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)1.2×10−5N directed towards wire X.✓ Correct
(B)1.2×10−4N directed away from wire X.
(C)1.2×10−4N directed towards wire X.
(D)2.4×10−5N directed towards wire X.
Answer: (A)
Solution
The force on wire Y due to the magnetic field created by wire X is given by:
F=2πrμ0I1I2ℓ1
where I1=2 A (current in X), I2=3 A (current in Y), ℓ1=0.5 m (length of X, which determines the region of interaction), and r=0.05 m.
F=2π×0.054π×10−7×2×3×0.5
F=2×0.054×10−7×6×0.5=0.112×10−7=1.2×10−5 N
Since currents flow in the same direction, the force is attractive (directed towards wire X).
JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option
