An LC circuit contains a 20 mH inductor and a `50 mu F` capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant at which the circuit which is closed be t=0. At what time the energy stored is completely magnetic ?

Options

A. t = 0
B. `t=1.54 ms`
C. `t=3.14 ms`
D. `t=6.28 ms`

Explanation

To solve the problem step by step, we will follow the outlined process:

### Step 1: Calculate the initial energy stored in the capacitor
The energy stored in a capacitor can be calculated using the formula:
\[ E = \frac{Q^2}{2C} \]
Where:
- \( Q = 10 \, \text{mC} = 10 \times 10^{-3} \, \text{C} \)
- \( C = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F} \)

Substituting the values:
\[ E = \frac{(10 \times 10^{-3})^2}{2 \times (50 \times 10^{-6})} \]
Calculating the numerator:
\[ (10 \times 10^{-3})^2 = 100 \times 10^{-6} = 10^{-4} \]
Now substituting back:
\[ E = \frac{10^{-4}}{2 \times 50 \times 10^{-6}} = \frac{10^{-4}}{100 \times 10^{-6}} = \frac{10^{-4}}{10^{-4}} = 1 \, \text{J} \]

### Step 2: Calculate the frequency of oscillation
The frequency of oscillation in an LC circuit is given by:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]
Where:
- \( L = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \)
- \( C = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F} \)

Substituting the values:
\[ f_0 = \frac{1}{2\pi\sqrt{(20 \times 10^{-3})(50 \times 10^{-6})}} \]
Calculating the product:
\[ LC = (20 \times 10^{-3})(50 \times 10^{-6}) = 1 \times 10^{-6} \, \text{H}\cdot\text{F} \]
Now substituting back:
\[ f_0 = \frac{1}{2\pi\sqrt{1 \times 10^{-6}}} = \frac{1}{2\pi \times 10^{-3}} = \frac{1000}{2\pi} \approx 159.15 \, \text{Hz} \]

### Step 3: Calculate the time period of oscillation
The time period \( T \) is the reciprocal of frequency:
\[ T = \frac{1}{f_0} \approx \frac{1}{159.15} \approx 0.00628 \, \text{s} \approx 6.28 \, \text{s} \]

### Step 4: Determine when the energy is completely magnetic
In an LC circuit, the energy oscillates between electric and magnetic forms. The energy is completely magnetic at specific intervals:
- The magnetic energy is maximum at \( t = \frac{T}{4}, \frac{3T}{4}, \frac{5T}{4}, \ldots \)

Calculating for the first instance:
\[ t = \frac{T}{4} = \frac{6.28}{4} \approx 1.57 \, \text{s} \]

### Final Answer
The time at which the energy stored in the circuit is completely magnetic is approximately:
\[ t \approx 1.57 \, \text{s} \]