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 :

Question

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 : ​​​​​​​, 1 . 2 × 10 - 5 N directed towards wire X ., 1 . 2 × 10 - 4 N directed away from wire X ., 1 . 2 × 10 - 4 N directed towards wire X ., 2 . 4 × 10 - 5 N directed towards wire X .

MARKS App · Accepted Answer

An attractive force will act normal to X towards Y because the direction of the currents in the wires is the same. Force exerted on length l due to the magnetic field is given as F = F X Y = F Y X = I 1 l 1 B 12 ⇒ F = μ 0 I 1 I 2 2 π r l 1 Given here, I 1 = 2 A , I 2 = 3 A , l 1 = 5 m , r = 5 cm . Putting the values, we have F = 4 π × 10 - 7 × 2 × 3 2 π × 5 × 10 - 2 × 50 × 10 - 2 = 1 . 2 × 10 - 5 N directed towards wire X .

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