Let is apply the concepts of motional emf and Lenz\'s law to a slide-wire genera
ID: 1448343 • Letter: L
Question
Let is apply the concepts of motional emf and Lenz's law to a slide-wire generator. Suppose the length L of the slide-wire rod in (Figure 1) is 0.10 m, its speed v is 2.5 m/s, the total resistance of the loop is 0.030 Ohm, and B is 0.60 T. Find the emf, epsilon, the induced current, the force acting on the rod, and the mechanical power needed to keep the rod moving at constant speed. To keep the rod moving at constant speed, a force equal in magnitude and opposite in direction to F_instant must act on the rod. Therefore, the mechanical power P needed to keep the rod moving is P = F_induced tau = (0.30 N) (2.5 m/s) = 0.75 W The expression for the emf is the same result we found from Faraday's law. The rate at which the induced emf delivers electrical energy to the circuit is P = cI = (0.15 V) (5.0 A) = 0.75 W This is equal to the mechanical power input, F_induced v, as we should expect. The system is converting mechanical energy (work) into electrical energy. Finally, the rate of dissipation of electrical energy in the circuit resistance is P = I^2 R = (50)^2(0.000 Ohm) = 0.75 W which is also to be expected. If L = 0.26 m, e = 4.8 m/s, the total resistance of the loop is 0.015 Ohm, and B = 0.50 T, find the rate at which the induced emf delivers electrical energy to the circuit. Express your answer to two significant figures and include appropriate units.Explanation / Answer
Now here, V = Bvl = 0.5*4.8*0.26 = 0.624 V
Now, Power = V2/R = 0.624*0.624/0.015 = 25.9584.
But, we need the answer in two significant figures, so, the answer is 26 W.
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