Question

Figure shows two long wires A and B, each carrying a current I, separated by a distance $l$. Find the magnetic induction at a point P located at a distance $l$ from both wires as shown.

Answer 1

$\frac{\mu_0 I}{2\pi l}$

Answer 2

The magnetic field at point P due to each wire (A and B) has the same magnitude, $B_0 = \frac{\mu_0 I}{2\pi l}$, because P is equidistant ($l$) from both wires.

By applying the right-hand thumb rule:

1. For wire A (current out of page), the magnetic field $B_A$ at P is perpendicular to AP and points upwards and to the left (at $150^\circ$ from horizontal).
2. For wire B (current into page), the magnetic field $B_B$ at P is perpendicular to BP and points upwards and to the right (at $30^\circ$ from horizontal).

When these two vectors are added, their horizontal components cancel out (due to symmetry: $B_0 \cos(150^\circ) = -B_0 \cos(30^\circ)$), and their vertical components add up.

The vertical component of $B_A$ is $B_0 \sin(150^\circ) = B_0 (1/2)$.

The vertical component of $B_B$ is $B_0 \sin(30^\circ) = B_0 (1/2)$.

The total magnetic field is $B_{net} = B_0/2 + B_0/2 = B_0 = \frac{\mu_0 I}{2\pi l}$, directed vertically upwards.