The diameter of the condenser plate is `4cm`. It is charged by an external current of `0.2 A`.The maximum magnetic field induced in the gap

Options

A. `2 muT`
B. `4 muT`
C. `6 muT`
D. `8 muT`

Explanation

To find the maximum magnetic field induced in the gap of a condenser plate, we can use the formula for the magnetic field around a current-carrying conductor. The relevant formula is:

\[ B = \frac{\mu_0 I}{2 \pi R} \]

Where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( I \) is the current (in Amperes),
- \( R \) is the radius of the condenser plate (in meters).

### Step-by-Step Solution:

1. **Identify the given values:**
- Diameter of the condenser plate = 4 cm
- Current \( I = 0.2 \, \text{A} \)

2. **Convert diameter to radius:**
- Radius \( R = \frac{\text{Diameter}}{2} = \frac{4 \, \text{cm}}{2} = 2 \, \text{cm} = 0.02 \, \text{m} \)

3. **Substitute the values into the formula:**
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( I = 0.2 \, \text{A} \)
- \( R = 0.02 \, \text{m} \)

4. **Calculate the magnetic field \( B \):**
\[ B = \frac{4\pi \times 10^{-7} \times 0.2}{2\pi \times 0.02} \]

5. **Simplify the expression:**
- The \( \pi \) cancels out:
\[ B = \frac{4 \times 10^{-7} \times 0.2}{2 \times 0.02} \]
- Calculate the denominator:
\[ 2 \times 0.02 = 0.04 \]
- Now substitute back:
\[ B = \frac{4 \times 10^{-7} \times 0.2}{0.04} \]

6. **Calculate the numerator:**
\[ 4 \times 0.2 = 0.8 \]
- Therefore:
\[ B = \frac{0.8 \times 10^{-7}}{0.04} \]

7. **Final calculation:**
\[ B = 0.8 \times 10^{-7} \div 0.04 = 0.8 \times 10^{-7} \times 25 = 20 \times 10^{-7} = 2 \times 10^{-6} \, \text{T} \]

### Conclusion:
The maximum magnetic field induced in the gap is:
\[ B = 2 \times 10^{-6} \, \text{T} \, \text{or} \, 2 \, \mu\text{T} \]