Question

2) Figure 2 illustrates a top view of a modified slide-bar generator. The x-y 1R. plane is also parallel to a horizontal plane. A uniform magnetic field, Figure 2 (magnitude B, direction is along positive z-axis) exists in the vicinity of a U-shaped conducting material (negligible resistance) comprising two parallel tracks (aligned with the x- axis) and an intersecting piece (aligned with the y-axis). A small section of the intersecting piece has been removed and replaced by a capacitor (C capacitance). A bar having length L, resistance R, and mass M is placed across the tracks such that it is a distance x from and parallel to the y-axis. At time, t 0, the bar is initially traveling along the x-axis with speed V and the capacitor is discharged. A pulling force, Fp, is applied by an external agent along the x-axis so as to maintain the bar in a state of constant motion. Determine expressions B, L, R, M, C, V, t for the following a) the current flowing through the bar when t20; 个 b) the pulling force, Fp, when t2 0; and c) the final charge of the capacitor

Answer 1

(a) Magnetic flux, phi = B middot A = BLx d phi/dt = BL dx/dt = BLV Therefore xi = d phi/dt = BLV equation (1) From Kirchoff�s Voltage Law, xi - IR - q/C = 0 xi - dq/dt R - q/C = 0 R dq/dt + q/C = xi q = C xi(1 - e^-t/RC) I = dq/dt = (xi/R)e^-t/RC equation (2) From (1) we can write I = (BLV/R)e^-t/RC equation (3) \r\n (b) current through the rod is flowing downward. Lorentz force, F = BIL is acting inward. In order to keep the rod move with constant velocity, F_p must be equal to F_L i.e. F_p = BIL = B(BLV/R)e^-t/RC times L (From (3)) F_p = B^2 L^2 V/R e^-t/RC (c) when capacitor is fully charge, we can consider t = infinity so that e^-t/RC rightarrow 0 From part (a) we have q = C xi(1 - e^-t/RC) q = C xi implies q = BLVC (From(1))