A rod of length l is perpendicular to the lines of induction of a uniform magnetic field of induction B. The rod revolves at an angular speed `omega` about an axis passing through the rod's end parallel to the lines of induction. Find the voltage across the rod's ends.

Answer

`trianglePsi=B omega l^2//2`.

Explanation

The rod moves perpendicularly to the lines of induction of the magnetic field, and in a small section of it an elementary e.m.f `triangle epsi=Bv triangle x` is established, `trianglex` being the length of the section and v Its velocity (see Fig. 30, 4a). The voltage across the rod is the sum of the elementary e.m.f..s. Since `v=omega r, triangle epsi=B omega x trianglex`. The result may be found by two methods:
(a) By integration. We have `triangle Psi=underset(0)overset(1)int B omega x dx=B omega [x^2/2]_(0)^l=1/2 Bwl^2`
(b) Graphically. Plot a graph of the strength of the induced field `E^*=triangleepsi //trianglex=B omega x`. Since this is a linear function, its graph is of the form shown in Fig 30.4 b. The voltage is numerically equal to the area under the graph : `triangle Psi=(lBwl)/(2)-(Bwl^2)/2`