The overhead horizontal power transmission line carries a current of 75 A from east to west direction. The magnetic field above it at a vertical distance of 1.2 m will be

$1.25 × 10^{-5} T$, towards north

The correct answer is Option (4) → $1.25 × 10^{-5} T$, towards north

Magnetic field due to a long straight current-carrying wire: $B = \frac{\mu_0 I}{2 \pi r}$

Given: $I = 75\ \text{A}$, $r = 1.2\ \text{m}$, $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$

$B = \frac{4\pi \times 10^{-7} \cdot 75}{2 \pi \cdot 1.2} = \frac{3 \times 10^{-5}}{2.4} \approx 1.25 \times 10^{-5}\ \text{T}$

Using the right-hand rule: current from east to west, magnetic field above the wire is directed towards north.

Magnetic field = 1.25 × 10⁻⁵ T, direction = north