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

Vedclass · Accepted Answer

$6 	imes 10^{-6} \ N$ towards $R$

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(B) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.For wire $Q$ (current $I_Q = 1 \ A$): $1$. Force due to wire $P$ ($I_P = 3 \ A$, $d_1 = 3 \ cm = 0.03 \ m$): Since currents are in opposite directions, the force is repulsive (away from $P$, i.e.,towards $R$).$F_{QP} = \frac{\mu_0 I_Q I_P}{2 \pi d_1} \ell = (2 	imes 10^{-7}) 	imes \frac{1 	imes 3}{0.03} 	imes 0.15 = 2 	imes 10^{-7} 	imes 100 	imes 0.15 = 3 	imes 10^{-6} \ N$ (towards $R$).$2$. Force due to wire $R$ ($I_R = 2 \ A$, $d_2 = 2 \ cm = 0.02 \ m$): Since currents are in the same direction, the force is attractive (towards $R$).$F_{QR} = \frac{\mu_0 I_Q I_R}{2 \pi d_2} \ell = (2 	imes 10^{-7}) 	imes \frac{1 	imes 2}{0.02} 	imes 0.15 = 2 	imes 10^{-7} 	imes 100 	imes 0.15 = 3 	imes 10^{-6} \ N$ (towards $R$).Total force $F_{net} = F_{QP} + F_{QR} = 3 	imes 10^{-6} + 3 	imes 10^{-6} = 6 	imes 10^{-6} \ N$ towards $R$.

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by $15 \ cm$ length of wire $Q$ is . . . . . . . $(\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A)$

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