HW07-solutions - Munoz (gm7794) HW07 TSOI (92515) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Munoz (gm7794) HW07 TSOI (92515) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points In the figure shown, the magnet is first moved downward toward the loop of wire, then with- drawn upward from the loop of wire. I I Counter- clockwise Clockwise induced current induced current S N down then up As viewed from above, the induced current in the loop is 1. for both cases clockwise with increasing magnitude. 2. for both cases counterclockwise with in- creasing magnitude. 3. first counter-clockwise, then clockwise. 4. first clockwise, then counter-clockwise. correct 5. for both cases clockwise with decreasing magnitude. 6. for both cases counterclockwise with de- creasing magnitude. Explanation: From Ohms law and Faradays law, the current in magnitude is I = V R =- 1 R d dt , where is the magnetic flux through the loop. We know the sign of the rate of change of the magnetic flux is changed when the magnet is withdrawn upward, as is the current direction according to the above eqaution. Using the right-hand-rule and from Lenzs law, we know that when the magnet is first moved downward toward the loop of wire, then withdrawn upward from the loop of wire, the current in the loop is first clockwise, then counter-clockwise , as viewed from above. 002 (part 1 of 2) 10.0 points In the arrangement shown in the figure, the resistor is 8 and a 2 T magnetic field is directed out of the paper. The separation between the rails is 7 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 2 m / s . Assume the bar and rails have negligible resistance and friction. m 1g 2 m / s 8 2 T 2 T I 7m Calculate the applied force required to move the bar to the right at a constant speed of 2 m / s. Correct answer: 49 N. Explanation: Motional emf is E = B v . Magnetic force on current is F = I B . Ohms Law is I = V R . The motional emf induced in the circuit is E = B v = (2 T) (7 m) (2 m / s) = 28 V . Munoz (gm7794) HW07 TSOI (92515) 2 From Ohms law, the current flowing through the resistor is I = E R = 28 V 8 = 3 . 5 A . Thus, the magnitude of the force exerted on the bar due to the magnetic field is F B = I B = (3 . 5 A)(7 m)(2 T) = 49 N . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = F B = 49 N . 003 (part 2 of 2) 10.0 points At what rate is energy dissipated in the resis- tor? Correct answer: 98 W. Explanation: The power dissipated in the resistor is P = I 2 R = (3 . 5 A) 2 (8 ) = 98 W ....
View Full Document

This note was uploaded on 06/10/2011 for the course PHY 302l taught by Professor Morrison during the Summer '08 term at University of Texas at Austin.

Page1 / 7

HW07-solutions - Munoz (gm7794) HW07 TSOI (92515) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online