A solenoid of length 1.5 m and 4 cm diameter possesses 10 turns per cm. A current of 5 A is flowing through it, the magnetic induction at axis inside the solenoid is
`(mu_0 = 4pi xx10^(-7)" weber amp"^(-1)m^(-1))`

Options

A. `4pixx10^(-5)` gauss
B. `2pixx10^(-5)` gauss
C. `4pixx10^(-5)` tesla
D. `2pixx10^(-5)` tesla

Explanation

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

\[ B = \mu_0 n I \]

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

### Step-by-step Solution:

**Step 1: Calculate the number of turns per meter (n)**
Given that the solenoid has 10 turns per cm, we need to convert this to turns per meter.

\[ n = 10 \, \text{turns/cm} \times 100 \, \text{cm/m} = 1000 \, \text{turns/m} \]

**Step 2: Identify the current (I)**
The current flowing through the solenoid is given as:

\[ I = 5 \, \text{A} \]

**Step 3: Substitute values into the formula**
Now we can substitute the values of \( \mu_0 \), \( n \), and \( I \) into the formula for magnetic induction.

\[ B = \mu_0 n I = (4\pi \times 10^{-7} \, \text{Wb/A/m}) \times (1000 \, \text{turns/m}) \times (5 \, \text{A}) \]

**Step 4: Calculate the magnetic induction (B)**
Calculating this gives:

\[ B = 4\pi \times 10^{-7} \times 1000 \times 5 \]

\[ B = 4\pi \times 5 \times 10^{-4} = 20\pi \times 10^{-4} \, \text{T} \]

**Step 5: Convert to a more standard form**
To express this in a more standard form:

\[ B = 2\pi \times 10^{-3} \, \text{T} = 2\pi \times 10^{-5} \, \text{T} \]

### Final Answer:
The magnetic induction at the axis inside the solenoid is:

\[ B = 2\pi \times 10^{-5} \, \text{T} \]