s06x2aan - NAME PHYS-1200 PHYSICS II QUIZ 2 SOLUTIONS PART...

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: NAME PHYS-1200 PHYSICS II QUIZ 2 MARCH 29, 2006 SOLUTIONS PART A. 1. D 6. A 2. B 7. B 3. D 8. B 4. B 5. C PART B . 1. TO THE BOTTOM OF THE PAGE The current in the wire produces a decreasing magnetic flux within the wire loop. According to Lenz’s law, the induced electric field would be in a direction to oppose that decrease. It must be counterclockwise about the loop, so at P , it is toward the bottom of the page. 2. a) TO THE LEFT The same direction as the current in the wires. b) OUT OF THE PAGE Based on the direction of the current and the right hand rule. 3. a) s t x v 04 . m . 1 = ∆ ∆ = v = 25 m/s b) The wavelength can be read directly from the graph. λ = 4.0 m c) m . 4 m/s 25 = = λ v f f = 6.2 Hz 1 NAME PART C . 1. a) F) 10 . 4 H)( 10 . 2 ( 2 1 2 1 7 3-- × × = = π π LC f f = 5.6 ×10 3 Hz = 5600 Hz b) dt di L = E . The easy way to get this is = + C q dt di L , so C q dt di L- = . Then, just after the switch is closed, C Q C q dt di L = = = E , so F 10 4 C 10 5 . 1 7 6...
View Full Document

This note was uploaded on 10/17/2011 for the course PHYS 1010 taught by Professor Schroeder during the Spring '07 term at Rensselaer Polytechnic Institute.

Page1 / 3

s06x2aan - NAME PHYS-1200 PHYSICS II QUIZ 2 SOLUTIONS PART...

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