A capacitor in LC oscillatory circuit has a maximum potential of V volts and maximum energy E. When the capacitor has a potential `V_(1)` volts and energy `E_(1)` joules, what is the emf across the inductor and energy stered in the magnetic field

Options

A. `V-V_(1),E-E_(1)`
B. `V-V_(1), E_(1)`
C. `V_(1),E_(1)`
D. `V_(1),E-E_(1)`

Answer

To solve the problem, we need to analyze the LC oscillatory circuit and apply the principles of energy conservation and the relationships between voltage, energy, and capacitance.

### Step-by-Step Solution:

1. **Understanding the LC Circuit**:
- In an LC circuit, energy oscillates between the capacitor and the inductor. The maximum energy stored in the capacitor when it is fully charged is given by \( E = \frac{1}{2} C V^2 \), where \( V \) is the maximum voltage across the capacitor.

2. **Energy Stored in the Capacitor**:
- When the capacitor has a voltage \( V_1 \), the energy stored in the capacitor at this moment is given by:
\[ E_1 = \frac{1}{2} C V_1^2 \]

3. **Total Energy Conservation**:
- The total energy \( E \) in the LC circuit is conserved. Therefore, the total energy can be expressed as:
\[ E = E_1 + E_L \]
where \( E_L \) is the energy stored in the inductor.

4. **Finding the Energy in the Inductor**:
- Rearranging the energy conservation equation gives us:
\[ E_L = E - E_1 \]

5. **Finding the EMF across the Inductor**:
- The potential difference (emf) across the inductor \( V_L \) is equal to the voltage across the capacitor at that moment, which is:
\[ V_L = V_1 \]

### Final Answers:
- The emf across the inductor is \( V_1 \).
- The energy stored in the magnetic field of the inductor is \( E - E_1 \).