A conduction electron in the pendulum will experience

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A conduction electron in the pendulum will experience a magnetic force opposite the di- rection of v × B , so using the right hand rule, we can determine that the motion of the elec- trons will be in the clockwise direction, which produces a counter-clockwise current. Because the magnetic field is pointing into the paper, the magnetic flux through the pen- dulum is increasing, so by Lenz’s law we know that the induced magnetic field is in the op- posite direction, or out of the paper. B B B B O e n t e ri n g fi e l d i F w Alternative Solution: Use the result of Part 1, and the right hand rule on the current flow to determine that the induced magnetic field must be directed out of the paper. 021 10.0 points A solenoid with circular cross section pro- duces a steadily increasing magnetic flux through its cross section. There is an octago- nally shaped circuit surrounding the solenoid as shown. B B B B X Y i i Figure 1: The increasing magnetic flux gives rise to a counterclockwise induced emf E . The circuit consists of two identical light bulbs of equal resistance, R , connected in series, leading to a loop equation E - 2 i R = 0. Now connect the points C and D with a wire CAD (as in figure 2). Label the currents i 1 , i 2 , and i 3 as indicated. B B B B X Y D C A i 2 i 1 i 3 i 3 Figure 2: To what do the corresponding loop and node equations lead? 1. E - 2 i 3 R = 0 : i 2 = i 3 3 , i 1 = 2 i 3 3 2. E - 2 i 3 R = 0 : i 2 = i 3 4 , i 1 = 3 i 3 4 3. E - i 3 R = 0 : i 2 = i 3 2 , i 1 = i 3 2 4. E - i 3 R = 0 : i 2 = i 3 3 , i 1 = 2 i 3 3 5. E - i 3 R = 0; i 2 = i 3 , i 1 = 0 correct 6. E - i 3 R = 0 : i 2 = i 3 4 , i 1 = 3 i 3 4
toupal (rgt374) – Homework 09 – Chiu – (58295) 13 7. E - i 3 R = 0 : i 2 = 0 , i 1 = i 3 8. E - 2 i 3 R = 0 : i 2 = i 3 2 , i 1 = i 3 2 9. E - 2 i 3 R = 0 : i 2 = i 3 , i 1 = 0 10. E - 2 i 3 R = 0 : i 2 = 0 , i 1 = i 3 Explanation: Consider the loop XCADX (where Y indi- cates the position of bulb Y and X indicates the position of bulb X ), with loop equation E - i 3 R = 0 . The loop XDY CX leads to E - i 3 R - i 1 R = 0 . Subtracting the former from the latter equa- tion we find i 1 = 0 . We also have one node equation; e.g. , at the junction C : i 3 = i 1 + i 2 i 3 = i 2 . 022 10.0 points A solenoid (with magnetic field B ) pro- duces a steadily increasing uniform magnetic flux through its circular cross section. A octagonal circuit surrounds the solenoid as shown in the figure. The wires connecting in the circuit are ideal, having no resistance. Two identical resistors with resistance R (la- beled X and Y ) are in the circuit. A wire connects points C and D . The ratio of the solenoid’s area A L left of the wire CD and the solenoid’s area A R right of the wire CD is 5. 4 E 5 - i 1 R = 0 and E 5 - i 2 R = 0 . 6. E 5 - i 1 R = 0 and 4 E 5 + i 2 R = 0 . 7. E 5 + i 1 R = 0 and 4 E 5 - i 2 R = 0 . 8. 4 E 5 + i 1 R = 0 and E 5 + i 2 R = 0 . B B B B R Y R X D C i 3 i 1 i 2 A L A R E = - d Φ dt Figure: Let i 1 , i 2 , and i 3 be defined as positive if the currents flow in the same direction as shown by the arrows in the figure (otherwise the currents i 1 , i 2 , and i 3 have negative values). Also, let the induced emf be defined as positive, E > 0 . What are the equations for the (right) loop CXDC and the (left) loop CDY C , respec- tively? Assume the induced emf for the closed loop octagonal CXDY C is E . 1. E 5 - i 1 R = 0 and 4 E 5 - i 2 R = 0 . 2. 4 E 5 - i 1 R = 0 and E 5 + i 2 R = 0 . 3. E 5 + i 1 R = 0 and 4 E 5 + i 2 R = 0 . cor- rect 4. 4 E 5 + i 1 R = 0 and E 5 - i 2 R = 0 .
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