The magnetic field between the plates of a capacitor when `r gt R` is given by-

Options

A. `(mu_(0)I_(D)r)/(2piR^(2))`
B. `(mu_(0)I_(D))/(2piR )`
C. `(mu_(0)I_(D))/(2pir)`
D. zero

Explanation

To solve the problem regarding the magnetic field between the plates of a capacitor when \( r > R \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a parallel plate capacitor with plates of radius \( R \).
- We are interested in the magnetic field at a distance \( r \) from the center of the capacitor plates, where \( r > R \).

2. **Applying Ampère's Law**:
- According to Ampère's Law, the magnetic field \( B \) around a current-carrying conductor can be calculated using the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
- Here, \( \mu_0 \) is the permeability of free space, \( I \) is the current flowing through the capacitor, and \( r \) is the distance from the center of the capacitor.

3. **Identifying the Region**:
- Since we are looking at the region outside the plates of the capacitor (where \( r > R \)), we can directly use the formula derived from Ampère's Law.

4. **Substituting Values**:
- For \( r > R \), the magnetic field \( B \) can be expressed as:
\[ B = \frac{\mu_0 I}{2 \pi r} \]

5. **Final Expression**:
- Therefore, the magnetic field between the plates of the capacitor when \( r > R \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]