The problem states that the energy changes from maximum to minimum in 5.0 ms. This corresponds to the current going from its maximum value to zero. In one complete cycle of the AC wave, the current goes from maximum to zero in a quarter of the cycle (T/4).

### Step 4: Relate the time to the period of the AC source
Since the time taken for the current to go from maximum to zero is T/4, we can set up the equation:
\[ \frac{T}{4} = 5.0 \, \text{ms} \]
From this, we can find the total period \( T \):
\[ T = 4 \times 5.0 \, \text{ms} = 20.0 \, \text{ms} \]

### Step 5: Calculate the frequency
The frequency \( f \) is the reciprocal of the period \( T \):
\[ f = \frac{1}{T} \]
Substituting the value of \( T \):
\[ f = \frac{1}{20.0 \, \text{ms}} = \frac{1}{20 \times 10^{-3} \, \text{s}} = \frac{1}{0.02 \, \text{s}} = 50 \, \text{Hz} \]

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

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