The magnituge of 'B' from a conductor carrying 35A at a perpendicular distance of 20cm from it, in tesla is

Options

A. `3.5xx10^(-5)`
B. `3.5xx10^(-3)`
C. `0.315xx10^(-5)`
D. `0.035xx10^(-5)`

Answer

To find the magnitude of the magnetic field \( B \) at a perpendicular distance from a straight conductor carrying current, we can use the formula:

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

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Current \( I = 35 \, \text{A} \)
- Distance \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)

2. **Substitute the values into the formula:**
\[ B = \frac{{\mu_0 I}}{{2 \pi r}} = \frac{{4\pi \times 10^{-7} \, \text{T m/A} \times 35 \, \text{A}}}{{2 \pi \times 0.2 \, \text{m}}} \]

3. **Simplify the expression:**
- The \( \pi \) in the numerator and denominator cancels out:
\[ B = \frac{{4 \times 10^{-7} \times 35}}{{2 \times 0.2}} \]

4. **Calculate the denominator:**
\[ 2 \times 0.2 = 0.4 \]

5. **Now substitute back into the equation:**
\[ B = \frac{{4 \times 35 \times 10^{-7}}}{{0.4}} \]

6. **Calculate \( 4 \times 35 \):**
\[ 4 \times 35 = 140 \]

7. **Now substitute this value back:**
\[ B = \frac{{140 \times 10^{-7}}}{{0.4}} \]

8. **Divide \( 140 \) by \( 0.4 \):**
\[ 140 \div 0.4 = 350 \]

9. **Now we have:**
\[ B = 350 \times 10^{-7} \, \text{T} \]

10. **Convert \( 350 \times 10^{-7} \) to scientific notation:**
\[ B = 3.5 \times 10^{-5} \, \text{T} \]

### Final Answer:
The magnitude of the magnetic field \( B \) at a perpendicular distance of 20 cm from the conductor carrying 35 A is:
\[ B = 3.5 \times 10^{-5} \, \text{T} \]