Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire Q is ____ .
(μₒ=4π× 10⁻⁷ T.m/A)

Solution

Physics sir · 3 steps

Both neighbours push Q towards R: P repels it and R pulls it, so the two forces add up.

1. Force between wires. Two parallel wires a distance d apart exert a force on each other. On a length L of either wire it is given below, with μ₀/2π = 2 × 10⁻⁷ T m A⁻¹. Like currents attract and opposite currents repel.

   F = (μ₀ I₁ I₂ L)/(2π d)

   Pair	Currents	Force on 15 cm of Q	Direction on Q
   P on Q, d = 0.03 m	3 A up, 1 A down: opposite, repel	(2 × 10⁻⁷ × 3 × 1 × 0.15)/0.03 = 3 × 10⁻⁶ N	away from P, so towards R
   R on Q, d = 0.02 m	2 A down, 1 A down: same, attract	(2 × 10⁻⁷ × 2 × 1 × 0.15)/0.02 = 3 × 10⁻⁶ N	towards R

2. Add the forces. Both forces point the same way, so they add: 3 × 10⁻⁶ + 3 × 10⁻⁶ = 6 × 10⁻⁶ N.
3. So. The force on 15 cm of Q is 6 × 10⁻⁶ N towards R, which is option (b).

The idea to keep. For a wire with two neighbours, find each force separately with (μ₀ I₁ I₂ L)/(2π d). Fix each direction from the rule that like currents attract and opposite currents repel, then add them as vectors.

Answer from the NTA final answer key. Solution by PrepWiser.