Three long, straight wires lie in the horizontal plane. The wires on the outside

Question

Three long, straight wires lie in the horizontal plane. The wires on the outside of the configuration are 10 cm apart. The wire on the inside is directly in the middle of the outer wires. The two outside wires carry a current of 10 A in the positive x-direction. The inside wire carries a current of 15 A in the negative x-direction. Find the force per unit length on the middle wire caused by the two outer wires. Find the force per unit length on one outer wire caused by the other outer wire and the inside wire. What is the magnetic field (magnitude and direction) at a point that is 15 cm from one of the outer wires?

Explanation / Answer

a) the force is given by F = 0 N. ( the force due to the two wires gets cancelled ) b) the force per unit length is F = 2*10^-7*(3- 1)*10*100 = 4*10^-4 N along the -ve y-axis. c) there will be two points which are 15 cm from the outer wires the magnetic field at the point nearer to the other outer wire is Bn = 2*10^-7* [ 2 + 2/3 -3/2 ] *100 = 2.33*10^-5 T the field at the farther point is Bf = 2*10^-7*100*[ 2/3 -1 +2/5 ] = 1.33*10^-6 T

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