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We assume that the wires are long compared to distance between them and the 15cm section.

B at the 15 cm section  due to current in A: \frac{Mu\ I}{2 π d} \\ \\
      = \frac{Mu 4 }{2π 0.03 T} \\

Field B is perpendicular to the length of wire B and to plane of A & B.

Force on the wire segment =
         = I L X B = 6A * 0.15 m * 21.22 Mu = 19.1 * Mu Newtons

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