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Questions  

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

a
1.4 ×10−4 N  towards right
b
1.4 ×10−4 N  towards left
c
2.6 ×10−4 N  towards right
d
2.6 ×10−4 N  towards left

detailed solution

Correct option is A

Force on wire Q due to wire P is FP= 10−7 × 2 × 30 × 100.1×0.1=  6 × 10−5N                                                      (Towards Left)Force on wire Q due to wire R isFR= 10−7 × 2 × 20 × 100.02×0.1=  20 × 10−5N                                                        (Towards Right)Hence  Fnet =FR−FP =14 × 10−5N= 1.4 × 10−4N                                                        (Towards Right)

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