A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the centre of the solenoid when a current of 2 A flows through it is

Options

A. `2.4 × 10^3 A m^(–1)`
B. `1.2 × 10^3 A m^(–1)`
C. `1 A m^(–1)`
D. `2.4 × 10^(–3) A m^(–1)`

Answer

To solve the problem, we need to find the magnetic intensity (H) at the center of a solenoid when a current flows through it. The formula for magnetic intensity in a solenoid is given by:

\[ H = n \cdot I \]

where:
- \( H \) is the magnetic intensity in amperes per meter (A/m),
- \( n \) is the number of turns per unit length (turns/m),
- \( I \) is the current in amperes (A).

### Step-by-Step Solution:

1. **Identify the given values**:
- Number of turns (N) = 1200 turns
- Length of the solenoid (L) = 2 m
- Current (I) = 2 A

2. **Calculate the number of turns per unit length (n)**:
\[ n = \frac{N}{L} = \frac{1200 \text{ turns}}{2 \text{ m}} = 600 \text{ turns/m} \]

3. **Substitute the values into the formula for magnetic intensity (H)**:
\[ H = n \cdot I = 600 \text{ turns/m} \cdot 2 \text{ A} = 1200 \text{ A/m} \]

4. **Express the result in scientific notation**:
\[ H = 1.2 \times 10^3 \text{ A/m} \]

### Final Answer:
The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is \( 1.2 \times 10^3 \text{ A/m} \).