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
Answer & explanation
Correct answer: option 3
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
$\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$
$B = \frac{4\pi \times 10^{-7} \times 304}{2\pi \times 0.06}$
$B = \frac{2 \times 10^{-7} \times 304}{0.06}$
$B = 1.013 \times 10^{-3}\ \text{T}$