Question

Two insulated current-carrying wires (wire 1 and wire 2) are bound together with wire ties to form a two-wire unit. The wires are 2.95 m long and are stretched out horizontally parallel to each other. Wire 1 carries a current of I, = 8.00 A and the other wire carries a current I, in the opposite direction. The two-wire unit is placed in a uniform magnetic field of magnitude 0.400 T such that the angle between the direction of I, and the magnetic field is 66.0°. While we don't know the current in wire 2, we do know that it is smaller than the current in wire 1. If the magnitude of the net force experienced by the two-wire unit is 3.50 N, determine the current in wire 2.

Answer 1

Magnitude of force on Wine & F = Force F = F B l gro - 8x0.4x 2.95* Sm66 - 80624 acting on wire thao @ Iz B e sono = Iz XO"U x 2.95 xSin 114 wir Wire (2) = 1;x1.08 ry' Brie B t₂ Across 'f & f acts on two wines on opposite direction, f, on wire 0 Vest Cal In Ward, while hon wire 2 Vest Call outward to paper- Corplane) So, Angle b/w them 1800 fnet = 1 F2-Fil 3.5= 112 x 1.08 -862) 3.5 =1081₂-8.62 3.5 +8.62 = 1:0812 0 12.12 = 100812 I, - 12:12 1.08 1102 A Ang