Question

Two parallel long (infinite for our purposes) wires are suspended by a system of thin strings of length L=3.2 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=1.75 G. As appropriate for the wires, their "weight" is quantified by the linear mass density, it is equal to 11 g/m.

The wires are found in mechanical equilibrium when each of the strings is at angle θ=7.45° 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=I₁ enabling the equilibrium when it is positive (I₁>0):
I₁= ______A.

Find current I=I₂ enabling the equilibrium when it is negative (I₂<0):
I₂= _____ A.

Two parallel long (infinite for our purposes) wires are suspended by a system of thin strings of length L=3.2 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=1.75 G. As appropriate for the wires, their "weight" is quantified by the linear mass density, it is equal to 11 g/m. B a The wires are found in mechanical equilibrium when each of the strings is at angle 0=7.45° 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=Iz enabling the equilibrium when it is negative (12<0): 12= A.

Answer 1

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11x103/4.8) 2074-45- HINo x2 (115x101) 47(0.032)Sin-t45° 440x1053-145x1011-0.0/५०१ -0 1:28.080) -20.81 10 4 28.089 I2 12081 14