C)-0.100 v D) 0.100 Vv E) 0.325 Vv 11. The 0.20 m-long rod in the figure is movi
ID: 1865336 • Letter: C
Question
C)-0.100 v D) 0.100 Vv E) 0.325 Vv 11. The 0.20 m-long rod in the figure is moving towards east at a rate of 25.0 m/'s in a magnetic field of of 0.30 T (vertically downward). The electromotive-force (emf) generated by the moving rod is: A) 1.5 V, with cloc B) 1.5 V, with counter-clockwise current C) 15 V, with clockwise current D) 15 V, with counter-clockwise current E) zero V current 12. In problem 11, the electric field in the rod: A) zero V/m B) 7.50 V/m, towards south B is dowaward c) 7.50 V/m, towards north D) 37.5 V/m, towards north 13. You are designing a generator to provide a rms voltage of 120.0 V. If the generator's coil has 40.00 turns and a eross-sectional area of 0.08500 m2, what would be the frequeney of the generator in a magnetic field of 0.1592 : A) 60 Hz B) 50 Hz C) 40 Hz D) 30 Hz E) 20 Hz 14. The coil of a generator has 24 loops and a cross-sectional area of 0.50 m2, What is the maximum emf gener by this generator if it is spinning with an angular velocity of 7.9 rad/s in a 2.1 T magnetic field? A) 50 V B) 100 V C) 200 V D) 300 V E) 400 V A coil produces an induced emf of 15.0 Vwhen the current in it changes from -2.00 A to +2.00 A in 8.0 seconds. The inductance of the coil is: A) 0.25 HExplanation / Answer
11)
EMF is given by:
E = BLv
= 0.3*0.2*25
= 1.5 V
12)
Electric field is given by:
E = V/L
= 1.5/0.2
= 7.5 V/m
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