Question #172441

The drawing shows a wire composed of three segments, AB, BC, and CD. There is a current of I = 2.1 A in the wire. There is also a magnetic field B (magnitude = 0.26 T) that is the same everywhere and points in the direction of +z axis. The lengths of the wire segments are LAB = 1.0 m, LBC = 0.5 m, and LCD = 0.5 m. Find the magnitude of the force that acts on each segment.



1
Expert's answer
2021-03-23T11:09:06-0400
FAB=ILBsinθ=(2.1)(1.0)(0.26)sin90=0.546 NFBC=ILBsinθ=(2.1)(0.5)(0.26)sin90=0.273 NFCD=ILBsinθ=(2.1)(0.5)(0.26)sin0=0 NF_{AB} = ILB\sin θ\\= (2.1)(1.0)(0.26)\sin 90=0.546\ N\\ F_{BC} = ILB\sin θ\\= (2.1)(0.5)(0.26)\sin 90=0.273\ N\\ F_{CD} = ILB\sin θ\\= (2.1)(0.5)(0.26)\sin 0=0\ N\\


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