A closed loop moves normal to a constant electric field between the plates of capacitor with plane of loop is normal to the field. When the loop is partially outside the plates of capacitor, the induced current in the loop

Options

A. depends on speed of loop
B. depends on size of loop
C. depends of intensity of electric field
D. zero

Answer

To solve the problem of the induced current in a closed loop moving in a constant electric field between the plates of a capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a parallel plate capacitor that creates a constant electric field between its plates.
- A closed loop is placed such that its plane is normal (perpendicular) to the electric field and it moves through the electric field, with part of the loop being outside the plates of the capacitor.

2. **Identify the Electric Field**:
- The electric field (E) between the plates of a capacitor is uniform and constant. This means that the electric field does not change with time.

3. **Calculate Electric Flux**:
- The electric flux (Φ_E) through the loop is given by the formula:
\[ \Phi_E = E \cdot A \]
where \(A\) is the area of the loop that is within the electric field. Since the loop is partially outside the capacitor, only the area inside the plates contributes to the flux.

4. **Determine Change in Flux**:
- Since the electric field is constant and the area of the loop that is inside the electric field does not change (as the loop moves), the electric flux through the loop remains constant.
- Therefore, the rate of change of electric flux (dΦ_E/dt) is zero:
\[ \frac{d\Phi_E}{dt} = 0 \]

5. **Apply Faraday's Law of Induction**:
- According to Faraday's law, the induced electromotive force (emf) in the loop is given by:
\[ \text{emf} = -\frac{d\Phi_E}{dt} \]
- Since we found that dΦ_E/dt = 0, it follows that:
\[ \text{emf} = 0 \]

6. **Conclusion about Induced Current**:
- The induced current (I) in the loop can be calculated using Ohm's law:
\[ I = \frac{\text{emf}}{R} \]
- Since the emf is zero, the induced current in the loop is also zero:
\[ I = 0 \]

### Final Answer:
The induced current in the loop is **zero**.

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