PHYSICS 240 Exam 3 -- Fall 2005 Solutions

PHYSICS 240 Exam 3 -- Fall 2005 Solutions - Physics 240...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 240 Fall 2005: Exam #3 Solutions Please print your name:___________________________________________________ Please list your discussion section number:___________________________________ Please list your discussion instructor:________________________________________ Form #1 Instructions 1. Fill in your name above 2. This will be a 1.5 hour, closed book exam. The exam includes 20 questions. 3. You may use a calculator, please do not share calculators 4. You may use three 3”x5” note cards (two from the earlier exams, plus one new one) with notes and equations you think may be useful. You can write on both sides of the card if you like. 5. You will be asked to show your University student ID card when you turn in your exam. ID checked by:___________________________________ Table of constants: ε 0 = 8.85 x 10 -12 C 2 /Nm 2 k = 1/(4 πε 0 ) q electron =-1.6 x 10 -19 C q proton =1.6 x 10 -19 C m electron =9.1 x 10 -31 kg m proton =1.67 x 10 -27 kg μ 0 = 4 π x 10 -7 Tm/A G = 6.67 x 10 -11 Nm 2 /kg 2
Background image of page 1

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

View Full DocumentRight Arrow Icon
1: A zero resistance rod slides to the right on zero resistance rails separated by 0.3 m. The rails are connected by a 25 Ω resistor. A constant magnetic field of 0.05 T permeates the region, coming straight out of the page. Find the speed at which the bar must be moved to produce a current of 0.01A in the resistor. a) 34 m/s b) *17 m/s c) 89 m/s d) 50 m/s e) 28 m/s 2: Four different loops of wire will move to the right with the same velocity from a region of zero field, into a region with constant field B. Rank them from least to greatest according to the maximum magnitude of the induced EMF they will feel. Relative heights and widths are listed below each loop. a) 1 is least, then 2 and 4 tie, and 3 is greatest b) 2 and 3 are tied for least, 1 and 4 are tied for greatest c) 3 is least, 4 is larger, and 1 and 2 are tied for greatest d) *3 is least, 2 and 4 are tied and larger, 1 is greatest e) 2 and 4 are tied for least, 3 is larger, 1 is greatest . v R = 25 Ω 0.3 m 4 2 1 3 3W x 2H 2W x 2H 3W x 1H 2W x 3H You know that the EMF will be EMF = d Φ B /dt = BL I = EMF / R = BLv / R v = RI / BL = 16.7 m/s This problem is very related to the above. The EMF in each loop will be d Φ B /dt = BHv. So if they move at the same speed the determining factor will be the height. This means the EMF in 3 is least, 2 and 4 are tied, and 1 is the most.
Background image of page 2
3: A small bar magnet is dropped from the same position below the centers of several different loops of wire, each of which is held fixed in place. Loop A has resistance 2R, loop B has resistance R, and loop C has a small break in it. Which of the following statements is correct? a) The magnet accelerates downward at the same rate in each case b) The magnet accelerates down away from loop C, but accelerates up towards loops A and B c) The magnet accelerates down away all loops, most rapidly from B, most slowly away from C d) The magnet accelerates down away slowly from loop A, down more rapidly from loop C, and up towards loop B
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 11

PHYSICS 240 Exam 3 -- Fall 2005 Solutions - Physics 240...

This preview shows document pages 1 - 4. Sign up to view the full document.

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