The capacity of a parallel plate condenser is `50pF`. A magnetic field of `4xx10^(-7)T` is produced at a distance of `10cm` from the axis of the gap. The charging current is

Options

A. `0.1 A`
B. `0.2 A`
C. `0.3 A`
D. `0.15 A`

Explanation

To solve the problem step by step, we will follow the concepts of capacitance, magnetic field, and the relationship between them.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Capacitance \( C = 50 \, \text{pF} = 50 \times 10^{-12} \, \text{F} \)
- Magnetic field \( B = 4 \times 10^{-7} \, \text{T} \)
- Distance from the axis \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Use the Formula for Capacitance:**
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( A \) is the area of the plates and \( d \) is the separation between them. For this problem, we will rearrange the formula to find \( A \):
\[ A = C \cdot d / \varepsilon_0 \]

3. **Assume the Plate Separation \( d \):**
Since \( d \) is not provided, we will assume a reasonable value for the separation. Let's assume \( d = 0.01 \, \text{m} \) (1 cm).

4. **Calculate Area \( A \):**
Using \( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \):
\[ A = \frac{50 \times 10^{-12} \times 0.01}{8.85 \times 10^{-12}} \approx 5.65 \times 10^{-2} \, \text{m}^2 \]

5. **Relate Magnetic Field to Current:**
The magnetic field \( B \) around a wire carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

6. **Rearrange to Find Current \( I \):**
Rearranging the equation gives:
\[ I = \frac{B \cdot 2 \pi r}{\mu_0} \]

7. **Substitute Values:**
Substituting the known values:
\[ I = \frac{(4 \times 10^{-7}) \cdot (2 \pi \cdot 0.1)}{4\pi \times 10^{-7}} \]

8. **Simplify the Equation:**
The \( 4\pi \) cancels out:
\[ I = \frac{(4 \times 10^{-7}) \cdot (0.2 \pi)}{4\pi \times 10^{-7}} = 0.2 \, \text{A} \]

### Final Answer:
The charging current \( I \) is \( 0.2 \, \text{A} \).