Two long parallel straight wires carry steady currents. They exert a force \(F\) per unit length on each other. The current in each wire is now doubled and the distance between the wires is halved. The force per unit length becomes
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(8F\)
Two long parallel wires carry currents \( I_a \) and \( I_b \) at a distance \( d \) apart. The force per unit length between them is \( F=\mu_0I_aI_b/2\pi d \). So \( F\propto I_aI_b/d \).
\( \displaystyle F=\frac{\mu_0I_aI_b}{2\pi d} \)
\( \begin{aligned} F^{\prime} &= \frac{\mu_0(2I_a)(2I_b)}{2\pi(d/2)} \\ &= 8\times\frac{\mu_0I_aI_b}{2\pi d} \\ &= 8F \end{aligned} \)
NCERT (2026-27):
Force Between Two Parallel Currents, the Ampere – “… We label this force as Fba, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by”
\( \displaystyle F_{ba}=I_bLB_a=\frac{\mu_0I_aI_b}{2\pi d}L \)\( \qquad (4.17) \)