An arrangement of three parallel straight wir | vedclass

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Question

(C) The force per unit length between two parallel wires carrying current $I$ separated by distance $d$ is given by $f = \frac{{\mu _0}{I^2}}{2{\pi d}

Vedclass · Accepted Answer

$\frac{{\mu _0}{I^2}}{{\sqrt 2 \pi d}}$

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(C) The force per unit length between two parallel wires carrying current $I$ separated by distance $d$ is given by $f = \frac{{\mu _0}{I^2}}{2{\pi d}}$.Since the currents are in the same direction,the force is attractive.Wire $B$ experiences an attractive force $F_{AB}$ towards wire $A$ and an attractive force $F_{BC}$ towards wire $C$.The magnitudes are $F_{AB} = F_{BC} = \frac{{\mu _0}{I^2}}{2{\pi d}}$ (per unit length).Since the wires are arranged at a $90^{\circ}$ angle,the vectors $\vec{F}_{AB}$ and $\vec{F}_{BC}$ are perpendicular to each other.The net force per unit length $f_{	ext{net}}$ is the vector sum: $f_{	ext{net}} = \sqrt{F_{AB}^2 + F_{BC}^2} = \sqrt{2} F_{BC}$.Substituting the value: $f_{	ext{net}} = \sqrt{2} \left( \frac{{\mu _0}{I^2}}{2{\pi d}} ight) = \frac{{\mu _0}{I^2}}{{\sqrt 2 \pi d}}$.

An arrangement of three parallel straight wires placed perpendicular to the plane of the paper carrying the same current $I$ along the same direction is shown in the figure. The magnitude of the force per unit length on the middle wire $B$ is given by:

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