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A horizontal overhead power line carries a current of 90 A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

(I) = 90 A is the magnitude of the current in the power line.

The point is placed at (r) = 1.5 m below the electrical cable.

As a result, the magnetic field at that location may be computed as follows:

$|\overline B | = \frac{{{\mu _0}2I}}{{4\pi r}}$

where,

${\mu _0} = $Permeability of free space 

$ = 4\pi  \times {10^{ – 7}}Tm{A^{ – 1}}$

Substituting values in the above equation, we get

$|\overline B | = \frac{{4\pi  \times {{10}^{ – 7}}}}{{4\pi }} \times \frac{{2 \times 90}}{{1.5}}$

$ = 1.2 \times {10^{ – 5}}T$

The movement of the current is from east to west. The point is located beneath the electrical cord.

As a result, the magnetic field is oriented southward according to Maxwell’s right-hand thumb rule.