2-8 - Create assignment, 00224, Homework 8, Apr 01 at 6:44...

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Unformatted text preview: Create assignment, 00224, Homework 8, Apr 01 at 6:44 pm 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Conducting Ring 31:01, trigonometry, multiple choice, < 1 min, fixed. 001 A constant current is flowing through a solenoid, creating a magnetic field. V I B conducting ring The force which the magnetic field exerts on a conducting ring positioned as shown is 1. There is no force, only a torque. 2. upward. 3. downward. 4. There is neither a force nor a torque. correct Explanation: The magnetic field within the conducting ring is constant since the current flowing through the solenoid is constant in time. In other words, there is no induced current in the conducting ring. Therefore, no force nor torque is exerted on the conducting ring. keywords: Applied Force on a Bar 03 31:02, trigonometry, multiple choice, > 1 min, wording-variable. 002 In the arrangement shown in the figure, the resistor is 5 Ω and a 4 T magnetic field is directed out of the paper. The separation between the rails is 3 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 2 m / s . Assume the bar and rails have negligible resistance and friction. m ¿ 1g 2 m / s 5Ω 4 T 4 T I 3m Calculate the applied force required to move the bar to the left at a constant speed of 2 m / s . Correct answer: 57 . 6 N. Explanation: Let : R = 5 Ω , B = 4 T , ‘ = 3 m , and v = 2 m / s . Motional emf: E = B ‘v . Magnetic force on current: ~ F = I ~ ‘ × ~ B . Ohm’s Law: I = V R . The motional emf induced in the circuit is E = B ‘v = (4 T) (3 m) (2 m / s) = 24 V . From Ohm’s law, the current flowing through the resistor is I = E R = 24 V 5 Ω = 4 . 8 A . Create assignment, 00224, Homework 8, Apr 01 at 6:44 pm 2 Thus, the magnitude of the force exerted on the bar due to the magnetic field is F B = I ‘B = (4 . 8 A)(3 m)(4 T) = 57 . 6 N . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = F B = 57 . 6 N . keywords: Square Copper Plate 02 31:02, trigonometry, multiple choice, > 1 min, wording-variable. 003 A square piece of copper is rotated about its center in a magnetic field B (into the page ⊗ , out of the page fl ). Shown below are dif- ferent charge configurations associated with this procedure. Select the figure with an acceptable charge distribution. 1. B B B B + + + + + + + +-------- ω ω correct 2. B B B B-------- + + + + + + + + ω ω 3....
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This note was uploaded on 04/08/2011 for the course PHYS 1402 taught by Professor Avram during the Spring '10 term at Austin Community College.

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2-8 - Create assignment, 00224, Homework 8, Apr 01 at 6:44...

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