Question

A copper rod of length 0.19m is moving with a uniform velocity of 10m/s parallel to a long straight wire carrying a current of 5 A . The rod itself is perpendicular to the wire with its ends at a distance of 0.01 m and 0.2 m from it . Calculate the emf induced in the rod .

Answer 1

Dear Student,

$$\begin{gathered} fieldatpointxfromthewire \\ B=\frac{{\mu }_{0}I}{2\mathrm{\pi x}} \\ \mathrm{take}\mathrm{dx}\mathrm{element}\mathrm{width}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{at}\mathrm{distance}\mathrm{x} \\ \mathrm{induced}\mathrm{EMF}\mathrm{in}\mathrm{this}\mathrm{element} \\ \int \mathrm{dE}={\int }_{0.01}^{0.2}\mathrm{Bvdx} \\ \mathrm{E}={\int }_{0.01}^{0.2}\frac{10{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi x}}\mathrm{dx} \\ \mathrm{E}=\frac{10{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi }}{\int }_{0.01}^{0.2}\frac{\mathrm{dx}}{\mathrm{x}} \\ \mathrm{E}=\frac{10{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi }}{\left|\mathrm{lnx}\right|}_{0.01}^{0.2}=\frac{10{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi }}\ln 20\mathrm{volt} \\ \mathrm{Regards} \end{gathered}$$