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
Answer & explanation
Correct answer: option 1
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: $B = 4 \times 10^{-8}\,T$