A solenoid is 3 m long and its inner diameter is 4.0 cm . It has three layers of windings of 2000 turns each and carries a current of 2.0 amperes. The magnetic flux for an cross-section of the solenoid is nearly

Options

A. 3.8 mwb
B. 12.8 mwb
C. 18.2 mwb
D. 6.4 mwb

Explanation

To find the magnetic flux through a cross-section of the solenoid, we can follow these steps:

### Step 1: Identify the given values
- Length of the solenoid (L) = 3 m
- Inner diameter of the solenoid (D) = 4.0 cm = 0.04 m
- Number of turns per layer = 2000 turns
- Number of layers = 3
- Current (I) = 2.0 A

### Step 2: Calculate the total number of turns (N)
The total number of turns (N) in the solenoid can be calculated as:
\[ N = \text{Number of turns per layer} \times \text{Number of layers} = 2000 \times 3 = 6000 \text{ turns} \]

### Step 3: Calculate the number of turns per unit length (n)
The number of turns per unit length (n) is given by:
\[ n = \frac{N}{L} = \frac{6000}{3} = 2000 \text{ turns/m} \]

### Step 4: Calculate the magnetic field (B) inside the solenoid
The magnetic field inside a solenoid is given by the formula:
\[ B = \mu_0 \cdot n \cdot I \]
Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space). Substituting the values:
\[ B = 4\pi \times 10^{-7} \cdot 2000 \cdot 2 \]
Calculating this gives:
\[ B = 4\pi \times 10^{-7} \cdot 4000 \]
\[ B = 16\pi \times 10^{-4} \, \text{T} \]

### Step 5: Calculate the cross-sectional area (A)
The cross-sectional area (A) of the solenoid can be calculated using the formula for the area of a circle:
\[ A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.04}{2}\right)^2 = \pi \left(0.02\right)^2 \]
\[ A = \pi \cdot 0.0004 \, \text{m}^2 \]
\[ A = 0.0004\pi \, \text{m}^2 \]

### Step 6: Calculate the magnetic flux (Φ)
The magnetic flux (Φ) through the cross-section of the solenoid is given by:
\[ \Phi = B \cdot A \]
Substituting the values we calculated:
\[ \Phi = (16\pi \times 10^{-4}) \cdot (0.0004\pi) \]
\[ \Phi = 16 \cdot 0.0004 \cdot \pi^2 \times 10^{-4} \]
\[ \Phi = 6.4 \times 10^{-7} \cdot \pi^2 \, \text{Wb} \]
Calculating this gives approximately:
\[ \Phi \approx 2.01 \times 10^{-6} \, \text{Wb} \]

### Step 7: Consider the total flux for 3 layers
Since there are 3 layers, the total magnetic flux is:
\[ \Phi_{\text{total}} = 3 \cdot \Phi = 3 \cdot 2.01 \times 10^{-6} \, \text{Wb} \]
\[ \Phi_{\text{total}} \approx 6.03 \times 10^{-6} \, \text{Wb} \]

### Final Answer
The magnetic flux for a cross-section of the solenoid is approximately:
\[ \Phi \approx 6.03 \times 10^{-6} \, \text{Wb} \]