HW #9-solutions - Brannon(blb675 HW#9 Shih(58250 This...

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Brannon (blb675) – HW #9 – Shih – (58250) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Two parallel rails having negligible resistance are connected by a resistor. The circuit also contains two metal rods with resistance, one with moving to the left and the other moving to the right. At t = 0 seconds the rods are 4 cm from the 2 Ω resistor. The rods are sliding along the rails and moving away from the 2 Ω resistor at constant velocities as shown in figure below. A uniform magnetic field is applied perpendicular to the plane of the rails. × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × 13 Ω 2 Ω 18 Ω B = 0 . 0152 T 5 m / s 2 m / s 10 cm 4 cm 4 cm Note: Drawing not to scale. Determine the magnitude of the current in the 2 Ω resistor at t = 6 s. Correct answer: 0 . 148919 mA. Explanation: R L E L R C R R E R I L I R I C Let : R L = 13 Ω , R C = 2 Ω , R R = 18 Ω , v L = 2 m / s , v R = 5 m / s , d = 10 cm , and B = 0 . 0152 T . Motional emf is E = v × B . Ohm’s Law is I = V R . The 4 cm distances and the time interval have no bearing on the solution to this prob- lem since the rods are traveling at a constant velocity. The motional emf induced in the two mov- ing rods are as following, respectively for the left-hand side, we have E L = B v L = (0 . 0152 T) (0 . 1 m) (2 m / s) = 0 . 00304 V , and for the right-hand side, we have E R = B v R = (0 . 0152 T) (0 . 1 m) (5 m / s) = 0 . 0076 V . The junction equation is I L + I C - I R = 0 . (1) Using the left-hand current loop (in a counter- clockwise direction), we have + R L I L - R C I C + 0 I R = E L . (2) Using the right-hand current loop (in a clock- wise direction), we have +0 I L + R C I C + R R I R = E R . (3) Solving these three linear equations using the determinant technique, we have I C = 1 0 - 1 R L E L 0 0 E R R R 1 1 - 1 R L - R C 0 0 R C R R = E L R R - E R R L - R R R C - R L R C - R L R R = - 0 . 04408 V Ω - 296 Ω 2 = 0 . 000148919 A = 0 . 148919 mA | I C | = 0 . 148919 mA .
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Brannon (blb675) – HW #9 – Shih – (58250) 2 where E L R R - E R R L = (0 . 00304 V) (18 Ω ) - (0 . 0076 V) (13 Ω ) = - 0 . 04408 V Ω , and - R R R C - R L R C - R L R R = - (18 Ω ) (2 Ω ) - (13 Ω ) (2 Ω ) - (13 Ω ) (18 Ω ) = - 296 Ω 2 . Incorrect Solution: In this problem, it’s important to know that the current in the 2 Ω resistor is the sum of I L and I R , which is due to the two emfs induced in the moving rods respectively. From Lenz’s Law, we know that I L has a direction di ff erent from that of I R , namely, I L is counter-clockwise while I R is clockwise. So the current should be I C = I R - I L . Applying Ohm’s Law, we get I L , I R I L = E L R + R L = 0 . 000202667 A . I R = E R R + R R = 0 . 00038 A . So the final current flow through 2 Ω resistor is I C = I R - I L = (0 . 00038 A) - (0 . 000202667 A) = 0 . 000177333 A = 0 . 177333 mA . Since the two sides of the circuit are coupled, they cannot be treated independently, conse- quently this incorrect solution is WRONG . 002 (part 2 of 2) 10.0 points The current in the 2 Ω resistor is flowing 1. upward.
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