A solenoid of length `50cm`, having `100` turns carries a current of `2*5A`. Find the magnetic field, (a) in the interior of the solenoid, (b) at one end of the solenoid.
Given `mu_0=4pixx10^-7WbA^-1m^-1`.

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 solve the problem step by step, we need to find the magnetic field inside the solenoid and at one end of the solenoid using the given parameters.

### Given Data:
- Length of the solenoid, \( L = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Number of turns, \( N = 100 \)
- Current, \( I = 2.5 \, \text{A} \)
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{Wb/A/m} \)

### Step 1: Calculate the number of turns per unit length (n)
The number of turns per unit length \( n \) can be calculated using the formula:
\[ n = \frac{N}{L} \]
Substituting the values:
\[ n = \frac{100}{0.5} = 200 \, \text{turns/m} \]

### Step 2: Calculate the magnetic field inside the solenoid (B)
The magnetic field inside a long solenoid is given by the formula:
\[ B = \mu_0 n I \]
Substituting the values:
\[ B = (4\pi \times 10^{-7}) \times (200) \times (2.5) \]
Calculating this step-by-step:
1. Calculate \( 4\pi \):
\[ 4\pi \approx 12.5664 \]
2. Multiply by \( 200 \):
\[ 12.5664 \times 200 = 2513.28 \]
3. Multiply by \( 2.5 \):
\[ 2513.28 \times 2.5 = 6283.2 \]
4. Finally, multiply by \( 10^{-7} \):
\[ B = 6283.2 \times 10^{-7} = 6.2832 \times 10^{-4} \, \text{T} \]

Thus, the magnetic field inside the solenoid is:
\[ B = 6.28 \times 10^{-4} \, \text{T} \]

### Step 3: Calculate the magnetic field at one end of the solenoid
The magnetic field at one end of the solenoid is given by:
\[ B = \frac{\mu_0 n I}{2} \]
Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \times (200) \times (2.5)}{2} \]
Using the previous calculation for \( 4\pi \times 200 \times 2.5 \):
\[ B = \frac{6283.2 \times 10^{-7}}{2} = 3141.6 \times 10^{-7} = 3.1416 \times 10^{-4} \, \text{T} \]

Thus, the magnetic field at one end of the solenoid is:
\[ B = 3.14 \times 10^{-4} \, \text{T} \]

### Final Answers:
(a) The magnetic field in the interior of the solenoid is \( 6.28 \times 10^{-4} \, \text{T} \).

(b) The magnetic field at one end of the solenoid is \( 3.14 \times 10^{-4} \, \text{T} \).