Three long,straight and parallel wires carryi | vedclass

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(A) 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

Vedclass · Accepted Answer

$1.4 	imes 10^{-4}\,N$ towards the right

Vedclass · Answer

(A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi r}$. The total force on a length $L$ is $F = \frac{\mu_0 I_1 I_2 L}{2\pi r} = 2 	imes 10^{-7} \frac{I_1 I_2 L}{r}$.$1$. Force on wire $Q$ due to wire $P$ $(I_P = 30\,A, I_Q = 10\,A, r = 0.1\,m, L = 0.1\,m)$:Since currents are in opposite directions,the force is repulsive (towards the left).$F_P = 2 	imes 10^{-7} 	imes \frac{30 	imes 10 	imes 0.1}{0.1} = 6 	imes 10^{-5}\,N$ (towards the left).$2$. Force on wire $Q$ due to wire $R$ $(I_R = 20\,A, I_Q = 10\,A, r = 0.02\,m, L = 0.1\,m)$:Since currents are in opposite directions,the force is repulsive (towards the right).$F_R = 2 	imes 10^{-7} 	imes \frac{20 	imes 10 	imes 0.1}{0.02} = 20 	imes 10^{-5}\,N$ (towards the right).$3$. Net force on wire $Q$:$F_{net} = F_R - F_P = 20 	imes 10^{-5} - 6 	imes 10^{-5} = 14 	imes 10^{-5}\,N = 1.4 	imes 10^{-4}\,N$ (towards the right).

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

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