This is a though long question, but Lifesaver rating will beprovided to the pers

Question

This is a though long question, but Lifesaver rating will beprovided to the person that answers it completely andcorrectly. Two parallel rails are connected together at one end by aresistance of R = 20 as shown below. Across these two rails,which are L = 45.0 cm apart, there lies a conducting metal bar. Themagnetic field is uniform, has a strength of B = 2.20 Tesla and isdirected out of the paper as shown. a force is applied to the metalbar so as to push the bar to the right with velocity v = 8.40m/s. a) What will be the resulting EMF in this circuit? b) What will be the direction of the resulting conventionalcurrent flowing through this circuit? c) What will be the magnitude of the resulting current? d) At what rate is electrical energy being generated? e) How much force is being applied to this bar? f) At what rate is mechanical energy being consumed?

Explanation / Answer

The resultant emf is given by, e=Blv e=2.2*0.45*8.4 e=8.316V The current is in the clockwise direction. The resultant current is given by, i=e/R=8.316/20=0.4158A P=i2R=0.41582*20=3.4577928W=3.4577928J/s F=Bl2v/R=2.2*0.452*8.4/20=0.18711N Hence we get by it. "Hope this help!Best of luck for the rest of yourcoursework."

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