CGPET 2016 Solved Paper

The wires P and Q carry current in same direction, therefore they attract each other. The force on R due to P is towards the wire P and is given by,

FRP=

μ0

4π

×

2iP×iR

rPR

×I
=

10−7×2×10×10

0.10

×0.15N
PRP=3×10−5N [towards left]
Similarly, wires Q and R attract each other as they also carry the current in the same direction. The force on R due to current in Q is towards right hand side. Therefore the force on R, due to Q, is given by
FQR=

μ0

4π

.

2iQiR

rQR

I=

10−7×2×20×10×0.15

0.10

FQR=6×10−5N [towards right]
∴ Net force on R is
F=6×10−5−3×10−5N
=3×10−5N [towards right (Q)]