A closely wound solenoid has 500 turns per metre length of the solenoid. A current of 2 A flows to it. The magnetic induction at the end of the solenoid on its axis is:

Options

A. `6 pi xx 10^(-4) T`
B. `2 pi xx 10^(-4) T`
C. `4pi xx 10^(-4) T`
D. `8 pi xx 10^(-4) T`

Explanation

To solve the problem of finding the magnetic induction at the end of a closely wound solenoid, we can follow these steps:

### Step 1: Identify the given values
- Number of turns per meter (n) = 500 turns/m
- Current (I) = 2 A
- Permeability of free space (μ₀) = 4π × 10⁻⁷ T·m/A

### Step 2: Use the formula for magnetic induction at the end of the solenoid
The magnetic induction (B) at the end of a solenoid on its axis can be calculated using the formula:
\[ B = \frac{\mu_0 \cdot n \cdot I}{2} \]

### Step 3: Substitute the known values into the formula
Substituting the values we have:
\[ B = \frac{(4\pi \times 10^{-7}) \cdot (500) \cdot (2)}{2} \]

### Step 4: Simplify the expression
The 2 in the numerator and denominator cancels out:
\[ B = (4\pi \times 10^{-7}) \cdot (500) \]

### Step 5: Calculate the product
Now, calculate:
\[ B = 4\pi \times 500 \times 10^{-7} \]
\[ B = 2000\pi \times 10^{-7} \]

### Step 6: Write in standard form
We can express this as:
\[ B = 2\pi \times 10^{-4} \, \text{Tesla} \]

### Step 7: Conclusion
Thus, the magnetic induction at the end of the solenoid on its axis is:
\[ B = 2\pi \times 10^{-4} \, \text{Tesla} \]

### Final Answer
The correct option is \( 2\pi \times 10^{-4} \, \text{Tesla} \).

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