Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is . If distance between wires is halved and currents on them are doubled, force on 10 cm length of wire P will be:
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
The force per unit length between two parallel wires is F/L = μ₀I₁I₂/(2πd).
F₁ = μ₀(10)(10)/(2π×0.05) × 0.10
For F₂, currents are doubled (20A each) and distance is halved (0.025m):
F₂ = μ₀(20)(20)/(2π×0.025) × 0.10
F₂/F₁ = (20×20/0.025)/(10×10/0.05) = (400/0.025)/(100/0.05) = 16000/2000 = 8
So F₂ = 8F₁. The current product quadruples and the distance halving doubles the force, giving an overall factor of 4×2 = 8.
F₁ = μ₀(10)(10)/(2π×0.05) × 0.10
For F₂, currents are doubled (20A each) and distance is halved (0.025m):
F₂ = μ₀(20)(20)/(2π×0.025) × 0.10
F₂/F₁ = (20×20/0.025)/(10×10/0.05) = (400/0.025)/(100/0.05) = 16000/2000 = 8
So F₂ = 8F₁. The current product quadruples and the distance halving doubles the force, giving an overall factor of 4×2 = 8.
[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]