If a long straight wire carries a current of 40 A, then the magnitude of the field B at a point 15 cm away from the wire is

Options

A. `5.34xx10^(-5)T`
B. `8.34xx10^(-5)T`
C. `9.6xx10^(-5)T`
D. `10.2xx10^(-5)T`

Explanation

To find the magnitude of the magnetic field \( B \) at a point 15 cm away from a long straight wire carrying a current of 40 A, we can use the formula for the magnetic field around a long straight conductor:

\[ B = \frac{\mu_0 I}{2 \pi r} \]

Where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( I \) is the current in amperes,
- \( r \) is the distance from the wire in meters.

### Step-by-Step Solution

1. **Identify the given values**:
- Current \( I = 40 \, \text{A} \)
- Distance \( r = 15 \, \text{cm} = 0.15 \, \text{m} \)

2. **Convert the distance from centimeters to meters**:
- Since \( 1 \, \text{cm} = 0.01 \, \text{m} \), we convert \( 15 \, \text{cm} \) to meters:
\[ r = 15 \, \text{cm} = 15 \times 0.01 = 0.15 \, \text{m} \]

3. **Substitute the values into the magnetic field formula**:
- Using the formula \( B = \frac{\mu_0 I}{2 \pi r} \):
\[ B = \frac{(4\pi \times 10^{-7} \, \text{T m/A}) \times 40 \, \text{A}}{2 \pi \times 0.15 \, \text{m}} \]

4. **Simplify the equation**:
- The \( \pi \) in the numerator and denominator cancels out:
\[ B = \frac{4 \times 10^{-7} \times 40}{2 \times 0.15} \]

5. **Calculate the values**:
- First, calculate the numerator:
\[ 4 \times 10^{-7} \times 40 = 1.6 \times 10^{-5} \]
- Then calculate the denominator:
\[ 2 \times 0.15 = 0.30 \]
- Now, divide the numerator by the denominator:
\[ B = \frac{1.6 \times 10^{-5}}{0.30} \approx 5.33 \times 10^{-5} \, \text{T} \]

6. **Final result**:
- The magnitude of the magnetic field \( B \) at a point 15 cm away from the wire is approximately:
\[ B \approx 5.33 \times 10^{-5} \, \text{T} \]