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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»


Solution :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`


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