A long straight conductor carries a current of 304 A. The magnitude of the magnetic field due to the current at a distance of 6 cm from the conductor is

1.013 mT

The correct answer is Option (3) → 1.013 mT

Given:

$I = 304\ \text{A}$

$r = 6\ \text{cm} = 0.06\ \text{m}$

Magnetic field due to a long straight conductor:

$B = \frac{\mu_0 I}{2\pi r}$

Substitute

$\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$

$B = \frac{4\pi \times 10^{-7} \times 304}{2\pi \times 0.06}$

$B = \frac{2 \times 10^{-7} \times 304}{0.06}$

$B = 1.013 \times 10^{-3}\ \text{T}$