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(B) 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

$32$

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(B) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi r}$.Wire $P$ carries $2 \ A$,wire $Q$ carries $4 \ A$,and wire $R$ carries $6 \ A$. The distance between $P$ and $Q$ is $1 \ m$,and the distance between $Q$ and $R$ is $1 \ m$.Since all currents are in the same direction,wire $P$ attracts wire $Q$ towards the left,and wire $R$ attracts wire $Q$ towards the right.Force per unit length on $Q$ due to $P$ is $f_{QP} = \frac{\mu_0 (4)(2)}{2 \pi (1)} = \frac{8 \mu_0}{2 \pi}$.Force per unit length on $Q$ due to $R$ is $f_{QR} = \frac{\mu_0 (4)(6)}{2 \pi (1)} = \frac{24 \mu_0}{2 \pi}$.The net force per unit length on $Q$ is $f_{	ext{net}} = f_{QR} - f_{QP} = \frac{\mu_0}{2 \pi} (24 - 8) = \frac{4 \pi 	imes 10^{-7}}{2 \pi} 	imes 16 = 2 	imes 10^{-7} 	imes 16 = 32 	imes 10^{-7} \ N/m$.

Three long straight wires are carrying current as shown in the diagram. The magnetic force per unit length on wire $Q$ is $..... \times 10^{-7} \ N/m$.

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