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(C) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by $f = \frac{\mu_0 I_

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(C) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi r}$.For a length $L$,the force is $F = \frac{\mu_0 I_1 I_2 L}{2\pi r} = 2 	imes 10^{-7} 	imes \frac{I_1 I_2 L}{r}$.$1$. Force on wire $C$ due to wire $D$ $(F_D)$:$I_D = 30\, A$,$I_C = 10\, A$,$r_{DC} = 3\, cm = 3 	imes 10^{-2}\, m$,$L = 25\, cm = 0.25\, m$.Since currents are in opposite directions,the force is repulsive (towards the right).$F_D = 2 	imes 10^{-7} 	imes \frac{30 	imes 10 	imes 0.25}{3 	imes 10^{-2}} = 2 	imes 10^{-7} 	imes \frac{75}{3 	imes 10^{-2}} = 5 	imes 10^{-4}\, N$ (towards right).$2$. Force on wire $C$ due to wire $G$ $(F_G)$:$I_G = 20\, A$,$I_C = 10\, A$,$r_{CG} = 2\, cm = 2 	imes 10^{-2}\, m$,$L = 0.25\, m$.Since currents are in opposite directions,the force is repulsive (towards the left).$F_G = 2 	imes 10^{-7} 	imes \frac{20 	imes 10 	imes 0.25}{2 	imes 10^{-2}} = 2 	imes 10^{-7} 	imes \frac{50}{2 	imes 10^{-2}} = 5 	imes 10^{-4}\, N$ (towards left).$3$. Net force on wire $C$:Since $F_D$ and $F_G$ are equal in magnitude and opposite in direction,the net force is $F_{net} = F_D - F_G = 5 	imes 10^{-4} - 5 	imes 10^{-4} = 0\, N$.

Three long,straight parallel wires carrying current are arranged as shown in the figure. The force experienced by a $25\, cm$ length of wire $C$ is

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