Question

Three very long current carrying wires are placed at the corners of an equilateral triangle as shown below. Wire 1 and wire 2 are fixed at their locations well above the ground using some anchors. Wire 3 with linear mass density 40 g/m is being levitated magnetically above the ground due to the other two wires. (a) Find the direction of current in the wire 3 to achieve this. (b) Find the magnitude of the current in wire 3 (I₃) to make it levitate above the ground if I₁ = I₂ = 75 A.

4.0 cm x Х I 12 13

Answer 1

SOLUTION 4.0cm 0 Iz 5 , Equilateral triangle . 0 = 60° F32 Iz Part (a) For given condition, current in wire 3 must be such that wire 1 and wire 2 attract wire 3. Now, as we know that if currents in wires are direction they attract each other. Hence, Iz has same direction as wire 1 and wire 2 have. in same Part (6) F31 = Force per unit due to wire 1 length on wire 3 Mo I, I3 [4AX 107] [75] [ Iz 21 [ 4x102] [37.5 x105] I3 N/m (a = 4cm - 4x10 m)

4x 10² kg/m F32 = Fonce per unit length on wire 3 due to wire 1. llo I2 I3 zad [4x107] [75] [ Iz 2[ 4x162] =[37.5x105] 53 Nm As, F31 and F32 are equal in magnitude. Let , & F32 = F - . Fnet ✓ Fg? + Fz 2 + 2 F3, F32 Cosbo ✓ F² + f² + 2 f² (4) [0=600 1 [37.5X105] 53 I 3 N/m = [64.952x10 5] Iz N/m W = Weight per un't length of wire 3 ug mass per unit length = (4x102) x g. 8 of we're 3 = 39.2 x 10-2 u = 40 g/m M m

i for given condition: Fnet w (64.952x10-5) 13 = 39-2x16 2 or, I3 = 39.2x 10-2 64. gS2x105 - 603.5226 A