The current flowing across the wire has a magnitude of (I) = 50 A.
The point B is 2.5 metres east of the wire.
As a result, the magnitude of the point’s distance from the wire (r) is 2.5 m.
The magnitude of the magnetic field at that place may be calculated using the following equation:
$|\overline B | = \frac{{{\mu _0}2I}}{{4\pi r}}$
where,
${\mu _0} = $Permeability of free space
$ = 4\pi \times {10^{ – 7}}Tm{A^{ – 1}}$
$|\overline B | = \frac{{4\pi \times {{10}^{ – 7}}}}{{4\pi }} \times \frac{{2 \times 50}}{{2.5}}$
$ = 4 \times {10^6}T$
At a distance of 2.5 m, the point is situated normal to the wire length. The current flow in the wire is vertically downward. As a result, the magnetic field at the given position is vertically upward, according to Maxwell’s right-hand thumb rule.