14. Three parallel wires carry currents, with magnitudes and in the directions shown in the figure below. The wires all lie in a plane, and are all 2 m long, with separations between them as shown in the figure. What is the total force on wire #3 due to wire #1 and wire #2? Wire 1 : 6A, 2.5 m from Wire 3 ---------------------gt;Wire 2: 3A, 1.0 m from Wire 3 lt;--------------------Wire 3: 3A ---------------------gt;Can't figure it out, any hints/suggestions?
The formula for force between wires is: (?/2π)(current1)(current2)(length/r) ? = 4πE-7 Tm/A r=distance between wires For this problem, you would just find the force on wire 3 of each one separately. Since wire 2 is going opposite direction as wire 3, they repel each other, while wire 1 and wire 3 attract each other since they are going same direction. I get Fwire1 to be 2.88E-6 N, and Fwire2 to be 3.6E-6N. Then subtract them since opposite forces, and since Fwire2 is stronger, the force would be 7.2E-7 N away/repelling.
Once you find the direction you then need to find the magnitude of each B-field. You can do this using the equation: B=( μ x I)/(2 x π x r) where μ= 4π x 10^-7.With this you will be able to find each B-field. Once you have this information, you can then find the force exerted from each wire with the equation F=I x L(Length) x B. I'm sure there are other methods of solving, but this is the way I would go. Hope this helps.
force between two current carrying wires per unit length = 10^-7*2i_1*i_2/a if current is in same direction force is attractive else repulsive.. now you use this and solve yourself.. as it is not clear from the data you provided.. whether the wire 3 is in middle or left end or right end.. but i have told you the concept.. i hope this helps.. if u need more help provide me with the image
