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(A) Given,length of wire $Q$,$L = 25\, cm = 0.25\, m$.The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated

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$3	imes10^{-4}\, N$ toward right

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(A) Given,length of wire $Q$,$L = 25\, cm = 0.25\, m$.The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi r}$.Force on wire $Q$ due to wire $R$ $(F_{QR})$:$I_Q = 10\, A$,$I_R = 20\, A$,$r = 0.05\, m$.Since currents are in opposite directions,the force is repulsive (towards left).$F_{QR} = \frac{\mu_0 I_Q I_R}{2\pi r} 	imes L = 2 	imes 10^{-7} 	imes \frac{10 	imes 20}{0.05} 	imes 0.25 = 20 	imes 10^{-5}\, N$ (towards left).Force on wire $Q$ due to wire $P$ $(F_{QP})$:$I_Q = 10\, A$,$I_P = 30\, A$,$r = 0.03\, m$.Since currents are in opposite directions,the force is repulsive (towards right).$F_{QP} = \frac{\mu_0 I_Q I_P}{2\pi r} 	imes L = 2 	imes 10^{-7} 	imes \frac{10 	imes 30}{0.03} 	imes 0.25 = 50 	imes 10^{-5}\, N$ (towards right).Net force $F_{	ext{net}} = F_{QP} - F_{QR} = 50 	imes 10^{-5} - 20 	imes 10^{-5} = 30 	imes 10^{-5}\, N = 3 	imes 10^{-4}\, N$ towards right.

Three straight parallel current-carrying conductors are shown in the figure. The force experienced by the middle conductor of length $25\, cm$ is

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