Question

A conducting rod moves with constant velocity u perpendicular to the long, straight wire carrying a current I as shown compute that the emf generated between the ends of the rod

• A. $\frac{{\mu }_0\nu Il}{\pi r}$
• B. $\frac{{\mu }_0\nu Il}{2\pi r}$
• C. $\frac{2{\mu }_0\nu Il}{\pi r}$
• D. $\frac{{\mu }_0\nu Il}{4\pi r}$

Answer 1

The correct option is B $\frac{{\mu }_0\nu Il}{2\pi r}$

The magnetic field near the conducting rod due to long current carrying wire is

$B=\frac{{\mu }_{0}I}{2\pi r}$ (using Ampere's law, $\int B.dl={\mu }_{0}I$)

Now , induced emf due to moving rod is $e=Blv=\frac{{\mu }_{0}I}{2\pi r}\times lv=\frac{{\mu }_{0}vIl}{2\pi r}$