The magnitude of the current flowing in the wire $A({I_A})$ is 8 A

The magnitude of the current flowing in wire $B({I_B})$ is 5 A

The distance between the two wires (r) is 0.04 m

The length of the section of wire A (L) is 0.1 m

The force exerted by the magnetic field on the length L is computed as follows:
$F = \frac{{{\mu _o}{I_A}{I_B}L}}{{2\pi r}}$

where,

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

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

Substituting the values, we get

$F = \frac{{4\pi  \times {{10}^{ – 7}} \times 8 \times 5 \times 0.1}}{{2\pi  \times 0.04}}$

$F = 2 \times {10^{ – 5}}N$

The magnitude of the force is $2 \times {10^{ – 5}}N$. Because the currents in the wires are in the same direction, this creates an attracting force normal to A towards B.