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 Ampere's circuital law. The formula for the magnetic field around a long straight conductor is given by:

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

where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (\( 4 \pi \times 10^{-7} \, \text{T m/A} \)),
- \( I \) 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 Distance to Meters:**
- Since \( R \) is given in centimeters, convert it to meters:
\[ R = 15 \, \text{cm} = 15 \times 10^{-2} \, \text{m} = 0.15 \, \text{m} \]

3. **Substitute the Values into the Formula:**
- Now substitute \( \mu_0 \), \( I \), and \( R \) into the formula:
\[ B = \frac{4 \pi \times 10^{-7} \times 40}{2 \pi \times 0.15} \]

4. **Simplify the Expression:**
- The \( \pi \) terms cancel out:
\[ B = \frac{4 \times 10^{-7} \times 40}{2 \times 0.15} \]
- Calculate the numerator:
\[ 4 \times 40 = 160 \]
- Calculate the denominator:
\[ 2 \times 0.15 = 0.30 \]
- Now substitute these values back into the equation:
\[ B = \frac{160 \times 10^{-7}}{0.30} \]

5. **Calculate the Final Value of B:**
- Performing the division:
\[ B = \frac{160}{0.30} \times 10^{-7} = 533.33 \times 10^{-7} = 5.33 \times 10^{-5} \, \text{T} \]

6. **Final Answer:**
- The magnitude of the magnetic field \( B \) at a point 15 cm away from the wire is:
\[ B \approx 5.34 \times 10^{-5} \, \text{T} \]