A solenoid is 1.5 m long and its inner diameter is 4.0 cm . It has three layers of windings of 1000 turns each and carries a current of 2.0 amperes. The magnetic flux for a cross-section of the solenoid is nearly

Options

A. `2.5 xx 10^(-7) "weber"`
B. `6.31 xx 10^(-6) "weber"`
C. `5.2 xx 10^(-5)"weber"`
D. `4.1 xx 10^(-5) "weber"`

Answer

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

### Step 1: Calculate the radius of the solenoid
The inner diameter of the solenoid is given as 4.0 cm. To find the radius, we divide the diameter by 2.

\[ \text{Radius} (r) = \frac{\text{Diameter}}{2} = \frac{4.0 \, \text{cm}}{2} = 2.0 \, \text{cm} = 0.02 \, \text{m} \]

### Step 2: Calculate the total number of turns (N)
The solenoid has three layers of windings, with each layer containing 1000 turns. Therefore, the total number of turns is:

\[ N = 3 \times 1000 = 3000 \, \text{turns} \]

### Step 3: Calculate the magnetic field (B) inside the solenoid
The magnetic field inside a solenoid can be calculated using the formula:

\[ B = \mu_0 \cdot \frac{N}{L} \cdot I \]

Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
- \( N = 3000 \, \text{turns} \)
- \( L = 1.5 \, \text{m} \)
- \( I = 2.0 \, \text{A} \)

Substituting the values:

\[ B = (4\pi \times 10^{-7}) \cdot \frac{3000}{1.5} \cdot 2.0 \]

Calculating \( \frac{3000}{1.5} \):

\[ \frac{3000}{1.5} = 2000 \]

Now substituting back into the equation for B:

\[ B = (4\pi \times 10^{-7}) \cdot 2000 \cdot 2.0 \]

Calculating \( 2000 \cdot 2.0 = 4000 \):

\[ B = (4\pi \times 10^{-7}) \cdot 4000 \]

Calculating \( 4 \cdot 4000 = 16000 \):

\[ B = 16000\pi \times 10^{-7} \, \text{T} \approx 5.0265 \times 10^{-3} \, \text{T} \]

### Step 4: Calculate the area (A) of the cross-section of the solenoid
The area of the cross-section of the solenoid can be calculated using the formula for the area of a circle:

\[ A = \pi r^2 \]

Substituting the radius:

\[ A = \pi (0.02)^2 = \pi (0.0004) \approx 1.25664 \times 10^{-3} \, \text{m}^2 \]

### Step 5: Calculate the magnetic flux (Φ)
The magnetic flux through a cross-section of the solenoid is given by the formula:

\[ \Phi = B \cdot A \]

Substituting the values of B and A:

\[ \Phi = (5.0265 \times 10^{-3}) \cdot (1.25664 \times 10^{-3}) \approx 6.316 \times 10^{-6} \, \text{Wb} \]

### Final Answer
The magnetic flux for a cross-section of the solenoid is approximately:

\[ \Phi \approx 6.32 \times 10^{-6} \, \text{Wb} \]

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