solve if therotatingrodis10cmlongandrotatingat0.5/sina1Tfield...

EXAMPLE 30.2 | Potential difference along a rotating bar A metal bar of length / rotates with angular velocity w about a pivot SOLVE Even though the bar is rotating, rather than moving in a at one end of the bar. A uniform magnetic field B is perpendicular straight line, the velocity of each charge carrier is perpendicular to to the plane of rotation. What is the potential difference between B. Consequently, the electric field created inside the bar is exactly the ends of the bar? that given in Equation 30.1, E = vB. But v, the speed of the charge VISUALIZE FIGURE 30.4 is a pictorial representation of the bar. carrier, now depends on its distance from the pivot. Recall that in The magnetic forces on the charge carriers will cause the outer end rotational motion the tangential speed at radius r from the center to be positive with respect to the pivot. of rotation is v = wr. Thus the electric field at distance r from the pivot is E = wrB. The electric field increases in strength as you FIGURE 30.4 Pictorial representation of a metal move outward along the bar. bar rotating in a magnetic field. The electric field E points toward the pivot, so its radial com- The electric field strength ponent is E, = -wrB. If we integrate outward from the center, the increases with r. .: potential difference between the ends of the bar is X X X . .... X AV = Vup - Vpivot = -|E, dr X X X X X B Angular velocity w X X X X X X - - [ (- worB ) dr = WB /rar = wl'B Pivot r X X X X X X AX X X ASSESS 2 w/ is the speed at the midpoint of the bar. Thus AV is The speed at distance r is v = wr. Vmid IB, which seems reasonable