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anonymous
 one year ago
A straight 1.2 m length iron rod is moved at 10 m/s through the earth's magnetic field at a location where it has magnitude 1.0 × 10^5 T. What is the magnitude of the induced emf between one end of the wire and the other end?
A. 1.2 V
B. 1.2 × 10^4 V
C. 1.0 × 10^4 V
D. 120 V
anonymous
 one year ago
A straight 1.2 m length iron rod is moved at 10 m/s through the earth's magnetic field at a location where it has magnitude 1.0 × 10^5 T. What is the magnitude of the induced emf between one end of the wire and the other end? A. 1.2 V B. 1.2 × 10^4 V C. 1.0 × 10^4 V D. 120 V

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I'm pondering...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2during the motion of that iron rod, an electric field will appear inside that iron rod, since all free electrons inside that rod are underwent to the Lorentz force. Now the motion of such free electrons will terminate when the Lorentz force is balanced with the resultant electrostatic force. The situation is as below: dw:1434343667110:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we can write: \[\Large evB = e\frac{V}{L}\] where V is the emf force, and v is the velocity of the iron rod after a simplification we get: \[\Large V = vBL\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2L is the length of the iron rod

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! what do we plug in?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here is the next step: \[\Large V = vBL = 10 \times {10^{  5}} \times 1.2 = ...volts\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! we get 1.2E4? so choice B is our solution? :O
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