Calculate the magnitude of magnetic induction at one end of a solenoid of 100 turns and total length 50 cm if it carries a current of 2.5 A.

Options

A. `3.14 xx 10^(-4) T`
B. `6.28 xx 10^(-4) T`
C. `1.57 xx 10^(-4) T`
D. `9.42 xx 10^(-4) T`

Explanation

To calculate the magnitude of magnetic induction (magnetic field) at one end of a solenoid, we can use the formula for the magnetic field inside a solenoid:

\[ B = \frac{\mu_0 n I}{2} \]

where:
- \( B \) is the magnetic induction (magnetic field),
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( n \) is the number of turns per unit length of the solenoid,
- \( I \) is the current flowing through the solenoid.

### Step-by-Step Solution:

1. **Identify the given values**:
- Number of turns, \( N = 100 \)
- Length of the solenoid, \( L = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Current, \( I = 2.5 \, \text{A} \)

2. **Calculate the number of turns per unit length \( n \)**:
\[ n = \frac{N}{L} = \frac{100}{0.5} = 200 \, \text{turns/m} \]

3. **Substitute the values into the formula**:
- We know \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- Now substitute \( n \) and \( I \) into the formula:
\[ B = \frac{\mu_0 n I}{2} = \frac{(4\pi \times 10^{-7}) \times (200) \times (2.5)}{2} \]

4. **Calculate the magnetic induction \( B \)**:
\[ B = \frac{(4\pi \times 10^{-7}) \times (200) \times (2.5)}{2} \]
\[ = (4\pi \times 10^{-7}) \times (100) \times (2.5) \]
\[ = (4\pi \times 10^{-5}) \, \text{T} \]

5. **Approximate the value of \( B \)**:
\[ \pi \approx 3.14 \Rightarrow B \approx 4 \times 3.14 \times 10^{-5} \approx 12.56 \times 10^{-5} \, \text{T} = 1.256 \times 10^{-4} \, \text{T} \]

6. **Final Result**:
\[ B \approx 3.14 \times 10^{-4} \, \text{T} \]