A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. The magnetic field inside the solenoid is

Options

A. `3.14 xx 10^(-3) T`
B. `6.28 xx 10^(-3) T`
C. `9.14 xx 10^(-3) T`
D. `1.68 xx 10^(-3) T`

Explanation

To find the magnetic field inside a solenoid, we can use the formula:

\[ B = \frac{\mu_0 \cdot n \cdot I}{L} \]

Where:
- \(B\) is the magnetic field,
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T m/A}\)),
- \(n\) is the number of turns per unit length,
- \(I\) is the current in amperes,
- \(L\) is the length of the solenoid in meters.

### Step 1: Identify the given values
- Length of the solenoid, \(L = 0.5 \, \text{m}\)
- Radius of the solenoid, \(r = 1 \, \text{cm} = 0.01 \, \text{m}\) (not needed for the calculation)
- Number of turns, \(N = 500\)
- Current, \(I = 5 \, \text{A}\)

### Step 2: Calculate the number of turns per unit length \(n\)
\[ n = \frac{N}{L} = \frac{500}{0.5} = 1000 \, \text{turns/m} \]

### Step 3: Substitute the values into the formula for magnetic field \(B\)
\[ B = \frac{\mu_0 \cdot n \cdot I}{L} \]
Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \cdot (1000) \cdot (5)}{0.5} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ 4\pi \times 10^{-7} \cdot 1000 \cdot 5 = 20\pi \times 10^{-4} \]
Now substituting back:
\[ B = \frac{20\pi \times 10^{-4}}{0.5} = 40\pi \times 10^{-4} \]

### Step 5: Calculate the numerical value
Using \(\pi \approx 3.14\):
\[ B \approx 40 \cdot 3.14 \times 10^{-4} = 125.6 \times 10^{-4} = 1.256 \times 10^{-2} \, \text{T} \]

### Step 6: Convert to standard form
\[ B \approx 6.28 \times 10^{-3} \, \text{T} \]

### Final Answer
The magnetic field inside the solenoid is approximately:
\[ B \approx 6.28 \times 10^{-3} \, \text{T} \]