JEE Main 2023 · 24 January · morning · Single correct · +4 / −1
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 F₁. If distance between wires is halved and currents on them are doubled, force F₂ on 10 cm length of wire P will be:
- (a) F₁/8
- (b) 10 F₁
- (c) F₁/10
- (d) 8 F₁
Answer and solution
Answer (d) 8 F₁
Solution
Physics sir · 4 steps
Force per length goes as I₁ I₂ / d: doubling both currents gives 4, halving the gap gives 2 more.
- Force formula. For two long parallel wires, the force on a length l of one wire depends on both currents and the gap d between them.
F = (μ₀ I₁ I₂ l)/(2π d)
- Apply the changes. Both currents double, so I₁ I₂ becomes 2 × 2 = 4 times larger. The distance halves, which doubles 1/d. The length stays 10 cm.
- Combine. F₂/F₁ = (2I)(2I)/(I × I) × d/(d/2) = 4 × 2 = 8.
- So. F₂ = 8F₁, which is option (d).
The idea to keep. Force between parallel wires goes as (I₁ I₂)/d. Scale each factor separately, and remember that both currents count, so doubling them gives a factor of 4.
Answer from the NTA final answer key. Solution by PrepWiser.