Two long and parallel straight conducting wires carrying currents of 5 A and 10 A in same direction are separated by a distance of 2cm. The force per unit length on one wire due to other is $(μ_0: 4π×10^{-7} NA^{-2})$

$5 × 10^{-4} N/m$; attractive force

The correct answer is Option (1) - $5 × 10^{-4} N/m$; attractive force

The force per unit length F/L between two long, parallel current carrying wires can be calculated using -

$\frac{F}{L}=\frac{μ_0I_1I_2}{2\pi d}$ [Ampere's law]

where,

F → force

L → length of wires

$I_1$ and $I_2$, currents in two wires = 5 A and 10 A

d, distance between wires = 2 cm = 0.02 m

$∴\frac{F}{L}=\frac{4π×10^{-7}×50}{2×0.02}$

$=5 × 10^{-4} N/m$

and attractive because the current is flowing in the same direction.