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Question

# Three long, straight and parallel wires carrying currents are arranged as shown in figure. The force experienced by 10 cm length of wire Q is

A
1.4×104N towards the right
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B
1.4×104N towards the left
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C
2.6×104N to the right
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D
2.6×104N to the left
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Solution

## The correct option is A 1.4×10−4N towards the right Force on wire Q due to R ; FR=10−7×2×20×10(2×10−2)×(10×10−2)=2×10−4m(Repulsive) Force on wire Q due to P ; FP=10−7×2×10×30(10×10−2)×(10×10−2)=0.6×10−4N (Repulsive) Hence net force Fnet=FR−FP=2×10−4−0.6×10−4=1.4×10−4N (towards right i.e. in the direction of →FR).

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