2. In the generator-rod of Sec 29-3, take B - 2.5 T, v -4 m/s, 30 cm, and Bl cur

Question

2. In the generator-rod of Sec 29-3, take B - 2.5 T, v -4 m/s, 30 cm, and Bl current 2.0 A, clockwise. Find (a) the increase in magnetic flux in one second, (b) the emf, (c) the magnetic back-force on the rod-the externalR2 force needed to keep it moving. (d) Calculate the rate (i) at which electrical energy is produced by the emf, and (ii) at which the external force does work, showing equality. (e) Find the resistance of the circuit [1,5 ?]. 2A In #2, now assume that a second, motor-rod is moving at 3.2 m/s, with all given data the same. Find (a) the magnetic force on the motor-rod and (b) the back-emf on it. (c) Find the net emf, and use this to find (d) the resistance r in this circuit. (e) Find the rate of production of mechanical energy and of heat, and check consistency with a preceding result. (f) What determines which rod is the generator and which the motor? What would happen if the two rods moved at the same speed? 3. QA

Explanation / Answer

#3:

A:

magnetic force F = i L B

F = 2* 0.3* 2.5

F = 1.5 N

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back emf e = l v B

e = 0.3* 3.2* 2.5

e = -2.4 Volts

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net emf e = rate of change in flux = 3 - 2.4 V = 0.6 V

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rate of electrical energy = e*I = 3*2 = 6 W

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direction of induced emf

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no induced emf if speed is same

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