A long solenoid is formed by winding 40` turns//cm `The current necessary for 40 mT inside the solenoid will be approximately equal to

Options

A. 8A
B. 4A
C. 2A
D. 1A

Answer

To solve the problem of finding the current necessary to induce a magnetic field of 40 mT inside a solenoid with 40 turns per cm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of turns per unit length (n) = 40 turns/cm
- Magnetic field (B) = 40 mT = 40 × 10^(-3) T
- Permeability of free space (μ₀) = 4π × 10^(-7) T·m/A

2. **Convert Turns per Centimeter to Turns per Meter:**
- Since 1 cm = 0.01 m, we convert 40 turns/cm to turns/m:
\[ n = 40 \, \text{turns/cm} \times 100 \, \text{cm/m} = 4000 \, \text{turns/m} \]

3. **Use the Formula for the Magnetic Field Inside a Solenoid:**
- The formula for the magnetic field inside a long solenoid is given by:
\[ B = \mu_0 n I \]
- Rearranging this formula to solve for current (I):
\[ I = \frac{B}{\mu_0 n} \]

4. **Substitute the Known Values into the Formula:**
- Plug in the values for B, μ₀, and n:
\[ I = \frac{40 \times 10^{-3}}{(4\pi \times 10^{-7}) \times 4000} \]

5. **Calculate the Denominator:**
- Calculate \(4\pi \times 10^{-7} \times 4000\):
\[ 4\pi \approx 12.56 \quad \text{(using } \pi \approx 3.14\text{)} \]
\[ 4\pi \times 4000 \approx 12.56 \times 4000 = 50240 \]
\[ 4\pi \times 10^{-7} \times 4000 \approx 50240 \times 10^{-7} = 5.024 \times 10^{-3} \]

6. **Calculate the Current (I):**
- Now substitute back into the equation for I:
\[ I = \frac{40 \times 10^{-3}}{5.024 \times 10^{-3}} \approx 7.95 \, \text{A} \]
- Rounding this gives approximately:
\[ I \approx 8 \, \text{A} \]

### Final Answer:
The current necessary to induce a magnetic field of 40 mT inside the solenoid is approximately **8 Amperes**.