Question

Two parallel long (infinite for our purposes) wires are suspended by a system of thin strings of length L=1.2 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 – I 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=2.35 G. As appropriate for the wires, their "weight" is quantified by the linear mass density, it is equal to 12.1 g/m. B L L -I The wires are found in mechanical equilibrium when each of the strings is at angle 0=6.85° 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): A. 12=

Answer 1

u = 12.19m, B = 2.356 = 2.35X104T L=12cm.,o=6.85° 7 I Sul B FB B FB [E = ug force due to magnetic field Fo= IB e IB = IX2.35x 104 e force direction on each wire is toward vertical axis. force due to the currents in each were F = Mo II 2X10 I2 2 T r -7 be r sino = r 2 - M=2L sino 2 r r = 2x1.2 sin (6.85) M= 2.4 x(0.11927) r = 0.286cm

Fa — 2x10 + ² 0,286 XIo 2 6.99x105 ?субS I2 Fc Fc Fo l 14, -М л=ға e L I coso l Trino e I caso = 9 Faces &og=Mg та — м2 ма Т-д У As they are in equilibrium, net force perunit length =0 FB ТЯ jи 0 = 0 9 l l Fc