Homework 10-solutions

# Homework 10-solutions - im(gi768 Homework 10 markert(55325...

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im (gi768) – Homework 10 – markert – (55325) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points In the arrangement shown in the figure, the resistor is 4 Ω and a 4 T magnetic field is directed into the paper. The separation between the rails is 1 m . An applied force moves the bar to the left at a constant speed of 8 m / s . m 1 g 8 m / s 4 Ω 4 T 4 T I 1 m Calculate the applied force required to move the bar to the left at a constant speed of 8 m / s. Assume the bar and rails have negligi- ble resistance and friction. Neglect the mass of the bar. Correct answer: 32 N. Explanation: The motional emf induced in the circuit is E = B ℓ v = (4 T) (1 m) (8 m / s) = 32 V . From Ohm’s law, the current flowing through the resistor is I = E R = 32 V 4 Ω = 8 A , so the magnitude of the force exerted on the bar due to the magnetic field is F B = I ℓ B = (8 A)(1 m)(4 T) = 32 N . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = F B = 32 N . 002(part2of2)10.0points At what rate is energy dissipated in the resis- tor? Correct answer: 256 W. Explanation: The power dissipated in the resistor is P = I 2 R = (8 A) 2 (4 Ω) = 256 W . 003 10.0points A 9 . 64 m long steel beam is accidentally dropped by a construction crane from a height of 9 . 08 m. The horizontal component of the Earth’s magnetic field over the region is 26 . 4 μ T. The acceleration of gravity is 9 . 8 m / s 2 . What is the induced emf in the beam just before impact with the Earth, assum- ing its long dimension remains in a horizontal plane, oriented perpendicularly to the hor- izontal component of the Earth’s magnetic field? Correct answer: 3 . 39509 mV. Explanation: We must find the speed of the beam just before impact. We use conservation of me- chanical energy 1 2 m v 2 = m g h, or v = radicalbig 2 g h = radicalBig 2 (9 . 8 m / s 2 )(9 . 08 m) = 13 . 3405 m / s . The magnitude of the induced field is |E| = B L v = (2 . 64 × 10 - 5 T)(9 . 64 m) × (13 . 3405 m / s) = 0 . 00339509 V = 3 . 39509 mV .

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im (gi768) – Homework 10 – markert – (55325) 2 004(part1of3)10.0points A rectangular loop of copper wire of resis- tance R has width a and length b . The loop is stationary in a uniform magnetic field. The magnetic field B at time t = 0 seconds is di- rected into the page as shown below. The uniform magnetic field varies with time t ac- cording to the relationship B = B 0 cos ω t , where ω and B 0 are positive constants and B is positive when the field is directed into the page. B B B B b a n turns The direction of the induced current in the loop when ω t = π 2 , after the magnetic field begins to oscillate is 1. undetermined, since the current is zero. 2. counter-clockwise. 3. clockwise. correct Explanation: When ω t = π 2 , B = B 0 cos π 2 = 0; i.e. , the field has been decreasing, and is about to change direction. The induced current will be in a direction to oppose this change; i.e. , clockwise.
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