JEE Main202229 Jul 2022Evening ShiftPhysicsMagnetic Effects of CurrentActual
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 :
Options
- A1 . 2 × 10 - 5   N directed towards wire X .
- B1 . 2 × 10 - 4   N directed away from wire X .
- C1 . 2 × 10 - 4   N directed towards wire X .
- D2 . 4 × 10 - 5   N directed towards wire X .
Correct answer
A. 1 . 2 × 10 - 5   N directed towards wire X .
Step-by-step solution
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 ×