2/25/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 1/2Master ItIn the figure below, the current in the long, straight wire is and the wire lies in the plane of therectangular loop, which carries a current The dimensions in the figure are and Find the magnitude and direction of the net force exerted on the loop by themagnetic field created by the wire.Part 1 of 4 ConceptualizeThere are forces in opposite directions on the loop as shown in the diagram below. The magnetic field is strongernear the straight wire than it is farther away. By symmetry, the forces exerted on sides 2 and 4 (the horizontalsegments of length a) are equal and opposite, adding to zero. The magnetic field in the plane of the loop isdirected into the page to the right of By the righthand rule, is directed toward the left for side 1of the loop. A smaller force is directed toward the right for side 3. We expect the net force to be to the left withmagnitude in the micronewton range.Part 2 of 4 CategorizeWe will use the equation for the magnetic force between two parallel wires applied to sides 1 and 3 of the loop tofind the net force resulting from these opposing force vectors.Part 3 of 4 AnalyzeFor the total magnetic force for sides 1 and 3 of the loop, we have the following.I1 = 6.10 AI2 = 10.0 A. c = 0.100 m,a = 0.150 m, = 0.685 m.I1. = I × F L B2/25/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 2/2Part 4 of 4 AnalyzeThe expression above evaluates toor a total magnetic force with magnitude 50.1 µN directed toward the -‐‑-‐‑-‐‑Select-‐‑-‐‑-‐‑ left.FinalizeThe net force is to the left and within the micronewton range as expected. The net force on the loop would be zeroif either current disappeared. The net force on the loop approaches zero as either dimension of the rectangleapproaches zero or as the distance of the wire from the loop approaches infinity. As the distance between the wireand the loop approaches infinity, the magnetic field becomes uniform. = 1 + 2 = − = = FF Fμ0I1I22π1c + a1cμ0I1I22π−ac(c + a)4π × 10−7 N/A2 61 A2 0.685 m2π− 0.15 m 0.025 m2 = − 5.01 × 10−5 N
View Full Document
Unlocking...