A long solenoid has 200 turns per cm and carries a current of 2.5 amps. The magnetic field at its centre is `(mu_(0)=4pixx10^(-7)"weber"//amp-m)`

Options

A. `3.14 xx 10^(-2) Wb//m^(2)`
B. `6.28 xx 10^(-6)Wb//m^(2)`
C. `9.42 xx 10^(-6) Wb//m^(2)`
D. `12.56 xx 10^(2) Wb//m^(2)`

Answer

To solve the problem of finding the magnetic field at the center of a long solenoid, we can follow these steps:

### Step 1: Identify the given values
- Number of turns per centimeter (n) = 200 turns/cm
- Current (I) = 2.5 A
- Permeability of free space (μ₀) = 4π × 10^(-7) T·m/A

### Step 2: Convert turns per centimeter to turns per meter
Since the number of turns is given in centimeters, we need to convert it to meters for consistency in units:
\[ n = 200 \, \text{turns/cm} = 200 \times 100 \, \text{turns/m} = 20000 \, \text{turns/m} \]

### Step 3: Use the formula for the magnetic field inside a solenoid
The formula for the magnetic field (B) inside a long solenoid is given by:
\[ B = \mu_0 n I \]
where:
- \(B\) is the magnetic field,
- \(\mu_0\) is the permeability of free space,
- \(n\) is the number of turns per unit length (in turns/m),
- \(I\) is the current in amperes.

### Step 4: Substitute the values into the formula
Now we can substitute the values we have into the formula:
\[ B = (4\pi \times 10^{-7} \, \text{T·m/A}) \times (20000 \, \text{turns/m}) \times (2.5 \, \text{A}) \]

### Step 5: Calculate the magnetic field
Calculating the expression:
1. First, calculate \(n \times I\):
\[ n \times I = 20000 \times 2.5 = 50000 \, \text{turns·A/m} \]
2. Now multiply by \(\mu_0\):
\[ B = 4\pi \times 10^{-7} \times 50000 \]
\[ B = 4 \times 3.14 \times 10^{-7} \times 50000 \]
\[ B = 6.28 \times 10^{-6} \, \text{T} \quad \text{(or Weber/m²)} \]

### Step 6: Final result
Thus, the magnetic field at the center of the solenoid is:
\[ B \approx 6.28 \times 10^{-6} \, \text{T} \]