A horizontal overheadpowerline is at height of `4 m` from the ground and carries a current of `100 A` from east to west. The magnetic field directly below it on the ground is
`( nu_(0) = 4 pi xx 10^(-7) Tm A^(-1)`

Options

A. ` 2.5 xx 10^(-7) T` southward
B. ` 5 xx 10^(-6) T` northward
C. ` 5 xx 10^(-6) T` southward
D. ` 2.5 xx 10^(-7) T` northward

Explanation

To find the magnetic field directly below a horizontal overhead power line, we can use the formula for the magnetic field due to a long straight conductor. The formula 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 (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{Tm/A} \)),
- \( I \) is the current (in Amperes),
- \( r \) is the distance from the wire to the point where the magnetic field is being calculated (in meters).

### Step-by-Step Solution:

1. **Identify the given values:**
- Height of the power line from the ground, \( h = 4 \, \text{m} \)
- Current flowing through the wire, \( I = 100 \, \text{A} \)
- Distance from the wire to the point directly below it on the ground, \( r = 4 \, \text{m} \) (since the wire is directly overhead).

2. **Substitute the values into the formula:**
\[ B = \frac{4 \pi \times 10^{-7} \, \text{Tm/A} \times 100 \, \text{A}}{2 \pi \times 4 \, \text{m}} \]

3. **Simplify the expression:**
- The \( \pi \) terms cancel out:
\[ B = \frac{4 \times 10^{-7} \times 100}{2 \times 4} \]
- Calculate the numerator:
\[ 4 \times 10^{-7} \times 100 = 4 \times 10^{-5} \]
- Calculate the denominator:
\[ 2 \times 4 = 8 \]
- Now substitute back:
\[ B = \frac{4 \times 10^{-5}}{8} \]

4. **Final calculation:**
\[ B = 0.5 \times 10^{-5} = 5 \times 10^{-6} \, \text{T} \]

Thus, the magnetic field directly below the power line on the ground is:

\[ B = 5 \times 10^{-6} \, \text{T} \]