An infinite wire carries a current `I`. An S- shaped conducting rod of two semicircles esch of radius `r` is placed at an angle `theta` to the wire. The center of the conductor is at a distance `d` from the wire. If the rod translates parallel to the wire with a velocity `v` as shown in Fig. .3.56, calculate the emf induced across the ends `Ob` of the rod.

Explanation

Join the end points `O` and `b` and replace the two semicircles bya straight rod of length `4r` (Fig. 3.57).
Induced emf in straight rod `OB` will be same as in the actual conductor. Now this straight rod can be replaced by a combination of two rods `OA` and `AB`. induced emf in `OB` will be same as the sum of induced emf in `OA` and induced emf in `AB`. But induced emf in `AB` will be zero because its velocity will be parallel to its length. Hence, induced emf in `OA` will be the net induced emf in actual conductor.
We now have a ros of effective length `4r cos theta` translating in a non-uniform magnetic field. consider a small element of the wire of length `dx` located at a distance `x` from the wire. The magnetic field at this element is `B = ((mu)(0)I)/(2pix)` ltbr. The potential difference across this element is `dV = (mu_(0)I)/(2pix)vdx`.
The potential difference across the ends of the rod can be calculated by integrating over the whole end. Therefore,
`V = int dV = int_(d + 2r cos theta)^(d - 2r cos theta)(mu_(0)I)/(2pix) vdx`
`=(mu_(0)Iv)/(2pi)1n[(d + 2r cos theta)/(d - 2r cos theta)]`
From the right hand rule, we see that electrons will accumulate at end `B`. Therefore, end `O` is at a higher potential than `B`.