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Fig. 6.05 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mOmega. Assume the field to be uniform. (a) Suppose K is open and the rod is moved with a speed of 12 cm s^(-1) in the direction shown. Give the polarity and magnitude of the induced emf. (b) Is there an excess charge built up at the ends of the rods when K is open ? What if K is closed ? (c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain. (d) What is the retarding force on the rod when'K is closed ? (e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s^(-1)) when Kis closed ? How much power is required when K is open ? (f) How much power is dissipated as heat in the closed circuit? What is the source of this power? (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular? |
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Answer» Solution :(a) `|varepsilon| = Blv sintheta.` Here `theta = 90^(@), l = 15 cm = 0.15 m, B = 0.50 T and V = 12 cm s^(-1)` Therefore, `lvarepsilol = Blv = 0.50 xx 0.15 xx 0.12 = 9.0 xx 10^(-3) V = 9.0 mV` Using left hand rule, the Lorentz.s force on electron is from P to RENDERING end P positive and end Q at negative potential. (b) Yes, an excess charge is build up at the ends of the rod when K is open, because of the induced emf set up between the ends. When K is closed, the excess charge is maintained by the CONTINUOUS flow of current due to induced emf. (c) The magnetic force `-e(vecvxxvecB)` is balanced by the electric force `-e vecE`. The electric force is set up due to the excess charges of opposite signs between the ends of the rod. (d) When K is closed the retarding force on the rod is `F = BIl = Bxxvarepsilon/Rxxl, and R = 9.0 MOMEGA = 9.0 xx 10^(-3)Omega` `therefore` The power required =`F v =7.5 xx 10^(-2) xx 10^(-3)W` The power dissipated `=varepsilo^(2)/R=((9.0xx10^(-3))^(2))/(9.0xx10^(-3))9.0 xx 10^(-3)W` The power is provided by the external agent SUPPLYING power. Since, `|varepsilon| = Blv sin theta and theta` in this case `0^(@)`, therefore the induced emf is zero. |
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