A solenoid of length `20cm` and radius `2cm` is closely wound with 200 turns. The magnetic field intensity at either end of the solenoid when the current in the winding is `5 amp`. Is

Options

A. `2500Amp//m`
B. `2000 Amp//m`
C. `1750 Amp//m`
D. `2940Amp//m`

Explanation

To find the magnetic field intensity at either end of the solenoid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the solenoid (L) = 20 cm = 0.2 m
- Radius of the solenoid (r) = 2 cm (not needed for this calculation)
- Number of turns (N) = 200
- Current (I) = 5 A

2. **Calculate the Number of Turns per Unit Length (n):**
\[ n = \frac{N}{L} = \frac{200 \text{ turns}}{0.2 \text{ m}} = 1000 \text{ turns/m} \]

3. **Use the Formula for Magnetic Field Intensity (H) at the Ends of the Solenoid:**
The magnetic field intensity at the ends of a long solenoid is given by:
\[ H = \frac{N \cdot I}{2 \cdot L} \]
where \(N\) is the total number of turns, \(I\) is the current, and \(L\) is the length of the solenoid.

4. **Substituting the Values into the Formula:**
\[ H = \frac{200 \text{ turns} \cdot 5 \text{ A}}{2 \cdot 0.2 \text{ m}} = \frac{1000}{0.4} = 2500 \text{ A/m} \]

5. **Conclusion:**
The magnetic field intensity at either end of the solenoid is:
\[ H = 2500 \text{ A/m} \]