Two long parallel wires A and B carry currents 8.0 A and 5.0 A in the same direction, separated by 4.0 cm. The force on a 10 cm section of wire A is approximately:

1. Force per unit length: $f = \dfrac{\mu_0 I_A I_B}{2\pi d}$, with $\dfrac{\mu_0}{2\pi}=2\times10^{-7}$. 2. Substitute $I_A = 8$, $I_B = 5$, $d = 0.04$ m: $f = \dfrac{2\times10^{-7}\times8\times5}{0.04}$. 3. Numerator $= 8\times10^{-6}$; so $f = 2\times10^{-4}$ N m⁻¹. 4. For $L = 0.10$ m: $F = fL = 2\times10^{-4}\times0.10 = 2\times10^{-5}$ N. 5. Same-direction currents attract, so the force is attractive. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.158_