JEE Main 2022 · 29 July · evening · Single correct · +4 / −1
A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is
- (a) 1.2 × 10⁻⁵ N directed towards wire X.
- (b) 1.2 × 10⁻⁴ N directed away from wire X.
- (c) 1.2 × 10⁻⁴ N directed towards wire X.
- (d) 2.4 × 10⁻⁵ N directed towards wire X.
Answer and solution
Answer (a) 1.2 × 10⁻⁵ N directed towards wire X.
Solution
Physics sir · 4 steps
Find the force on short X in Y's field, then flip it for Y using Newton's third law.
- Which length to use. Y is long, so its field is the same all along X, at B = (μ₀ I_Y)/(2π d). X is only 50 cm long, so its field is not uniform along Y. The easy route is to find the force on X first.
F = (μ₀ I_X I_Y)/(2π d) L_X
- Put in numbers. F = 2 × 10⁻⁷ × (2 × 3)/0.05 × 0.5 = 2 × 10⁻⁷ × 120 × 0.5 = 1.2 × 10⁻⁵ N.
- Third law and direction. Y pushes on X and X pushes back on Y with an equal and opposite force, so Y also feels 1.2 × 10⁻⁵ N. Currents in the same direction attract, so the force on Y points towards X.
- So. The force on Y is 1.2 × 10⁻⁵ N directed towards wire X, which is option (a).
The idea to keep. When the two parallel wires have different lengths, only the overlap matters: use the shorter wire's length. Like currents attract, unlike currents repel.
Answer from the NTA final answer key. Solution by PrepWiser.