The metal wire in FIGURE moves with speed $v$parallel to a straight wire that is carrying current $I$. The distance between the two wires is$d$. Find an expression for the potential difference between the two ends of the moving wire.

A field of force is generated by a current-carrying wire, and the Biot-Savart law allows us to calculate the magnitude and direction of this flux. Any moment the force field varies as a function of the segment$\Delta \overrightarrow{s}$of current-carrying wire is given by equation

$B=\frac{{\mu }_o}{2\pi }\frac{I}{x}$

The induced emf, as per Faraday's law, is the change in magnetic force inside the loop, and it's computed using an equation within the shape.

$\epsilon =\left|\frac{d{\Phi }_{\mathrm{m}}}{dt}\right|$

Where${\Phi }_{\mathrm{m}}$is that the flux through the wire which is that the quantity of magnetic flux that flows through a wire of area $A$and it's given by

$d{\Phi }_{\mathrm{m}}=d\left(BA\right)=BdA$

$=\left(\frac{{\mu }_o}{2\pi }\frac{I}{x}\right)\left(Ldx\right)$

$=\frac{{\mu }_{0}L}{2\pi }\frac{I}{x}dx$

The flux changes from point $x=d$to point $x=d+l$where$l$is the length of the wire. So, we integrate the flux by

$\int d{\Phi }_{\mathrm{m}}=\frac{{\mu }_{o}LI}{2\pi }{\int }_d^{d+l}\frac{1}{x}dx$

${\Phi }_{\mathrm{m}}=\frac{{\mu }_{o}L1}{2\pi }[\ln x{]}_d^{d+l}$

${\Phi }_{\mathrm{m}}=\frac{{\mu }_{o}LI}{2\pi }\left[\ln \right(d+l)-\ln (d\left)\right]$

${\Phi }_{\mathrm{m}}=\frac{{\mu }_{o}LI}{2\pi }\left(1+\frac{l}{d}\right)$

Use this expression of ${\Phi }_{\mathrm{m}}$into equation to urge$\epsilon$by

$\epsilon =\frac{d}{dt}\left(\frac{{\mu }_{o}LI}{2\pi }\left(1+\frac{l}{d}\right)\right)$

$=\frac{{\mu }_{0}I}{2\pi }\left(1+\frac{l}{d}\right)\frac{dL}{dt}$

$=\frac{{\mu }_{o}Iv}{2\pi }\left(1+\frac{l}{d}\right)$

Where $v$is that the speed of the loop