final 01 - Version 002/AAAAC – final 01 – Turner –...

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: Version 002/AAAAC – final 01 – Turner – (59070) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Given: Assume the bar and rails have neg- ligible resistance and friction. In the arrangement shown in the figure, the resistor is 6 Ω and a 7 T magnetic field is directed into the paper. The separation between the rails is 2 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 5 m / s . m ≪ 1g 5 m / s 6Ω 7 T 7 T I 2m At what rate is energy dissipated in the resistor? 1. 2508.8 2. 5512.5 3. 648.0 4. 10035.2 5. 432.0 6. 400.0 7. 816.667 8. 6.25 9. 81.0 10. 8.0 Correct answer: 816 . 667 W. Explanation: Basic Concept: Motional E E = B ℓv . Ohm’s Law I = V R . Solution: The motional E induced in the circuit is E = B ℓv = (7 T) (2 m) (5 m / s) = 70 V . From Ohm’s law, the current flowing through the resistor is I = E R = B ℓv R = (7 T) (2 m) (5 m / s) R = 11 . 6667 A . The power dissipated in the resistor is P = I 2 R = B 2 ℓ 2 v 2 R 2 R = B 2 ℓ 2 v 2 R = (7 T) 2 (2 m) 2 (5 m / s) 2 (6 Ω) = 816 . 667 W . Note: Second of four versions. 002 10.0 points What is the longest wavelength that can be observed in the fifth-order spectrum using a diffraction grating with 4800 slits per cen- timeter? 1. 416.667 2. 540.541 3. 454.545 4. 408.163 5. 606.061 6. 512.821 7. 434.783 8. 666.667 9. 500.0 10. 487.805 Correct answer: 416 . 667 nm. Explanation: Version 002/AAAAC – final 01 – Turner – (59070) 2 Let : N = 4800cm − 1 = 4 . 8 × 10 5 m − 1 . The interference maxima are at angles given by d sin θ = mλ, and for a diffraction grating, d = 1 N . Therefore the longest wavelength that can be observed in the fifth-order spectrum is λ = d sin90 ◦ 5 = 1 5 N = 1 5 (4 . 8 × 10 5 m − 1 ) · 10 9 nm m = 416 . 667 nm . 003 10.0 points Consider this particular situation, which is spherically symmetric. We have a charge q 1 on a metallic ball at the center, inside of a conducting shell of inner radius R 2 and outer radius R 3 . There is a total charge of q 2 on the shell. O q 2 q 1 A B C a b c R 1 , q 1 R 2 , q ′ 2 R 3 , q ′′ 2 Find E at C, where OC = c . 1. E C = 0 2. E C = k 3 q 1 c 2 3. E C = k q 1 + q 2 b 2 4. E C = k q 1 − q 2 2 a 2 5. E C = k q 1 c 2 6. E C = k q 1 + q 2 c 2 correct 7. E C = k q 1 − q 2 c 2 8. E C = k q 1 2 b 2 9. E C = k q 1 a 2 10. E C = k 4 q 1 c 2 Explanation: Here the Gaussian surface is a sphere cen- tered at the point charge q 1 and of radius c . The enclosed charge in this sphere is all the charge, or q 1 + q 2 . The electric field at C is E C = k q 1 + q 2 c 2 ....
View Full Document

This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 20

final 01 - Version 002/AAAAC – final 01 – Turner –...

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