A solenoid of 1.5 metre length and 4.0 cm diameter possesses 10 turn per cm. A current of 10 A is flowing through it . The magnetic induction at axis inside the solenoid is

Options

A. `4pixx10^-3T`
B. `2pixx10^-5T`
C. `2pixx10^-2` gauss
D. `2pi xx10^-5` gauss

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 field inside the solenoid,
- \( \mu_0 \) 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 (in turns per meter),
- \( I \) is the current flowing through the solenoid (in amperes).

### Step 1: Calculate the number of turns per meter
The given number of turns is 10 turns per centimeter. To convert this to turns per meter:

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

### Step 2: Substitute the values into the formula
Now we can substitute the values into the formula for \( B \):

\[ B = \mu_0 n I \]

Substituting the known values:

- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( n = 1000 \, \text{turns/m} \)
- \( I = 10 \, \text{A} \)

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

### Step 3: Simplify the expression
Calculating the above expression step by step:

\[ B = 4\pi \times 10^{-7} \times 10000 \]
\[ B = 4\pi \times 10^{-3} \, \text{T} \]

### Step 4: Calculate the numerical value
Using the approximate value of \( \pi \approx 3.14 \):

\[ B \approx 4 \times 3.14 \times 10^{-3} \approx 12.56 \times 10^{-3} \, \text{T} \approx 0.01256 \, \text{T} \]

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

\[ B \approx 0.01256 \, \text{T} \text{ or } 12.56 \, \text{mT} \]

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