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 produce a magnetic field of 40 mT inside a long solenoid with 40 turns per centimeter, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for the magnetic field inside a solenoid**:
The magnetic field \( B \) inside a long solenoid is given by the formula:
\[ B = \mu_0 n I \]
where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( n \) is the number of turns per unit length (in turns/meter),
- \( I \) is the current in amperes.

2. **Convert the number of turns per centimeter to turns per meter**:
Given that the solenoid has 40 turns per centimeter:
\[ n = 40 \, \text{turns/cm} = 40 \times 100 = 4000 \, \text{turns/m} \]

3. **Convert the magnetic field from milliTesla to Tesla**:
The magnetic field \( B \) is given as 40 mT:
\[ B = 40 \, \text{mT} = 40 \times 10^{-3} \, \text{T} \]

4. **Substitute the known values into the formula**:
Now we can substitute \( B \), \( \mu_0 \), and \( n \) into the formula:
\[ 40 \times 10^{-3} = (4\pi \times 10^{-7}) \times (4000) \times I \]

5. **Rearranging the formula to solve for \( I \)**:
Rearranging gives:
\[ I = \frac{40 \times 10^{-3}}{4\pi \times 10^{-7} \times 4000} \]

6. **Calculate the denominator**:
First, calculate \( 4\pi \):
\[ 4\pi \approx 12.56 \]
Then calculate:
\[ 4\pi \times 10^{-7} \times 4000 = 12.56 \times 10^{-7} \times 4000 = 5.024 \times 10^{-3} \]

7. **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 to the nearest whole number gives:
\[ I \approx 8 \, \text{A} \]

### Final Answer:
The current necessary for 40 mT inside the solenoid will be approximately equal to **8 A**.

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