The number of turns per unit length of a solenoid is 10. If its average radius is 5 cm and it carries a current of 10A, then the ratio of flux densities obtained at the centre and at the end on the axis will be -

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `1:4`

Answer

To solve the problem, we need to find the ratio of the magnetic flux densities at the center and at the end of a solenoid. Let's break this down step by step.

### Step 1: Identify the given values
- Number of turns per unit length (n) = 10 turns/m
- Average radius (r) = 5 cm = 0.05 m
- Current (I) = 10 A

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

Substituting the values:
\[ B_C = \mu_0 n I = (4\pi \times 10^{-7}) \times (10) \times (10) \]
\[ B_C = (4\pi \times 10^{-7}) \times 100 \]
\[ B_C = 4\pi \times 10^{-5} \, \text{T} \]

### Step 3: Calculate the magnetic flux density at the end of the solenoid
At the end of the solenoid, the magnetic field is given by:
\[ B_E = \frac{\mu_0 n I}{2} \]

Substituting the values:
\[ B_E = \frac{(4\pi \times 10^{-7}) \times (10) \times (10)}{2} \]
\[ B_E = \frac{(4\pi \times 10^{-7}) \times 100}{2} \]
\[ B_E = 2\pi \times 10^{-5} \, \text{T} \]

### Step 4: Calculate the ratio of the magnetic flux densities
Now, we can find the ratio of the magnetic flux densities at the center and at the end:
\[ \text{Ratio} = \frac{B_C}{B_E} = \frac{4\pi \times 10^{-5}}{2\pi \times 10^{-5}} \]

The \( \pi \) and \( 10^{-5} \) terms cancel out:
\[ \text{Ratio} = \frac{4}{2} = 2 \]

### Final Answer
The ratio of the magnetic flux densities obtained at the center and at the end on the axis of the solenoid is \( 2:1 \).

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