question_answer

A rod PQ of length L moves with a uniform velocity v parallel to a long straight wire carrying a current i, the end P remaining at a distance r from the wire. The emf induced across the rod is           

A) \[\frac{{{\mu }_{0}}i{{v}^{2}}}{2\pi }ln\,\left( \frac{r+L}{R} \right)\]

B)        \[\vec{B}\]

C) \[\frac{{{\mu }_{0}}iv}{2\pi }ln\,\left( \frac{r+L}{R} \right)\]

D)        \[\frac{{{\mu }_{0}}iv}{2\pi }ln\,\left( \frac{{{r}^{2}}+{{L}^{2}}}{{{L}^{2}}} \right)\]

Correct Answer: C

Solution :

[c] Consider a small element of length dx of the rod at a distance x and (x+dx) from the wire. The emf induced across the element \[de=Bvdx\]                   ?.(i) We know that magnetic field B at a distance x from a wire carrying a current; is given by \[B=\frac{{{\mu }_{0}}}{2\pi }.\frac{i}{x}\]                  ?. (ii) From eqs. (i) and (ii), \[de=\frac{{{\mu }_{0}}i}{2\pi x}vdx\]              ...(iii) The emf induced in the entire length of the rod PQ is given by \[e=\int{de=\int_{P}^{Q}{\frac{{{\mu }_{0}}}{2\pi }\frac{i}{x}v\,dx}}\] \[=\int_{r}^{r+L}{\frac{{{\mu }_{0}}}{2\pi }\frac{i}{x}v\,dx=\frac{{{\mu }_{0}}}{2\pi }i\,v\int_{r}^{r+L}{\frac{dx}{x}}}\] \[=\frac{{{\mu }_{0}}iv}{2\pi }{{\log }_{e}}\left( \frac{r+L}{r} \right)\]