A solenoid 1.5 metre and 4.0 cm in diameter possesses 10 turns/cm. A current of 5.0 A is flowing throught it. Calculate the magnetic induction (i) inside are (ii) At one end on the axis of solenoid respectively

Options

A. `2pi xx 10^(-3)T, pi xx 10^(-3)T`
B. `pi xx 10^(-3) T, 2pi xx 10^(-3)T`
C. `2pi xx 10^(-3) T, 2pi xx 10&(-3) T`
D. `pi xx 10^(-3)T, pi xx 10^(-3) T`

Answer

To solve the problem, we need to calculate the magnetic induction (magnetic field) inside the solenoid and at one end on the axis of the solenoid. We will use the formulas for magnetic induction in a solenoid.

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Length of the solenoid, \( L = 1.5 \, \text{m} \)
- Diameter of the solenoid, \( d = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Radius of the solenoid, \( R = \frac{d}{2} = \frac{0.04}{2} = 0.02 \, \text{m} \)
- Number of turns per unit length, \( n = 10 \, \text{turns/cm} = 10 \times 100 = 1000 \, \text{turns/m} \)
- Current flowing through the solenoid, \( I = 5.0 \, \text{A} \)

**Step 2: Calculate the magnetic induction inside the solenoid.**
The formula for the magnetic induction \( B \) inside a solenoid is given by:
\[ B = \mu_0 n I \]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)

Substituting the values:
\[ B = (4\pi \times 10^{-7}) \times (1000) \times (5) \]
Calculating this:
\[ B = 4\pi \times 5 \times 10^{-4} = 20\pi \times 10^{-7} \, \text{T} \]
Simplifying further:
\[ B = 20\pi \times 10^{-4} \, \text{T} = 2\pi \times 10^{-3} \, \text{T} \]

**Step 3: Calculate the magnetic induction at one end on the axis of the solenoid.**
The formula for the magnetic induction \( B \) at one end on the axis of the solenoid is given by:
\[ B = \frac{1}{2} \mu_0 n I \]
Using the previously calculated value of \( \mu_0 n I \):
\[ B = \frac{1}{2} (2\pi \times 10^{-3}) = \pi \times 10^{-3} \, \text{T} \]

### Final Answers:
(i) Magnetic induction inside the solenoid: \( B = 2\pi \times 10^{-3} \, \text{T} \)

(ii) Magnetic induction at one end on the axis of the solenoid: \( B = \pi \times 10^{-3} \, \text{T} \)

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