The magnitude of magnetic induction at a distance 4 cm due to straight conductor carrying a current of 10 A is

Options

A. `5 xx 10^(-6) Wb//m^(2)`
B. `5 xx 10^(-5)` N/Am
C. `5 xx 10^(-5)` gauss
D. `5 xx 10^(-6)` tesla

Answer

To find the magnitude of magnetic induction (magnetic field strength) at a distance of 4 cm from a straight conductor carrying a current of 10 A, we can use the formula for the magnetic field around a long straight conductor:

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

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Current, \( I = 10 \, \text{A} \)
- Distance, \( r = 4 \, \text{cm} = 0.04 \, \text{m} \) (convert cm to m)

2. **Substitute the values into the formula:**
\[ B = \frac{4\pi \times 10^{-7} \, \text{T m/A} \times 10 \, \text{A}}{2 \pi \times 0.04 \, \text{m}} \]

3. **Simplify the equation:**
- The \( \pi \) cancels out:
\[ B = \frac{4 \times 10^{-7} \times 10}{2 \times 0.04} \]
- Calculate the denominator:
\[ 2 \times 0.04 = 0.08 \]
- Now substitute back:
\[ B = \frac{4 \times 10^{-6}}{0.08} \]

4. **Calculate \( B \):**
\[ B = 5 \times 10^{-5} \, \text{T} \]

5. **Final answer:**
The magnitude of magnetic induction at a distance of 4 cm due to a straight conductor carrying a current of 10 A is:
\[ B = 5 \times 10^{-5} \, \text{T} \]