metal rod of resistance of 15ohm is moved to the right at a constant speed 60 cm/s along two parallel conducting rails 2 cm apart and shorted at one end. A magnetic field of mangitude 0.35 T points into the page. Calculate the induced emf.
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Explanation:
Length of the rod, l = 15 cm = 0.15 m Magnetic field strength, B = 0.50 T Resistance of the closed loop, R = 9 mΩ = 9 × 10−3 Ω (a) Induced emf = 9 mV; polarity of the induced emf is such that end P shows positive while end Q shows negative ends. Speed of the rod, v = 12 cm/s = 0.12 m/s Induced emf is given as: e = Bvl = 0.5 × 0.12 × 0.15 = 9 × 10−3 v = 9 mV The polarity of the induced emf is such that end P shows positive while end Q shows negative ends. (b) Yes; when key K is closed, excess charge is maintained by the continuous flow of current. When key K is open, there is excess charge built up at both ends of the rods. When key K is closed, excess charge is maintained by the continuous flow of current. (c)Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod. There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods. (d) Retarding force exerted on the rod, F = IBl Where, I = Current flowing through the rod (e) 9 mW; no power is expended when key K is open. Speed of the rod, v = 12 cm/s = 0.12 m/s Hence, power is given as: When key K is open, no power is expended. (f) 9 mW; power is provided by an external agent. Power dissipated as heat = I2 R = (1)2 × 9 × 10−3 = 9 mW The source of this power is an external agent. (g) Zero In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.Read more on Sarthaks.com - https://www.sarthaks.com/18612/figure-shows-metal-resting-smooth-rails-positioned-between-poles-permanent-magnet-rails