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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 $d$ is given by $f = \frac{{\mu _0}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by $f = \frac{{\mu _0}{I_1}{I_2}}{2\pi d}$.Since all wires carry the same current $I$,the force per unit length exerted by wire $A$ on $B$ is $f_A = \frac{{\mu _0}{I^2}}{2\pi d}$ (directed towards $A$).The force per unit length exerted by wire $C$ on $B$ is $f_C = \frac{{\mu _0}{I^2}}{2\pi d}$ (directed towards $C$).Since the wires $A$,$B$,and $C$ are arranged such that the angle between the directions of forces $f_A$ and $f_C$ is $90^\circ$,the net force per unit length on wire $B$ is:$f_{net} = \sqrt{f_A^2 + f_C^2} = \sqrt{\left(\frac{{\mu _0}{I^2}}{2\pi d}\right)^2 + \left(\frac{{\mu _0}{I^2}}{2\pi d}\right)^2}$$f_{net} = \sqrt{2 \left(\frac{{\mu _0}{I^2}}{2\pi d}\right)^2} = \sqrt{2} \frac{{\mu _0}{I^2}}{2\pi d} = \frac{{\mu _0}{I^2}}{\sqrt{2}\pi d}$.

An arrangement of three parallel wires placed perpendicular to the plane of the paper,each carrying the same current $I$ in 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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