Question

Please solve and explain how its done.

Two parallel long (infinite for our purposes) wires are suspended by a system of thin strings of length L=4.6 cm each as illustrated in the figure below showing the plane perpendicular to the wires. These wires are part of the electric circuit and run the same (but unknown) current I in the opposite directions. We treat the current I algebraically assigning the positive values to the current running out of the screen towards us (hence, currents I and -1 next to the wires' position in the figure). The wires are in the region of the uniform magnetic field B as shown with the magnitude B=2.85 G. As appropriate for the wires, their "weight" is quantified by the linear mass density, it is equal to 13.75 g/m. B 010 L L -1 The wires are found in mechanical equilibrium when each of the strings is at angle 9=5.650 from the vertical as shown in the figure. This equilibrium can be realized, however, for both directions of the current in the wires. Find current I=11 enabling the equilibrium when it is positive (11>0): 11= A. Find current I=12 enabling the equilibrium when it is negative (12<0): 12= A.

Answer 1

Apply Newton's and law, Those = ing - Fo+ Tsino -Fi Tsino = finfo @ mgtano - F- FB fi= M0 I²4 - MoI²4 21 (alsine) Yi Lsine D) głane = Hom - IB Yll LSino

XI2 I (2 85x104) 13 75 x103 (9.8) tans 65 - 40x107 4TXO 046xsins 65 2.2x105 I ² -2.85x10 45 -0.0133 20. I = 319A, -18.99 I = 31.9A I 2 = 18.94 A