A long wire with a small current element of length 1 cm is placed at the origin and carries a current of 10 A along the x-axis. The magnitude of the magnetic field, due to the element, on the y-axis at a distance 0.5 m from it, would be

$4 × 10^{-8} T$

The correct answer is Option (1) → $4 × 10^{-8} T$

Given:

Current, $I = 10\,A$

Length of element, $dl = 1\,cm = 0.01\,m$

Distance from element, $r = 0.5\,m$

Magnetic field due to a current element:

$dB = \frac{\mu_0}{4\pi} \frac{I\,dl\,\sin\theta}{r^2}$

Since point lies on y-axis and current is along x-axis, $\theta = 90^\circ$, $\sin\theta = 1$.

$dB = \frac{(4\pi \times 10^{-7})}{4\pi} \times \frac{10 \times 0.01}{(0.5)^2}$

$dB = 10^{-7} \times \frac{0.1}{0.25}$

$dB = 10^{-7} \times 0.4 = 4 \times 10^{-8}\,T$

Magnetic field magnitude: $B = 4 \times 10^{-8}\,T$