In the figure below, the current in the long, straight wire is I, = 6.20 A and the wire lies in the plane of the rectangular loop, which carries a current I, = 10.0 A. The dimensions in the figure are c = 0.100 m, a = 0.150 m, and e = 0.410 m. Find the magnitude anddirection of the net force exerted on the loop by the magnetic field created by the wire.magnitudeUNdirection|---Select---

Given: Current I1=6.20A Curren I2=10.0A c=0.100m a=0.150m l=0.410m The forces on the two horizontal parts of the loop in above diagram is zero. F3=F4=0The magnetic force due to current I1 is given by, F1=μ0I1I2(l)2πc Plugging all given values F1=4π*10-7*6.20*10(0.410)2π*0.100 F1=508.4*10-7N Magnetic force due to current I2 is given by, F2=μ0I1I2(l)2π(c+a) Plugging all given values in above equation F2=4π*10-7*6.20*10*0.4102π(0.100+0.150) F2=203.36*10-7N Net force is given by, F=F1-F2 Since F1 and F2 are in opposite direction. F=508.4*10-7-203.36*10-7 F=305.04*10-7N Thus the net magnetic force is 305*10-7 N. Since F1>F2, Direction of net magnetic force is towards the left.

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Advanced Physics

In the figure below, the current in the long, straight wire is I, = 6.20 A and the wire lies in the plane of the rectangular loop, which carries a current I, = 10.0 A direction of the net force exerted on the loop by the magnetic field created by the wire. magnitude UN

In the figure below, the current in the long, straight wire is I, = 6.20 A and the wire lies in the plane of the rectangular loop, which carries a current I, = 10.0 A direction of the net force exerted on the loop by the magnetic field created by the wire. magnitude UN