A solenoid of 2.5 m length and 2.0 cm diameter possesses 10 turns per cm. A current of 0.5 A is flowing through it . The magnetic induction at axis inside the solenoid is

Options

A. `2pixx10^(-4)T`
B. `2pixx10^(-5)T`
C. `2pixx10^(-6)T`
D. `2pixx10^(-7)T`

Answer

To find the magnetic induction (magnetic field) at the axis inside a solenoid, we can use the formula:

\[ B = \mu_0 \cdot n \cdot I \]

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

### Step 1: Identify the given values
- Length of the solenoid, \( L = 2.5 \, \text{m} \)
- Diameter of the solenoid, \( d = 2.0 \, \text{cm} = 0.02 \, \text{m} \)
- Number of turns per centimeter, \( N = 10 \, \text{turns/cm} = 10 \times 100 = 1000 \, \text{turns/m} \)
- Current, \( I = 0.5 \, \text{A} \)

### Step 2: Calculate the magnetic field \( B \)
Using the formula for the magnetic field inside the solenoid:

\[ B = \mu_0 \cdot n \cdot I \]

Substituting the values:

\[ B = (4\pi \times 10^{-7} \, \text{T m/A}) \cdot (1000 \, \text{turns/m}) \cdot (0.5 \, \text{A}) \]

### Step 3: Perform the calculations
First, calculate the product of \( \mu_0 \) and \( n \):

\[ B = 4\pi \times 10^{-7} \cdot 1000 \cdot 0.5 \]

Calculating \( 4\pi \):

\[ 4\pi \approx 12.5664 \]

Now substituting this back into the equation:

\[ B = 12.5664 \times 10^{-7} \cdot 1000 \cdot 0.5 \]

Calculating \( 12.5664 \cdot 1000 \cdot 0.5 \):

\[ B = 12.5664 \times 500 \times 10^{-7} \]

\[ B = 6283.2 \times 10^{-7} \, \text{T} \]

\[ B = 6.2832 \times 10^{-4} \, \text{T} \]

### Step 4: Final answer
Thus, the magnetic induction at the axis inside the solenoid is:

\[ B \approx 2\pi \times 10^{-4} \, \text{T} \]